335 free, high-quality past-paper-style questions on Nuclear physics for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Atomic structure, isotopes, fission and fusion; Radioactivity, half-life and safety. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,072 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IGPC-039-00-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A sodium nucleus is represented as 2311Na. How many neutrons are in this nucleus?
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The nucleon number (23) is the total number of protons and neutrons. The proton number is 11, so the number of neutrons is 23 − 11 = 12. Option 11 is the proton number, 23 is the nucleon number, and 34 comes from adding the two numbers instead of subtracting.
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Answer B
Answer: B
Exam tip: Top number = protons + neutrons. Bottom number = protons. Neutrons = top − bottom.
2IGPC-039-00-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which statement about two isotopes of the same element is correct?
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Isotopes are atoms of the same element, so they have the same proton number. They differ only in the number of neutrons, so their nucleon numbers are different. Their neutral atoms have equal numbers of electrons because the number of electrons equals the number of protons.
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Answer C
Answer: C
Common error: Swapping protons and neutrons in the definition. The number of protons decides which element an atom is, so it cannot change between isotopes.
3IGPC-039-00-03Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A neutral atom loses one electron. What is formed?
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After losing a negative electron, the atom has one more proton than electrons. It now has an overall positive charge, so it is a positive ion. The nucleus is unchanged, so the element and the isotope stay the same.
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Answer B
Answer: B
Common error: Thinking that losing an electron makes the atom negative. Losing negative charge leaves the particle positive.
4IGPC-039-00-07Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which statement describes nuclear fusion?
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Fusion is the joining of two light nuclei to form a heavier nucleus, and energy is released. This is the energy source of stars. Splitting a large nucleus is fission. Emitting an alpha particle is radioactive decay.
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Answer B
Answer: B
Exam tip: Fusion joins nuclei together; fission splits a nucleus apart. Both release energy.
5IGPC-039-00-08Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
What is meant by the nucleon number of a nucleus?
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Nucleons are the particles found in the nucleus, which are protons and neutrons. The nucleon number is their total. The number of protons alone is the proton number. Electrons are outside the nucleus, so they are not nucleons.
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Answer C
Answer: C
Common error: Including electrons in the nucleon number. Electrons are not in the nucleus.
6IGPC-039-00-11Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
A lithium atom is represented as 73Li.
(a)
Name the two types of particle found in the nucleus.
(1)
(b)
State the number of protons and the number of neutrons in this nucleus.
(2)
(c)
The neutral lithium atom gains one electron. State the name of the particle formed and the sign of its overall charge.
(2)
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Part (a)
protons and neutrons (both needed)
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Part (b)
protons = 3 ; neutrons = 7 − 3 = 4
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Part (c)
an ion ; negative charge
✓ 2
Common error: Giving 7 as the number of neutrons. The nucleon number includes the protons, so subtract the proton number.
7IGPC-039-01-01Radioactivity, half-life and safetyPaper CEasy1 mark
Which type of radiation produces the most ionisation as it passes through air?
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α-particles are relatively heavy and carry a charge of +2. They interact strongly with air molecules and lose their energy over a few centimetres, so they produce the most ionisation. γ-rays are the most penetrating, but they are the least ionising. β-particles lie between the two.
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Answer A
Answer: A
Common error: Penetrating ability and ionising ability are linked the opposite way round. The most ionising radiation (α) is the least penetrating.
8IGPC-039-01-02Radioactivity, half-life and safetyPaper CEasy1 mark
A radioactive source emits only one type of radiation. The radiation passes through a sheet of paper. It is stopped by a sheet of aluminium 5 mm thick. Which type of radiation does the source emit?
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β-particles pass through paper but are absorbed by a few millimetres of aluminium. α-particles would be stopped by the paper. γ-rays would pass through 5 mm of aluminium with little reduction and need thick lead or concrete to absorb them. X-rays are not emitted by a radioactive nucleus and also penetrate aluminium.
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Answer B
Answer: B
Exam tip: Learn the three absorbers: paper stops α, a few mm of aluminium stops β, and thick lead only reduces γ.
9IGPC-039-01-03Radioactivity, half-life and safetyPaper CEasy1 mark
The corrected count rate from a radioactive sample is 640 counts/min. The half-life of the isotope is 3.0 hours. What is the corrected count rate after 9.0 hours?
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9.0 hours is 9.0 ÷ 3.0 = 3 half-lives. The count rate halves three times: 640 → 320 → 160 → 80 counts/min. 213 counts/min comes from dividing by 3 instead of halving. 160 counts/min stops after two half-lives. 40 counts/min uses four half-lives.
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Answer B
Answer: B
Common error: Find the number of half-lives first, then halve that many times. Do not divide the count rate by the number of half-lives.
10IGPC-039-01-08Radioactivity, half-life and safetyPaper CEasy1 mark
A laboratory technician uses a source that emits γ-rays. Which action reduces her exposure to the radiation most effectively?
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γ-rays are very penetrating. Only dense, thick materials such as lead absorb a large fraction of them. A lead-lined box between the source and the technician (shielding) greatly reduces her exposure. Cardboard and paper have almost no effect on γ-rays. Temperature does not change the rate of radioactive decay.
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Answer B
Answer: B
Exam tip: The three ways to reduce exposure are: reduce the time spent near the source, increase the distance from it, and use suitable shielding.
11IGPC-039-01-12Radioactivity, half-life and safetyPaper CEasy4 marks
This question is about the three types of nuclear radiation.
Measured count rate for each absorber. The dashed line shows the background count rate.
(a)
State what an α-particle consists of.
(1)
(b)
State the charge on a γ-ray.
(1)
(c)
A student places different absorbers between a source and a detector. The background count rate is 22 counts/min. Her results are: no absorber: 250 counts/min sheet of paper: 248 counts/min 5 mm of aluminium: 22 counts/min Deduce which type or types of radiation the source emits. Explain your answer.
(2)
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Part (a)
two protons and two neutrons (a helium nucleus)
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Part (b)
zero / it is uncharged
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Part (c)
The source emits β-particles only . The paper causes almost no drop, so no α is emitted, and 5 mm of aluminium reduces the count to background, so no γ is emitted .
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Exam tip: Compare each reading with the background count rate. A reading equal to the background means no radiation from the source is getting through.
12IGPC-039-01-11Radioactivity, half-life and safetyPaper CEasy5 marks
A student uses a Geiger–Müller tube connected to a counter to investigate background radiation in a laboratory.
(a)
State two sources of background radiation.
(2)
(b)
With no source present, the student measures the count rate three times. She obtains 18, 22 and 20 counts/min. Explain why the readings are not all the same.
(1)
(c)
Calculate the mean background count rate.
(1)
(d)
A radioactive source is placed near the tube. The reading is now 164 counts/min. Calculate the count rate due to the source alone.
(1)
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Part (a)
Any two different sources from: radon gas in the air; rocks or building materials; cosmic rays; food or drink; medical sources (e.g. radioactive tracers); fallout from nuclear weapons tests or accidents.
—
First correct source
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Second, different correct source
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Part (b)
Radioactive decay is random (and spontaneous), so the number of decays detected in each minute varies
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Part (c)
(18 + 22 + 20) ÷ 3 = 20 counts/min
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Part (d)
164 − 20 = 144 counts/min
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Exam tip: A corrected count rate always means the measured count rate minus the background count rate.
13IGPC3-5-05c-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A water-treatment engineer notes that a chlorine atom is written 3517Cl. How many neutrons are in the nucleus of this atom?
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Neutrons = nucleon number − proton number = 35 − 17 = 18, so B is correct. A is the number of protons. C is the nucleon number (protons plus neutrons). D adds the two numbers instead of subtracting.
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Answer B
Answer: B
Common error: The top number is the total of protons and neutrons, so subtract the bottom number to find neutrons.
14IGPC3-5-05c-02Radioactivity, half-life and safetyPaper CEasy1 mark
A student has three radioactive sources. Which type of radiation is completely stopped by a single sheet of paper?
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Alpha particles are strongly ionising and lose their energy within a few centimetres of air, so a sheet of paper absorbs them: A is correct. Beta particles need a few millimetres of aluminium (B wrong). Gamma rays are only reduced by thick lead or concrete (C wrong). D is wrong because beta and gamma pass through paper.
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Answer A
Answer: A
Exam tip: Remember the order of penetration: paper stops alpha, aluminium stops beta, thick lead only reduces gamma.
15IGPC3-5-05c-05Radioactivity, half-life and safetyPaper CEasy1 mark
A hospital technician regularly handles a sealed gamma-emitting source. Which action does most to reduce the radiation she receives?
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Dense lead absorbs much of the gamma radiation, so storing the source in a lead-lined container cuts exposure: B is correct. Paper stops only alpha particles (A wrong). Hand-washing helps against contamination by unsealed sources, not irradiation by a sealed one (C wrong). Holding the source close increases the dose (D wrong).
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Answer B
Answer: B
Exam tip: Reduce exposure by limiting time, increasing distance and using shielding that suits the radiation type.
16IGPC3-5-05c-07Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which statement describes nuclear fusion in the core of the Sun?
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Fusion is the joining of light nuclei (such as hydrogen) into a heavier nucleus with energy released: A is correct. B describes fission. C describes alpha decay. D confuses a nuclear process with a chemical reaction.
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Answer A
Answer: A
Common error: Fusion joins small nuclei and fission splits large ones; do not mix them up.
17IGPC3-5-05c-09Atomic structure, isotopes, fission and fusionPaper CEasy7 marks
A piece of aircraft metal contains aluminium-27 atoms, written 2713Al.
(a)
State the number of protons, neutrons and electrons in a neutral atom of aluminium-27. [3]
(3)
(b)
Another isotope of aluminium has 12 neutrons in its nucleus. Calculate its nucleon number. [2]
(2)
(c)
State one way in which atoms of two isotopes of the same element are the same, and one way they differ. [2]
(2)
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Part (a)
13 protons
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14 neutrons
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13 electrons
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Part (b)
nucleon number = protons + neutrons
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13 + 12 = 25
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Part (c)
same: number of protons (or electrons)
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different: number of neutrons (or nucleon number)
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Common error: The nucleon number counts protons and neutrons together, so it is never equal to the number of neutrons alone.
18IGPC3-5-05c-11Radioactivity, half-life and safetyPaper CEasy6 marks
A student investigates background radiation using a Geiger–Müller counter.
(a)
State two natural sources of background radiation. [2]
(2)
(b)
With no source present the counter records 45 counts in 3.0 minutes. Calculate the background count rate in counts/min. [2]
(2)
(c)
A source is placed near the counter, which records 420 counts in 2.0 minutes. Calculate the corrected count rate caused by the source alone. [2]
(2)
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Part (a)
any one from: radon gas in the air; cosmic rays from space; rocks and soil; food and drink
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a second, different source from the same list
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Part (b)
45 ÷ 3.0
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15 counts/min
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Part (c)
total rate = 420 ÷ 2.0 = 210 counts/min
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210 − 15 = 195 counts/min
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Exam tip: Convert counts to a rate (counts per minute) before subtracting the background rate.
19IGPC3-5-05c-14Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
In the core of a star, hydrogen nuclei undergo nuclear fusion.
(a)
State what happens in nuclear fusion. [2]
(2)
(b)
Explain why fusion happens only at very high temperatures. [2]
(2)
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Part (a)
light nuclei join together
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to form a heavier nucleus, releasing energy
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Part (b)
nuclei are positively charged so they repel each other
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high temperature gives them enough kinetic energy (speed) to get close enough to fuse
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Exam tip: For fusion questions, link the temperature to the kinetic energy of the nuclei and the repulsion between like charges.
20IGPC3-5-05r5-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Iodine-131 is used in pharmacology to treat an overactive thyroid. An atom of iodine-131 has nucleon number 131 and proton number 53. How many neutrons are in one nucleus of iodine-131?
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Neutrons = nucleon number − proton number = 131 − 53 = 78. 53 is only the proton number. 131 is the nucleon number, which counts protons and neutrons together. 184 comes from adding the two numbers instead of subtracting.
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Answer B
Answer: B
Common error: The nucleon number counts protons plus neutrons, so subtract the proton number to find neutrons.
21IGPC3-5-05r5-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Four nuclides are described. W: 6 protons, 6 neutrons X: 6 protons, 8 neutrons Y: 7 protons, 7 neutrons Z: 8 protons, 8 neutrons Which two nuclides are isotopes of the same element?
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Isotopes have the same proton number but different neutron numbers: W and X both have 6 protons. X and Y share a nucleon number of 14 but have different proton numbers. X and Z share a neutron number of 8 but are different elements. W and Y each have equal protons and neutrons, which says nothing about being the same element.
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Answer C
Answer: C
Exam tip: The proton number decides the element. Isotopes keep it fixed and change only the neutron number.
22IGPC3-5-05r5-03Radioactivity, half-life and safetyPaper CEasy1 mark
A water company wants to find a leak in a buried pipe by adding a radioactive tracer to the water and scanning the ground above with a detector. Which type of radiation should the tracer emit?
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Gamma rays are very penetrating, so enough get through the pipe wall and soil to reach the detector. Alpha particles are strongly ionising but are stopped by a few centimetres of air or a thin layer of material. Beta particles are stopped by a few millimetres of metal. Large mass makes alpha penetrate least, not most.
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Answer D
Answer: D
Common error: Strong ionising power does not mean strong penetration. The two properties go in opposite directions.
23IGPC3-5-05r5-09Radioactivity, half-life and safetyPaper CEasy5 marks
Iodine-131 is a radioactive isotope used to treat an overactive thyroid gland. A nucleus of iodine-131 has nucleon number 131 and proton number 53.
(a)
State the number of protons and the number of neutrons in a nucleus of iodine-131.
(2)
(b)
Explain why a gamma emitter, rather than an alpha emitter, is chosen when the radiation must be detected outside the body.
(1)
(c)
A dose has an initial activity of 4.0 MBq. The half-life of iodine-131 is 8.0 days. Calculate the activity of the dose after 24 days.
(2)
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Part (a)
53 protons
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78 neutrons
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Part (b)
gamma is penetrating and passes out of the body to the detector, whereas alpha would be absorbed in the tissue
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Part (c)
24 days is 3 half-lives, so activity = 4.0 ÷ 23
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= 0.5 MBq
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Exam tip: For half-life questions, first work out how many half-lives have passed, then halve that many times.
24IGPC3-5-05r5-16Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
Fluoride compounds are added to some water supplies. A neutral atom of fluorine-19 has nucleon number 19 and proton number 9.
(a)
State the number of electrons in a neutral atom of fluorine-19 and explain why.
(2)
(b)
Fluorine-18 is an isotope of fluorine-19 and is used in medical scans. State what is the same about the nuclei of the two isotopes.
(1)
(c)
Calculate the number of neutrons in a nucleus of fluorine-18.
(1)
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Part (a)
9 electrons
✓ 1
the number of electrons equals the number of protons so the positive and negative charges cancel
✓ 1
Part (b)
they have the same number of protons, 9
✓ 1
Part (c)
18 − 9 = 9
✓ 1
Exam tip: Isotopes are the same element, so they always have the same proton number and the same number of electrons when neutral.
25IGPC3-5-07r5-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Cobalt-60 is used as a gamma source in cancer treatment. A nucleus of cobalt-60 is written 6027Co. How many neutrons are in one nucleus of cobalt-60?
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Neutrons = nucleon number − proton number = 60 − 27 = 33, so B is correct. A is the proton number. C is the nucleon number (protons plus neutrons). D adds the two numbers instead of subtracting.
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Answer B
Answer: B
Common error: The larger number in the nuclide symbol is the total of protons and neutrons, not the number of neutrons.
26IGPC3-5-07r5-02Radioactivity, half-life and safetyPaper CEasy1 mark
A radioactive tracer used in a hospital has a half-life of 6.0 hours. What percentage of the original radioactive nuclei remains after 18 hours?
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18 hours is 3 half-lives (18 ÷ 6.0), so the amount is halved three times: 100 → 50 → 25 → 12.5 %. C is correct. A uses 6 ÷ 18 as if decay were linear. B uses 1/6 from the number of hours. D stops after two halvings, miscounting the half-lives.
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Answer C
Answer: C
Exam tip: Divide the total time by the half-life first to get the number of halvings, then halve that many times.
27IGPC3-5-07r5-09Atomic structure, isotopes, fission and fusionPaper CEasy6 marks
Iodine-131, 13153I, is a radioactive isotope given to patients to treat an overactive thyroid gland.
(a)
State the number of protons and the number of electrons in a neutral atom of iodine-131.
(2)
(b)
Calculate the number of neutrons in a nucleus of iodine-131.
(1)
(c)
Iodine-127 is a stable form of iodine. Explain why iodine-131 and iodine-127 are isotopes of the same element.
(2)
(d)
Iodine-131 emits beta particles. State what a beta particle is.
(1)
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Part (a)
53 protons
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53 electrons
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Part (b)
131 − 53 = 78 neutrons
✓ 1
Part (c)
same number of protons (same proton number)
✓ 1
different number of neutrons (different nucleon number)
✓ 1
Part (d)
a fast-moving electron (emitted from the nucleus)
✓ 1
Exam tip: Isotopes of one element always share a proton number; only the neutron count and nucleon number differ.
28IGPC3-5-07r5-10Radioactivity, half-life and safetyPaper CEasy5 marks
A water company adds a radioactive tracer to water in a pipe to study how the water flows. The tracer has a half-life of 5.0 hours. A sample of the water initially has an activity of 960 counts per second.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the activity of the sample after 20 hours.
(2)
(c)
Explain why a tracer with a half-life of 5.0 hours is more suitable than one with a half-life of 5.0 years.
(2)
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Part (a)
the time taken for the activity (or number of undecayed nuclei) of a sample to fall to half of its initial value
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Part (b)
20 h is 4 half-lives
✓ 1
960 → 480 → 240 → 120 → 60 counts per second
✓ 1
Part (c)
the 5.0-year tracer would stay radioactive in the water for years, which is a hazard to people using it
✓ 1
the 5.0-hour tracer decays away within days but lasts long enough to complete the test
✓ 1
Exam tip: Show each halving of the activity in turn; it makes the working easy to follow and check.
29IGPC3-5-07r5-11Radioactivity, half-life and safetyPaper CEasy6 marks
A hospital pharmacy prepares capsules that contain a gamma-emitting radioactive substance.
(a)
State two precautions that staff should take when handling the capsules.
(2)
(b)
Explain why a source that emits only alpha particles is a smaller hazard than a gamma source while it is outside the body, but a greater hazard if it is swallowed.
(3)
(c)
State one harmful effect of ionising radiation on living cells.
(1)
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Part (a)
one suitable precaution, for example handling with tongs or at a distance
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a second different precaution, for example storing in a lead-lined container, limiting the time near the capsules or wearing a dosimeter badge
✓ 1
Part (b)
outside the body alpha particles are absorbed by air or the outer layer of skin
✓ 1
gamma rays are penetrating and pass through the body
✓ 1
if swallowed, alpha particles are absorbed by nearby tissue and their strong ionisation damages cells
✓ 1
Part (c)
damage to DNA, mutation, cancer or cell death (any one)
✓ 1
Common error: Alpha particles are only a low hazard outside the body; do not say alpha radiation is harmless in general.
30IGPC3-5-05r8-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A hospital uses a technetium isotope whose nucleus contains 43 protons and has a nucleon number of 99. How many neutrons are in one nucleus of this isotope?
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Neutrons = nucleon number − proton number = 99 − 43 = 56. 43 is the proton number only; 99 is the nucleon number; 142 comes from adding the two numbers instead of subtracting.
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Answer B
Answer: B
Exam tip: Neutron number = nucleon number − proton number. Subtract; never add.
31IGPC3-5-05r8-02Radioactivity, half-life and safetyPaper CEasy1 mark
In a hospital physics laboratory a radioactive source is placed in front of a detector. A sheet of paper placed in between hardly changes the count rate, but a 3 mm thick aluminium plate reduces it to the background level. Which radiation does the source emit?
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Beta passes through paper but is stopped by a few mm of aluminium, so beta only is correct. Alpha would be stopped by the paper. Gamma would pass through 3 mm of aluminium with little change. Alpha and gamma fails for both reasons.
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Answer C
Answer: C
Exam tip: Order of penetration: alpha (paper) < beta (few mm of aluminium) < gamma (thick lead or concrete reduces it).
32IGPC3-5-05r8-06Radioactivity, half-life and safetyPaper CEasy1 mark
A hospital technician must handle a radioactive source. Which action gives the technician the smallest radiation dose?
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Distance and time both reduce dose, so long tongs and a short time is correct. Holding the source close increases dose, taking twice as long doubles the exposure time, and cotton gloves do not absorb ionising radiation.
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Answer C
Answer: C
Exam tip: Three ways to reduce dose: increase distance, reduce time, use shielding such as lead.
33IGPC3-5-05r8-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
An iodine-131 nucleus is used in a hospital treatment. It contains 53 protons and has a nucleon number of 131.
(a)
State what is meant by isotopes.
(2)
(b)
State the number of electrons in a neutral iodine atom.
(1)
(c)
Calculate the number of neutrons in an iodine-131 nucleus.
(1)
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Part (a)
Atoms of the same element / same number of protons
✓ 1
with different numbers of neutrons (different nucleon number)
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Part (b)
53
✓ 1
Part (c)
131 − 53 = 78
✓ 1
Exam tip: A neutral atom has equal numbers of protons and electrons.
34IGPC3-5-05r8-13Radioactivity, half-life and safetyPaper CEasy4 marks
A radiographer works with sealed radioactive sources in a hospital.
(a)
State two precautions the radiographer should take when handling the sources.
(2)
(b)
State what is meant by background radiation.
(1)
(c)
Name one natural source of background radiation.
(1)
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Part (a)
Use tongs / handle at a distance, or keep time of exposure short
✓ 1
Store sources in a lead-lined container, or wear a dosimeter badge (any second valid precaution)
✓ 1
Part (b)
Radiation that is always present from the surroundings, not from the source being tested
✓ 1
Part (c)
Any one of: rocks / radon gas / cosmic rays / food
✓ 1
Exam tip: Name precautions that match the hazard: distance, time and shielding.
35IGPC3-5-06r8-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A neutral atom of aluminium, used in aircraft skin, has a proton number of 13 and a nucleon number of 27. How many electrons does the atom have?
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A neutral atom has as many electrons as protons: 13. 14 is the neutron number, 27 is the nucleon number, and 40 adds protons and nucleons.
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Answer C
Answer: C
Exam tip: Neutral atom: electrons = protons.
36IGPC3-5-06r8-02Radioactivity, half-life and safetyPaper CEasy1 mark
A narrow beam of radiation from a source used to follow fertiliser in soil passes between two charged metal plates. The beam is not deflected. Which statement is correct?
Diagram not to scale
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Only uncharged radiation passes undeflected through an electric field, so the beam is gamma rays. Alpha and beta are charged and are deflected; mass or speed does not stop deflection.
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Answer A
Answer: A
Common error: Deflection depends on charge. Gamma rays are uncharged, so are not deflected by electric or magnetic fields.
37IGPC3-5-06r8-06Radioactivity, half-life and safetyPaper CEasy1 mark
An aircraft exit sign contains a radioactive source. Which statement defines the half-life of the source?
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Half-life is the time for the activity (or number of undecayed nuclei) to halve. All nuclei never decay at a definite time, so B fails; a nucleus does not lose half its nucleons, and background radiation is independent of the source.
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Answer C
Answer: C
Exam tip: Quote 'activity (or number of undecayed nuclei) falls to half' in definitions.
38IGPC3-5-06r8-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
Old aircraft instrument dials were painted with radium-226 paint. A radium-226 nucleus has 88 protons and a nucleon number of 226.
(a)
State the two types of particle found in the nucleus.
(2)
(b)
State the number of protons in a nucleus of radium-228.
(1)
(c)
Calculate the number of neutrons in a radium-226 nucleus.
(1)
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Mark
Notes
Part (a)
Protons
✓ 1
Neutrons
✓ 1
Part (b)
88
✓ 1
Part (c)
226 − 88 = 138
✓ 1
Exam tip: Isotopes of one element always have the same proton number.
39IGPC3-5-06r8-10Radioactivity, half-life and safetyPaper CEasy4 marks
A grain store uses a sealed radioactive source to monitor the level of grain in a silo.
(a)
State one harmful effect of ionising radiation on living cells.
(1)
(b)
State two safety precautions for a farm worker who must work near the source.
(2)
(c)
Name one natural source of background radiation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Any one of: damages or kills cells / causes cancer / causes mutations in DNA
✓ 1
Part (b)
Keep the source in a lead-lined container or use shielding
✓ 1
Keep time near the source short, or stay as far away as possible
✓ 1
Part (c)
Any one of: rocks / radon gas / cosmic rays / food
✓ 1
Exam tip: Give precautions that reduce time, increase distance or add shielding.
40IGPC3-5-05r10-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which pair of nuclides are isotopes of the same element?
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Marking point
Mark
Notes
B is correct: both have 6 protons (same element) but different numbers of neutrons (6 and 8). A has the same nucleon number but different proton numbers, so they are different elements (matching nucleon number does not make isotopes). C and D differ in both proton number and nucleon number, so they are different elements.
✓ 1
Answer B
Answer: B
Common error: Isotopes share the proton number, not the nucleon number.
41IGPC3-5-05r10-02Radioactivity, half-life and safetyPaper CEasy1 mark
Which statement correctly describes the half-life of a radioactive isotope?
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Marking point
Mark
Notes
B is correct: half-life is the time for half of the unstable nuclei present to decay. A is wrong because the mass of a sample barely changes as the decay products remain in it. C is wrong because decay is random and a sample never fully decays in a fixed time. D is wrong because the count rate keeps falling but never reaches zero in a finite time.
✓ 1
Answer B
Answer: B
Common error: Half-life does not mean the sample has gone after two half-lives; half of what is left decays each time.
42IGPC3-5-05r10-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
Iodine-131, 13153I, is used in medicine. A stable isotope of iodine is iodine-127.
(a)
State the number of protons in a nucleus of iodine-131.
(1)
(b)
Calculate the number of neutrons in a nucleus of iodine-131.
(1)
(c)
Describe how a nucleus of iodine-127 differs from a nucleus of iodine-131, and state what they have in common.
(2)
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Marking point
Mark
Notes
Part (a)
53
✓ 1
Part (b)
131 - 53 = 78
✓ 1
Part (c)
iodine-127 has fewer neutrons (74 compared with 78)
✓ 1
same number of protons / same proton number of 53
✓ 1
Common error: The nucleon number is protons plus neutrons, so subtract the proton number to find the neutrons.
43IGPC3-5-05r10-16Radioactivity, half-life and safetyPaper CEasy4 marks
A shopping centre has a Geiger-Müller counter that records the background radiation. It records 126 counts in 6.0 minutes.
(a)
Calculate the background count rate in counts/min.
(1)
(b)
State two sources of background radiation.
(2)
(c)
A sample is placed near the counter and the counter records 96 counts/min. Calculate the count rate due to the sample alone.
(1)
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Marking point
Mark
Notes
Part (a)
126 / 6.0 = 21 counts/min
✓ 1
Part (b)
any one from: radon gas / rocks and soil / cosmic rays / food and drink / buildings / medical sources
✓ 1
a different second source from the same list
✓ 1
Part (c)
96 - 21 = 75 counts/min
✓ 1
Common error: Always subtract the background count rate to find the count rate from the source alone.
44IGPC3-5-06r10-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Archaeologists date wooden objects using carbon-14, 146C. How many neutrons are in one nucleus of carbon-14?
Show mark scheme
Marking point
Mark
Notes
Neutrons = nucleon number - proton number = 14 - 6 = 8 (B). A is the number of protons. C is the nucleon number (protons plus neutrons). D adds the protons and nucleons together.
✓ 1
Answer B
Answer: B
Common error: The nucleon number counts protons and neutrons together.
45IGPC3-5-06r10-02Radioactivity, half-life and safetyPaper CEasy1 mark
The smoke detectors in a data centre contain a source of alpha particles. Which statement about alpha particles is correct?
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Mark
Notes
B is correct: alpha particles are strongly ionising and are absorbed by a few centimetres of air, which is why a small gap in a detector can be used. A describes gamma rays. D is wrong because paper stops alpha. C is wrong as many heavy nuclei emit alpha particles, not only lead.
✓ 1
Answer B
Answer: B
Exam tip: Ranking penetration: alpha is stopped by paper, beta by a few millimetres of aluminium, gamma is reduced only by thick lead or concrete.
46IGPC3-5-06r10-09Atomic structure, isotopes, fission and fusionPaper CEasy3 marks
A carbon nucleus has 6 protons and 8 neutrons. A second carbon nucleus has 6 protons and 6 neutrons.
(a)
Explain why the two nuclei are isotopes of the same element.
(2)
(b)
Calculate the nucleon number of the nucleus with 8 neutrons.
(1)
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Mark
Notes
Part (a)
they have the same number of protons
✓ 1
but different numbers of neutrons
✓ 1
Part (b)
6 + 8 = 14
✓ 1
Common error: Isotopes of one element always have the same proton number.
47IGPC3-5-06r10-16Radioactivity, half-life and safetyPaper CEasy4 marks
Some old bicycle lamps use a glass tube containing tritium, a radioactive isotope of hydrogen, to glow in the dark. The half-life of tritium is 12 years. A new tube has an activity of 800 Bq.
(a)
Calculate the number of half-lives in 36 years.
(1)
(b)
Calculate the activity of the tube after 36 years.
(2)
(c)
Suggest why the sealed glass tube makes the lamp safer to handle.
(1)
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Mark
Notes
Part (a)
36 / 12 = 3
✓ 1
Part (b)
800 / 23
✓ 1
= 100 Bq
✓ 1
Part (c)
the radiation (low-energy beta particles) cannot pass through the glass so it stays inside the tube
✓ 1
Exam tip: Write the number of half-lives first; it keeps the halving calculation clear.
48IGPC3-5-06r11-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A marine biologist tags crabs with strontium-90. Each atom has 38 protons and 52 neutrons in its nucleus. What is the nucleon number of this isotope?
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Notes
The nucleon number is the total number of protons and neutrons: 38 + 52 = 90. A is the proton number. B is the neutron number. D is the difference between the numbers of neutrons and protons.
✓ 1
Answer C
Answer: C
Common error: Nucleon number counts protons and neutrons together; proton number counts protons only.
49IGPC3-5-06r11-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Driftwood samples are dated using carbon-12 and carbon-14. Which statement correctly describes these two isotopes?
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Mark
Notes
Isotopes of one element have the same proton number but different neutron numbers, so different nucleon numbers. B swaps protons and neutrons. C is wrong because the nucleon numbers 12 and 14 differ. D describes ions, not isotopes.
✓ 1
Answer A
Answer: A
Exam tip: The element is set by the proton number; isotopes differ only in neutron number.
50IGPC3-5-06r11-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
Wooden beams in an old harbour warehouse are dated using carbon-14. A neutral atom of carbon-14 has nucleon number 14 and proton number 6.
(a)
Determine the number of neutrons in a nucleus of carbon-14.
(1)
(b)
State the number of electrons in a neutral atom of carbon-14.
(1)
(c)
Carbon-12 is another isotope of carbon. Explain why carbon-12 and carbon-14 are isotopes of the same element.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
14 − 6 = 8
✓ 1
Part (b)
6
✓ 1
Part (c)
they have the same number of protons (proton number 6)
✓ 1
they have different numbers of neutrons
✓ 1
Exam tip: In a neutral atom the number of electrons equals the number of protons.
51IGPC3-5-06r11-10Radioactivity, half-life and safetyPaper CEasy4 marks
Aluminium foil used to package frozen cargo is checked on a production line using a radioactive source.
(a)
State which of alpha, beta and gamma radiation is the most strongly ionising.
(1)
(b)
State which of the three types of radiation is electromagnetic.
(1)
(c)
A beta source is used to monitor the thickness of the foil. Explain why beta is more suitable than alpha or gamma.
(2)
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Mark
Notes
Part (a)
alpha
✓ 1
Part (b)
gamma
✓ 1
Part (c)
beta is partly absorbed by thin foil, so the detected count rate changes with the thickness
✓ 1
alpha would be stopped completely and gamma would hardly be absorbed, so neither would show changes in thickness
✓ 1
Exam tip: Match the radiation to the absorber: the material should reduce the count rate partly, not completely and not at all.
52IGPC3-5-06r11-11Radioactivity, half-life and safetyPaper CEasy5 marks
A marine biologist injects fish with a radioactive tracer that has a half-life of 8.0 days. The initial activity of the tracer in one fish is 640 Bq.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the activity of the tracer after 24 days.
(2)
(c)
Calculate the time taken for the activity to fall to 20 Bq.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the time taken for the activity (or number of undecayed nuclei) to fall to half its initial value
✓ 1
Part (b)
24 days is 3 half-lives
✓ 1
640 → 320 → 160 → 80 Bq
✓ 1
Part (c)
640 → 320 → 160 → 80 → 40 → 20 is 5 half-lives
✓ 1
5 × 8.0 = 40 days
✓ 1
Exam tip: Count the number of halvings first, then multiply by the half-life.
53IGPC3-5-07r19-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
An archaeologist analyses a bronze artefact that contains copper-63, written 6329Cu. Which row gives the particles in one neutral atom of this nuclide?
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Mark
Notes
A is correct: the proton number is 29, the neutron number is 63 − 29 = 34, and a neutral atom has 29 electrons. B treats the nucleon number as the number of neutrons. C swaps the numbers of protons and neutrons. D treats the nucleon number as the number of protons.
✓ 1
Answer A
Answer: A
Common error: The top number is the nucleon number (protons + neutrons), not the number of neutrons.
54IGPC3-5-07r19-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Wood from an archaeological site contains atoms of carbon-12 and carbon-14. Which statement about these two nuclides is correct?
Show mark scheme
Marking point
Mark
Notes
B is correct: isotopes of an element have the same proton number but different numbers of neutrons (6 and 8 here). A reverses protons and neutrons. C is wrong because isotopes differ in nucleon number. D is wrong because both are carbon, so both have 6 protons and the same chemical properties.
✓ 1
Answer B
Answer: B
Exam tip: Isotopes are atoms of the same element, so the proton number must be the same.
55IGPC3-5-07r19-03Radioactivity, half-life and safetyPaper CEasy1 mark
A technician in a hospital handles a sealed radioactive source. Which action would NOT reduce the technician's exposure to radiation?
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Notes
C is correct because it brings the source closer, so the radiation received increases. A increases the distance, B absorbs radiation in the shielding and D reduces the exposure time, so all three reduce exposure.
✓ 1
Answer C
Answer: C
Exam tip: Remember the three ways to reduce exposure: more distance, more shielding, less time.
56IGPC3-5-07r19-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
A data centre fire alarm contains americium-241, written 24195Am. Americium-243 is another nuclide of the same element.
(a)
State the number of protons and the number of neutrons in one nucleus of americium-241.
(2)
(b)
State one way in which a nucleus of americium-243 is the same as, and one way in which it is different from, a nucleus of americium-241.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
95 protons
✓ 1
241 − 95 = 146 neutrons
✓ 1
Part (b)
same: both have 95 protons (accept same proton number)
✓ 1
different: americium-243 has 148 neutrons / two more neutrons / greater nucleon number
✓ 1
Common error: Isotopes differ in neutron number; the proton number defines the element.
57IGPC3-5-07r19-10Radioactivity, half-life and safetyPaper CEasy5 marks
A conservator tests three unknown sources, P, Q and R, using a counter. The results are: Source P: the count rate falls to the background level when a sheet of paper is placed in front of it. Source Q: paper has no effect, but 3 mm of aluminium reduces the count rate to the background level. Source R: aluminium of thickness 3 mm has almost no effect, but a thick block of lead reduces the count rate a lot.
(a)
Identify the type of radiation emitted by each of P, Q and R.
(3)
(b)
State which of the three types of radiation is the most strongly ionising.
(1)
(c)
State one harmful effect of ionising radiation on living cells.
(1)
Show mark scheme
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Mark
Notes
Part (a)
P is alpha
✓ 1
Q is beta
✓ 1
R is gamma
✓ 1
Part (b)
alpha
✓ 1
Part (c)
any one from: damages / mutates DNA; causes cancer; kills cells; causes cell damage
✓ 1
Exam tip: Remember the order of penetration: alpha (paper), beta (thin aluminium), gamma (thick lead).
58IGPC3-5-07r21-01Radioactivity, half-life and safetyPaper CEasy1 mark
A sample of sea-floor sediment contains the nuclide strontium-90, written 9038Sr. How many neutrons are in one nucleus of this nuclide?
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Marking point
Mark
Notes
Neutrons = nucleon number − proton number = 90 − 38 = 52, so B is correct. A is the proton number. C is the nucleon number, which includes protons. D adds the two numbers instead of subtracting.
✓ 1
Answer B
Answer: B
Exam tip: In nuclide notation the top number is protons plus neutrons and the bottom number is protons only.
59IGPC3-5-07r21-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A marine-biology laboratory compares four nuclides. The numbers of protons and nucleons (protons plus neutrons) are: P: 6 protons, 12 nucleons Q: 7 protons, 14 nucleons R: 6 protons, 14 nucleons S: 8 protons, 16 nucleons Which two nuclides are isotopes of the same element?
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Marking point
Mark
Notes
Isotopes have the same number of protons but different numbers of neutrons. P and R both have 6 protons (6 and 8 neutrons), so C is correct. Q and R share a nucleon number but have different proton numbers, so they are different elements. R and S both have 8 neutrons but different proton numbers. P and S each have equal numbers of protons and neutrons, but their proton numbers differ.
✓ 1
Answer C
Answer: C
Common error: Isotopes are matched by proton number, not by nucleon number or neutron number.
60IGPC3-5-07r21-09Atomic structure, isotopes, fission and fusionPaper CEasy6 marks
Carbon-14, written 146C, is found in shells on the sea floor.
(a)
State the number of protons and the number of electrons in a neutral atom of carbon-14.
(2)
(b)
Calculate the number of neutrons in one nucleus of carbon-14.
(1)
(c)
Carbon-12 and carbon-14 are isotopes of carbon. State what is meant by isotopes.
(2)
(d)
Carbon-14 is a beta emitter. State what a beta particle is.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
6 protons
✓ 1
6 electrons
✓ 1
Part (b)
14 − 6 = 8
✓ 1
Part (c)
atoms (nuclei) with the same number of protons
✓ 1
but different numbers of neutrons (different nucleon numbers)
✓ 1
Part (d)
a fast-moving electron emitted from the nucleus
✓ 1
Common error: Isotopes of an element have the same proton number; it is the neutron number (and so the nucleon number) that differs.
61IGPC3-5-07r21-15Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
In an experiment, alpha particles are fired at a very thin sheet of gold foil in a vacuum, and a detector records where they emerge. Gold has the nuclide 19779Au.
Diagram not to scale
(a)
State what was observed for most of the alpha particles, and what this shows about the atom.
(2)
(b)
A very small number of alpha particles bounce back. Explain what this shows about the nucleus.
(2)
(c)
Calculate the number of neutrons in one gold nucleus.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
most alpha particles pass straight through without deflection
✓ 1
so most of the atom is empty space
✓ 1
Part (b)
alpha particles are repelled by a concentrated positive charge, so the nucleus is positively charged
✓ 1
the nucleus is very small and contains most of the mass of the atom
✓ 1
Part (c)
197 − 79 = 118
✓ 1
Exam tip: Pair each observation with the conclusion it supports: no deflection means empty space, rebound means a small dense nucleus.
62IGPC3-5-07r22-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A cobalt nuclide used to sterilise food has proton number 27 and nucleon number 60. How many neutrons are in one nucleus of this nuclide?
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Marking point
Mark
Notes
Neutrons = nucleon number − proton number = 60 − 27 = 33. 27 is only the number of protons. 60 is the total number of nucleons, not the neutrons alone. 87 comes from adding the two numbers instead of subtracting.
✓ 1
Answer B
Answer: B
Common error: Nucleon number counts protons and neutrons together, so subtract the proton number to find the neutrons.
63IGPC3-5-07r22-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Four nuclides are described by their numbers of protons and neutrons. P: 19 protons, 20 neutrons Q: 20 protons, 19 neutrons R: 19 protons, 22 neutrons S: 18 protons, 22 neutrons Which two nuclides are isotopes of the same element?
Show mark scheme
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Mark
Notes
Isotopes of one element have the same number of protons but different numbers of neutrons. P and R both have 19 protons and differ in neutron number (20 and 22). P and Q have the same nucleon number (39) but different proton numbers, so they are different elements. Q and S have different proton numbers. R and S have equal neutron numbers but different proton numbers.
✓ 1
Answer C
Answer: C
Common error: Isotopes share the proton number (the element); matching nucleon number or matching neutron number does not make nuclides isotopes.
64IGPC3-5-07r22-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
Sodium-24 is a radioactive isotope that food technologists add in tiny amounts to a liquid to trace its flow through a processing pipe. A nucleus of sodium-24 has proton number 11 and nucleon number 24. Stable sodium-23 has the same proton number.
(a)
Explain what is meant by the term isotopes.
(2)
(b)
State the number of electrons in a neutral atom of sodium-24.
(1)
(c)
Calculate the number of neutrons in a nucleus of sodium-24.
(2)
Show mark scheme
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Mark
Notes
Part (a)
atoms (nuclei) of the same element / with the same number of protons
✓ 1
but different numbers of neutrons
✓ 1
Part (b)
11
✓ 1
Part (c)
neutrons = nucleon number − proton number
✓ 1
24 − 11 = 13
✓ 1
Common error: Neutral atoms have equal numbers of protons and electrons; the neutrons do not affect this.
65IGPC3-5-07r22-10Radioactivity, half-life and safetyPaper CEasy5 marks
A student studies a sample of iodine-131. The table shows the count rate from the sample after the background count rate has been subtracted. Time / days: 0, 8, 16, 24 Corrected count rate / counts per minute: 960, 480, 240, 120
(a)
Determine the half-life of iodine-131.
(2)
(b)
Predict the corrected count rate after 40 days.
(2)
(c)
Explain why the background count rate is measured and subtracted.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
time taken for the corrected count rate to halve, e.g. 960 to 480
✓ 1
8 days
✓ 1
Part (b)
40 days = 5 half-lives
✓ 1
960 ÷ 25 = 30 counts per minute
✓ 1
ECF from (a)
—
Part (c)
background radiation is present even with no sample, so it would make the sample seem more active than it is
✓ 1
Exam tip: Write down each halving step or the number of half-lives so that method marks are clear.
66IGPC3-5-07r22-14Radioactivity, half-life and safetyPaper CEasy6 marks
A school science department stores small radioactive sources that emit alpha, beta or gamma radiation.
(a)
State which of alpha, beta and gamma radiation is stopped by a sheet of paper.
(1)
(b)
State which of the three types of radiation is the most penetrating.
(1)
(c)
Give two safety precautions for storing or using a radioactive source in a school.
(2)
(d)
Name two sources of background radiation.
(2)
Show mark scheme
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Mark
Notes
Part (a)
alpha
✓ 1
Part (b)
gamma
✓ 1
Part (c)
one sensible precaution, e.g. store in a lead-lined container, handle with tongs, keep a distance
✓ 1
a second, different precaution, e.g. limit the time of use, label with the hazard symbol, keep locked away
✓ 1
Part (d)
one natural or artificial source, e.g. radon gas, rocks and soil, cosmic rays
✓ 1
a second, different source, e.g. food and drink, medical sources, nuclear fallout
✓ 1
Common error: Background radiation includes natural sources; do not describe it as coming only from nuclear power stations.
67IGPC3-5-09r27-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A sample of gas from a volcanic vent contains radon. One nuclide of radon is written 22286Rn. How many neutrons are in the nucleus of this nuclide?
Show mark scheme
Marking point
Mark
Notes
Neutron number = nucleon number − proton number = 222 − 86 = 136. 86 is only the number of protons. 222 is the nucleon number, which counts protons and neutrons together. 308 comes from adding the two numbers instead of subtracting.
✓ 1
Answer B
Answer: B
Exam tip: In nuclide notation the top number is the nucleon number and the bottom number is the proton number; neutrons = top − bottom.
68IGPC3-5-09r27-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Two atoms of sulfur are 3216S and 3416S. Which statement about these atoms is correct?
Show mark scheme
Marking point
Mark
Notes
Both nuclides have 16 protons, so both are sulfur and they are isotopes; they have 16 and 18 neutrons. The statement with equal neutrons is wrong because it is the neutron number that differs. The nucleon numbers are 32 and 34, so they are not equal. An element is defined by its proton number, not its mass, so a different mass does not make a different element.
✓ 1
Answer D
Answer: D
Common error: Isotopes are not different elements: the proton number fixes the element, and only the neutron number changes.
69IGPC3-5-09r27-04Radioactivity, half-life and safetyPaper CEasy1 mark
A dock worker has to move a sealed container holding a gamma-emitting source. Which action best reduces the radiation dose she receives?
Show mark scheme
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Mark
Notes
Distance and dense shielding both reduce exposure to gamma radiation, so long tongs and lead are the best choice. Cotton cloth absorbs almost no gamma radiation. A plastic bag stops contamination, but gamma rays pass straight through it. Washing removes contamination only; it does nothing about the radiation received while the source is held close.
✓ 1
Answer C
Answer: C
Exam tip: To cut exposure to a source: increase distance, reduce time, and use shielding such as lead.
70IGPC3-5-09r27-09Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
A theatre's smoke detector contains a small sample of americium-241, written 24195Am. Americium-243 is another form of americium.
(a)
Explain why americium-241 and americium-243 are isotopes of the same element.
(2)
(b)
State the number of electrons in a neutral atom of americium-241.
(1)
(c)
Calculate the number of neutrons in a nucleus of americium-241.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
both have the same number of protons (95)
✓ 1
they have different numbers of neutrons
✓ 1
Part (b)
95
✓ 1
Part (c)
241 − 95 = 146
✓ 1
Common error: Do not say isotopes have different 'atomic numbers'; the proton number is the same and the neutron number is different.
71IGPC3-5-09r27-10Radioactivity, half-life and safetyPaper CEasy4 marks
A geologist samples a radioactive gas from a volcanic vent. After correcting for background, the count rate from the gas was 640, 320, 160 and 80 counts/min at times of 0, 4, 8 and 12 days after collection.
(a)
Determine the half-life of the gas.
(2)
(b)
Predict the corrected count rate 20 days after collection.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the count rate halves from 640 to 320 counts/min in 4 days
✓ 1
half-life = 4 days
✓ 1
Part (b)
20 days is 5 half-lives, so the count rate halves 5 times
✓ 1
640 / 32 = 20 counts/min
✓ 1
Exam tip: Show the halving steps (640, 320, 160, ...) to secure the method mark even if the final halving is slipped.
72IGPC3-5-09r28-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A neutral atom of chlorine-35 has 17 protons and 18 neutrons. Which nucleus is an isotope of this atom?
Show mark scheme
Marking point
Mark
Notes
Isotopes of an element have the same number of protons but different numbers of neutrons, so 17 protons with 20 neutrons is correct. 18 protons with 17 neutrons has the same nucleon number (35) but is a different element, so it is not an isotope. 17 protons with 18 neutrons is identical to the original atom, not a different isotope. 19 protons with 20 neutrons has a different number of protons, so it is another element.
✓ 1
Answer B
Answer: B
Common error: Isotopes share the same proton number, not the same nucleon number. Check the protons first.
73IGPC3-5-09r28-02Radioactivity, half-life and safetyPaper CEasy1 mark
A nucleus decays by emitting a beta-minus particle. How do the proton number and the nucleon number of the nucleus change?
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Mark
Notes
In beta-minus decay a neutron changes into a proton and an electron, which is emitted. The proton number rises by 1 and the nucleon number stays the same. A fall of 2 in proton number with a fall of 4 in nucleon number describes alpha decay. A fall of 1 in proton number confuses beta-minus with the opposite process of a proton becoming a neutron. An unchanged proton number with one fewer nucleon would mean a neutron was simply lost, which does not happen in beta decay.
✓ 1
Answer D
Answer: D
Common error: The emitted electron comes from the nucleus: a neutron becomes a proton, so the nucleon number stays fixed.
74IGPC3-5-09r28-06Radioactivity, half-life and safetyPaper CEasy1 mark
A technician handles a gamma-emitting source in a hospital laboratory. Which procedure is the best way to reduce the radiation dose received?
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Notes
Lead absorbs gamma radiation, tongs increase the distance from the source and a short time of exposure reduces the dose, so these three measures together are best. A thin plastic bag absorbs almost no gamma radiation. Carrying the source in a pocket keeps it as close as possible to the body. A glass jar gives almost no shielding and leaves the source unprotected.
✓ 1
Answer A
Answer: A
Exam tip: Link each safety precaution to a reason: shielding, distance or time.
75IGPC3-5-09r28-08Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which statement describes nuclear fusion, the process that releases energy in the Sun?
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Notes
Fusion is the joining of light nuclei, such as hydrogen isotopes, to form a heavier nucleus such as helium, with energy released. Splitting after absorbing a neutron describes fission. Emitting an electron as a neutron becomes a proton describes beta-minus decay. Fusion cannot happen gently at room temperature because the positively charged nuclei repel strongly, so very high temperature and pressure are needed.
✓ 1
Answer B
Answer: B
Common error: Do not mix up fission (a heavy nucleus splits) and fusion (light nuclei join).
76IGPC3-5-09r28-09Atomic structure, isotopes, fission and fusionPaper CEasy7 marks
Iodine-131 is a radioactive isotope used in the treatment of thyroid conditions. A nucleus of iodine-131 is written 13153I. Iodine-127 is a stable isotope of iodine.
(a)
State what is meant by isotopes.
(2)
(b)
State the number of electrons in a neutral atom of iodine-131.
(1)
(c)
Calculate the number of neutrons in a nucleus of iodine-131.
(2)
(d)
Iodine-131 decays by beta-minus emission. Describe what happens inside the nucleus when a beta particle is emitted.
(2)
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Part (a)
atoms (nuclei) of the same element / same number of protons
✓ 1
with different numbers of neutrons (different nucleon numbers)
✓ 1
Part (b)
53
✓ 1
Part (c)
M1 neutrons = nucleon number − proton number = 131 − 53
✓ 1
A1 78
✓ 1
Part (d)
a neutron changes into a proton
✓ 1
and an electron (the beta particle) is emitted from the nucleus
✓ 1
Exam tip: Give the nucleon number minus the proton number for neutrons; do not subtract the electron count.
77IGPC3-5-08r29-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Phosphorus-32, 3215P, is used as a tracer to follow how crops take up fertiliser. How many protons, neutrons and electrons are in a neutral atom of this nuclide?
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The proton number is 15, so there are 15 protons and, in a neutral atom, 15 electrons. The neutron number is 32 − 15 = 17. The row with 32 neutrons uses the nucleon number as the neutron number, the row with 17 protons swaps the proton and neutron numbers, and the row with 17 electrons gives a charged atom.
✓ 1
Answer C
Answer: C
Common error: Neutrons = nucleon number − proton number. The electron number equals the proton number only in a neutral atom.
78IGPC3-5-08r29-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A soil scientist studies carbon-12 and carbon-14 in a field sample. Which statement about these two isotopes of carbon is correct?
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Notes
Isotopes are atoms of the same element, so they have the same proton number, and they differ in neutron number and hence in nucleon number. The statements with different proton numbers describe different elements, and neutral atoms of the same element have equal numbers of electrons.
✓ 1
Answer A
Answer: A
Exam tip: The element is fixed by the number of protons; isotopes only differ in the number of neutrons.
79IGPC3-5-08r29-03Radioactivity, half-life and safetyPaper CEasy1 mark
A radioactive source emits radiation that passes through a sheet of paper but is almost completely stopped by a 3 mm sheet of aluminium. What is the radiation?
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Beta particles pass through paper but are stopped by a few millimetres of aluminium. Alpha particles would be stopped by the paper. Gamma rays would pass through the aluminium with only a small reduction. Ultraviolet radiation is not emitted by radioactive nuclei and would be absorbed by the paper.
✓ 1
Answer D
Answer: D
Exam tip: Remember the order of penetration: paper stops alpha, aluminium stops beta, thick lead or concrete reduces gamma.
80IGPC3-5-08r29-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
Aircraft cabins contain smoke detectors. Each detector has a small source of americium-241, 24195Am, which emits alpha particles.
(a)
State what an alpha particle is.
(1)
(b)
Another isotope is americium-243. Explain what is meant by isotopes.
(2)
(c)
State the number of protons and the number of neutrons in a nucleus of americium-241.
(2)
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Part (a)
A helium nucleus (2 protons and 2 neutrons)
✓ 1
Part (b)
Same number of protons (same proton number)
✓ 1
Different number of neutrons
✓ 1
Part (c)
95 protons
✓ 1
146 neutrons
✓ 1
Exam tip: Neutron number is nucleon number minus proton number.
81IGPC3-5-09r29-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A shop sells ornaments made of copper. One atom of copper-63 is written 6329Cu. How many neutrons are in the nucleus of this atom?
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Notes
The nucleon number is the total of protons and neutrons, so neutrons = 63 − 29 = 34. 29 is the number of protons, 63 is the whole nucleon number, and 92 comes from wrongly adding the two numbers.
✓ 1
Answer B
Answer: B
Common error: The lower number is the proton number and the upper number is the nucleon number; neutrons are found by subtracting.
82IGPC3-5-09r29-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
In a scattering experiment, alpha particles are fired at a very thin gold foil. Almost all of them pass straight through, but a tiny fraction are deflected through large angles. Which conclusion is supported by these results?
Diagram not to scale
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Notes
Most alpha particles pass undeflected, so most of the atom is empty space; the rare large deflections show a small, dense, positive nucleus. A solid positive sphere would deflect nearly every particle slightly. A nucleus filling most of the atom would deflect many particles, not very few. Electrons are far lighter than alpha particles and cannot cause large deflections.
✓ 1
Answer D
Answer: D
Exam tip: Link each observation to a conclusion: 'most pass through' means mostly empty space; 'a few deflected strongly' means a small, dense, charged nucleus.
83IGPC3-5-09r29-07Radioactivity, half-life and safetyPaper CEasy1 mark
A technician in a hospital must handle a gamma-emitting source. Which set of actions gives the lowest exposure to radiation?
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Exposure is reduced by increasing distance (tongs), reducing time and using shielding (lead). Handling by hand, slow movements and keeping the source close all increase exposure, and leaving the shield open lets gamma rays escape.
✓ 1
Answer D
Answer: D
Exam tip: Safety answers usually need two or three of: shorter time, larger distance, shielding.
84IGPC3-5-09r29-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
A shop sells lithium batteries. A neutral atom of lithium-7 is written 73Li.
(a)
State the number of protons and the number of neutrons in one nucleus of lithium-7.
(2)
(b)
A different atom is written 63Li. Explain why it is an isotope of lithium-7.
(2)
(c)
A lithium atom loses one electron to form a positive ion. State the number of electrons in this ion.
(1)
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Part (a)
3 protons
✓ 1
4 neutrons
✓ 1
Part (b)
same number of protons / same proton number
✓ 1
different number of neutrons (3 instead of 4)
✓ 1
Part (c)
3 − 1 = 2
✓ 1
Common error: An isotope has the same proton number but a different nucleon number; the element is fixed by the number of protons.
85IGPC3-5-09r29-11Radioactivity, half-life and safetyPaper CEasy4 marks
A factory makes thin aluminium foil. A radioactive source on one side of the foil and a detector on the other side are used to monitor the thickness of the foil as it is made.
Diagram not to scale
(a)
State which one of alpha, beta and gamma radiation is stopped by a sheet of paper.
(1)
(b)
State which one of alpha, beta and gamma radiation is not deflected by an electric field.
(1)
(c)
State which type of radiation is most suitable for monitoring the thickness of the foil and give a reason for your choice.
(2)
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Part (a)
alpha
✓ 1
Part (b)
gamma
✓ 1
Part (c)
beta
✓ 1
reason: alpha would all be absorbed / gamma would pass almost unchanged, but beta is partly absorbed so the count rate changes with the thickness
✓ 1
Exam tip: For thickness-gauge questions, a suitable radiation is one that is partly absorbed; explain why each of the other two would not work.
86IGPC3-5-08r30-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A neutral atom of sodium is written 2311Na. How many protons, neutrons and electrons does it contain?
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The proton number is 11, so there are 11 protons; neutrons = nucleon number − proton number = 23 − 11 = 12; a neutral atom has as many electrons as protons, so 11. The choice with 23 neutrons uses the nucleon number as the neutron count. The choice with 12 protons and 11 neutrons swaps the proton and neutron numbers. The choice with 12 electrons gives the atom a negative charge.
✓ 1
Answer A
Answer: A
Common error: Neutron number = nucleon number (top) − proton number (bottom).
87IGPC3-5-08r30-06Radioactivity, half-life and safetyPaper CEasy1 mark
A school technician has to move a sealed radioactive source for a demonstration. Which action is the safest?
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Notes
Long tongs keep the source far from the body, and pointing it away from people reduces exposure to anyone nearby. Carrying it in a pocket keeps it close to the body for a long time. Leaving it out of its shielded box gives unnecessary exposure. An unlabelled cupboard means people cannot know of the hazard.
✓ 1
Answer B
Answer: B
Exam tip: Safe handling means keeping a large distance, for a short time, with shielding.
88IGPC3-5-08r30-07Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
An atom of aluminium, 2713Al, loses three electrons and becomes an ion. How many protons and electrons does the ion contain?
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Notes
Losing electrons does not change the nucleus, so there are still 13 protons, and the number of electrons falls from 13 to 10. 10 protons and 13 electrons removes the charge from the protons instead. 16 protons and 13 electrons adds three protons. 13 protons and 16 electrons gains three electrons instead of losing them.
✓ 1
Answer D
Answer: D
Common error: Ions form by gaining or losing electrons only; the number of protons never changes.
89IGPC3-5-08r30-15Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
A theatre’s prop store holds an old stage clock whose dial is painted with a radioactive paint. The paint emits alpha, beta and gamma radiation.
(a)
State what an alpha particle is.
(1)
(b)
State what a beta particle is.
(1)
(c)
State what gamma radiation is.
(1)
(d)
State which of the three types of radiation is the most ionising and which is the most penetrating.
(2)
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Notes
Part (a)
a helium nucleus / two protons and two neutrons
✓ 1
Part (b)
a high-speed electron (from the nucleus)
✓ 1
Part (c)
electromagnetic radiation of very high frequency
✓ 1
Part (d)
alpha is the most ionising
✓ 1
gamma is the most penetrating
✓ 1
Common error: A beta particle is a fast electron, not a gamma ray; gamma rays are waves with no mass or charge.
90IGPC3-5-07r38-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A marine laboratory keeps a sealed sample of strontium-90. Each nucleus contains 38 protons and has a nucleon number of 90. How many neutrons are in one nucleus?
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Notes
Neutrons = nucleon number − proton number = 90 − 38 = 52. The value 38 is the number of protons, 90 is the total number of nucleons, and 128 comes from adding the two numbers instead of subtracting.
✓ 1
Answer B
Answer: B
Common error: Do not confuse nucleon number (protons + neutrons) with proton number; the number of neutrons is their difference.
91IGPC3-5-07r38-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Timber from a shipwreck is dated by comparing the carbon-14 and carbon-12 in it. Which statement correctly compares atoms of carbon-12 and carbon-14?
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Notes
Both are carbon, so each nucleus has 6 protons; the isotopes differ only in neutron number (6 and 8). Different proton numbers would make them different elements, the neutron numbers are not the same, and a neutral atom of either isotope has 6 electrons.
✓ 1
Answer D
Answer: D
Exam tip: Isotopes of one element always share the same proton number; the difference is in the neutron number.
92IGPC3-5-07r38-06Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A coastal power station generates electricity from uranium-235. Which statement describes nuclear fission in the reactor?
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Fission is the splitting of a large nucleus after neutron absorption, with energy and further neutrons released. Joining of small nuclei is fusion, emission of an alpha particle is radioactive decay rather than fission, and absorbing a proton to become lead is not what happens in a reactor.
✓ 1
Answer C
Answer: C
Common error: Fission splits a heavy nucleus; fusion joins light nuclei. Do not swap them.
93IGPC3-5-07r38-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
A smoke detector in an office building contains a small sample of americium-241. An atom of this isotope has a proton number of 95 and a nucleon number of 241.
(a)
State the number of electrons in a neutral atom of americium-241.
(1)
(b)
Calculate the number of neutrons in one nucleus of americium-241.
(2)
(c)
Americium-243 is another isotope of americium. State one similarity and one difference between atoms of americium-241 and americium-243.
(2)
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Notes
Part (a)
95
✓ 1
Part (b)
M1 neutrons = nucleon number − proton number
✓ 1
A1 146
✓ 1
Part (c)
B1 similarity: same number of protons (or electrons)
✓ 1
B1 difference: americium-243 has more neutrons (two more)
✓ 1
Exam tip: Isotopes are the same element, so they share the proton number and differ only in the neutron number.
94IGPC3-5-07r38-13Atomic structure, isotopes, fission and fusionPaper CEasy4 marks
An energy company is comparing nuclear fission and nuclear fusion for supplying power to a new city district.
(a)
Describe what happens to a uranium-235 nucleus during nuclear fission.
(2)
(b)
State the two conditions needed for nuclear fusion to take place.
(2)
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Part (a)
B1 the nucleus absorbs a neutron
✓ 1
B1 it splits into two smaller nuclei, releasing energy and further neutrons
✓ 1
Part (b)
B1 a very high temperature
✓ 1
B1 a very high pressure (or density)
✓ 1
Exam tip: Give both conditions for fusion; high temperature alone earns only one of the two marks.
95IGPC3-5-08r39-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Potassium-40 is used to date volcanic rock. A neutral atom of 4019K contains which numbers of particles?
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Notes
The proton number is 19, so there are 19 protons. Neutrons = 40 − 19 = 21. A neutral atom has equal numbers of protons and electrons, so 19 electrons. 40 neutrons confuses the nucleon number with the neutron number. The 21 protons option swaps protons and neutrons. 21 electrons wrongly sets electrons equal to neutrons.
✓ 1
Answer A
Answer: A
Common error: Nucleon number counts protons plus neutrons; subtract the proton number to find the neutrons.
96IGPC3-5-08r39-02Radioactivity, half-life and safetyPaper CEasy1 mark
A shipyard checks welds in a steel hull 25 mm thick by placing a radioactive source on one side and a detector on the other. Which radiation is suitable for this test?
Diagram not to scale
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Notes
Gamma rays are highly penetrating and can pass through 25 mm of steel with enough getting through to detect changes caused by a flaw. Alpha particles are stopped by a sheet of paper or a few centimetres of air. Beta particles are stopped by a few millimetres of aluminium. The three types differ greatly in penetration, so they are not equally suitable.
✓ 1
Answer C
Answer: C
Exam tip: Match the radiation to the job: thin materials use beta, thick materials use gamma, and alpha is stopped almost at once.
97IGPC3-5-08r39-06Radioactivity, half-life and safetyPaper CEasy1 mark
A technician must handle a sealed radioactive source. Which set of precautions is correct?
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Notes
Distance, shielding and short exposure time all reduce the dose, so tongs, a lead-lined box and limited time are correct. A pocket keeps the source close to the body for long periods and plastic does not shield. Paper gloves do not stop penetrating radiation and holding the source close raises the dose. A glass jar on an open shelf gives little shielding to anyone nearby.
✓ 1
Answer A
Answer: A
Exam tip: Safety answers should mention shielding, distance or time; give a specific action for each.
98IGPC3-5-08r39-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
Argon-40 is formed in volcanic rock when potassium-40 decays, which allows the rock to be dated. The proton number of argon is 18.
(a)
Calculate the number of neutrons in a nucleus of 4018Ar.
(2)
(b)
Potassium has isotopes with nucleon numbers 39 and 40. State what is meant by isotopes and how these two isotopes differ.
(2)
(c)
State where the electrons are found in an atom.
(1)
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Notes
Part (a)
neutron number = nucleon number − proton number
✓ 1
= 40 − 18 = 22
✓ 1
Part (b)
Isotopes are atoms of the same element, with the same number of protons
✓ 1
They have different numbers of neutrons, so different nucleon numbers
✓ 1
Part (c)
Outside the nucleus, in orbits or shells around it
✓ 1
Common error: Isotopes of one element have the same proton number; only the neutron number differs.
99IGPC3-5-08r39-10Radioactivity, half-life and safetyPaper CEasy6 marks
A theatre's smoke detector contains a small americium-241 source that emits alpha particles into an air gap between two electrodes. A small steady current flows across the gap.
Diagram not to scale
(a)
State what an alpha particle is.
(1)
(b)
Explain how the detector sounds an alarm when smoke enters the gap.
(3)
(c)
Explain why a gamma-emitting source would be unsuitable for this detector.
(2)
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Notes
Part (a)
A helium nucleus, made of two protons and two neutrons
✓ 1
Part (b)
Alpha particles ionise the air, producing ions that carry the current
✓ 1
Smoke particles absorb or stop alpha particles, so fewer ions are produced
✓ 1
The current falls, which triggers the alarm circuit
✓ 1
Part (c)
Gamma rays are only weakly ionising, so very little current would flow
✓ 1
Most gamma rays would pass straight through smoke, so the current would hardly change
✓ 1
Exam tip: Alpha particles are strongly ionising but have very short range, which makes them suitable for detectors in air.
100IGPC3-5-07r43-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A hospital uses iodine-123 and iodine-131 in different tests. The two are isotopes of iodine. Which statement about the nuclei of these isotopes is correct?
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Notes
Isotopes are atoms of the same element, so they have the same number of protons (the proton number defines the element). They differ in the number of neutrons, which gives different nucleon numbers. The claim of the same neutron count with different protons would describe different elements. Same nucleon number with different proton numbers describes different elements too, not isotopes. Electrons are not found in the nucleus, so the last statement is wrong.
✓ 1
Answer B
Answer: B
Common error: Isotopes differ in neutron number only; the proton number is what identifies the element.
101IGPC3-5-07r43-02Radioactivity, half-life and safetyPaper CEasy1 mark
A factory that makes foil for shops monitors the foil thickness using a radioactive source. The radiation from the source passes through a sheet of paper, is stopped by 5 mm of aluminium and is deflected by a magnetic field. Which type of radiation is it?
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Notes
Beta particles pass through paper, are stopped by a few millimetres of aluminium and, being charged, are deflected by a magnetic field. Alpha particles would be stopped by the paper. Gamma rays pass through 5 mm of aluminium easily and, having no charge, are not deflected. X-rays are also uncharged electromagnetic waves, are not deflected and penetrate thin aluminium.
✓ 1
Answer D
Answer: D
Exam tip: Use two clues together: penetration tells you the range, and deflection in a field tells you whether the radiation is charged.
102IGPC3-5-07r43-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
A wooden beam used in a biomass power plant contains atoms of carbon-12 and carbon-14. A neutral atom of carbon-14 has nucleon number 14 and proton number 6.
(a)
Describe the structure of a neutral atom of carbon-14, stating where the protons, neutrons and electrons are found.
(2)
(b)
Carbon-12 and carbon-14 are isotopes of carbon. State one way in which their atoms are the same and one way in which they differ.
(2)
(c)
State the number of neutrons in one nucleus of carbon-14.
(1)
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Notes
Part (a)
protons and neutrons are in the nucleus (at the centre)
✓ 1
electrons move around / orbit the nucleus
✓ 1
Part (b)
same number of protons (and electrons)
✓ 1
different number of neutrons (so different nucleon number)
✓ 1
Part (c)
A1 8
✓ 1
Common error: Neutrons = nucleon number − proton number; do not give the nucleon number as the neutron count.
103IGPC3-5-07r43-16Radioactivity, half-life and safetyPaper CEasy5 marks
A shop sells granite worktops. A student suspects the granite emits more radiation than the surroundings and tests it with a radiation detector.
(a)
State two sources of background radiation.
(2)
(b)
State why the student measures the count rate with no worktop nearby before starting the test.
(1)
(c)
With the detector beside the worktop the count rate is 45 counts/min. The background count rate is 18 counts/min. Calculate the count rate due to the worktop.
(2)
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Notes
Part (a)
any two valid sources, e.g. radon gas / rocks and soil / cosmic rays / food and drink / medical sources / nuclear waste or fallout
✓ 2
Part (b)
to find the background count rate, so it can be subtracted
104IGPC3-5-08r48-01Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
A neutral atom of plutonium-238 (nucleon number 238, proton number 94) is the heat source in the power unit of a space probe. How many neutrons are in the nucleus of one atom?
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Notes
Number of neutrons = nucleon number − proton number = 238 − 94 = 144. 94 is the number of protons, not neutrons. 238 is the total number of nucleons, which includes the protons. 332 comes from adding the two numbers instead of subtracting.
✓ 1
Answer C
Answer: C
Common error: The nucleon number counts protons and neutrons together, so subtract the proton number to find the neutrons.
105IGPC3-5-08r48-02Atomic structure, isotopes, fission and fusionPaper CEasy1 mark
Which pair of atoms are isotopes of the same element?
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Notes
Isotopes have the same number of protons but different numbers of neutrons, which is true of the first pair. The second pair have the same nucleon number (40) but different proton numbers, so they are different elements. The third pair have the same number of neutrons but different proton numbers. The fourth pair each have equal numbers of protons and neutrons, but their proton numbers differ, so they are different elements.
✓ 1
Answer A
Answer: A
Exam tip: Compare proton numbers first: if they differ, the atoms are different elements, whatever the other numbers are.
106IGPC3-5-08r48-09Atomic structure, isotopes, fission and fusionPaper CEasy5 marks
Plutonium-238 (nucleon number 238, proton number 94) is used as the heat source in the power unit of a deep-space probe.
(a)
State the number of protons, neutrons and electrons in a neutral atom of plutonium-238.
(3)
(b)
Plutonium-239 has a nucleon number of 239. State one way in which atoms of plutonium-238 and plutonium-239 are the same and one way in which they differ.
(2)
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Notes
Part (a)
94 protons
✓ 1
144 neutrons (238 − 94)
✓ 1
94 electrons
✓ 1
Part (b)
same: both have 94 protons (and 94 electrons)
✓ 1
different: plutonium-239 has one more neutron (145 against 144)
✓ 1
Exam tip: Isotopes of one element have the same proton number but different neutron numbers.
107IGPC3-5-08r48-13Radioactivity, half-life and safetyPaper CEasy6 marks
A bridge engineer is told about the three types of radiation from radioactive sources.
Diagram not to scale
(a)
State what each of the following is: an alpha particle, a beta particle and a gamma ray.
(3)
(b)
State which of these three radiations is the most strongly ionising.
(1)
(c)
The engineer wants to inspect the inside of a thick steel girder with a radioactive source and a film on the other side. State which radiation should be used and give a reason.
(2)
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Notes
Part (a)
alpha particle: a helium nucleus (two protons and two neutrons)
✓ 1
beta particle: a fast-moving electron
✓ 1
gamma ray: an electromagnetic wave
✓ 1
Part (b)
alpha
✓ 1
Part (c)
gamma (rays)
✓ 1
most penetrating of the three, so some passes through the thick steel to reach the film (alpha and beta would be absorbed)
✓ 1
Exam tip: Link each radiation to what stops it: paper for alpha, a few millimetres of aluminium for beta, thick lead for gamma.
108IGPC3-5-08r48-16Radioactivity, half-life and safetyPaper CEasy5 marks
A team inspects a bridge using a sealed gamma source.
(a)
State two ways in which the workers can reduce their exposure to the radiation from the source.
(2)
(b)
Explain why the source is stored in a lead-lined box when it is not being used.
(1)
(c)
Explain why an alpha source is more hazardous inside the body than 1 m outside the body.
(2)
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Notes
Part (a)
any one from: increase the distance from the source, reduce the time spent near it, use shielding
✓ 1
a second different method
✓ 1
Part (b)
lead absorbs gamma radiation, so little escapes to the surroundings
✓ 1
Part (c)
alpha particles are stopped by a few centimetres of air, so they do not reach a person 1 m away
✓ 1
inside the body they are absorbed by cells and strongly ionise them, causing damage
✓ 1
Common error: Alpha radiation is harmless outside the body only because it is easily absorbed; if the source is swallowed or inhaled it is very dangerous.
109IGPC-039-00-12Atomic structure, isotopes, fission and fusionPaper CMedium5 marks
Chlorine has two common isotopes, 3517Cl and 3717Cl.
(a)
Explain what is meant by the term isotopes.
(2)
(b)
State the number of neutrons in each isotope.
(2)
(c)
A neutral atom of each isotope contains the same number of electrons. Suggest why.
(1)
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Notes
Part (a)
atoms (nuclei) of the same element / with the same number of protons ; with different numbers of neutrons / different nucleon numbers
both atoms have 17 protons, and a neutral atom has equal numbers of protons and electrons
✓ 1
Exam tip: A full definition of isotopes needs two points: the same proton number and a different neutron number.
110IGPC-039-01-04Radioactivity, half-life and safetyPaper CMedium1 mark
A detector placed near a radioactive source gives a reading of 340 counts/min. The background count rate is 20 counts/min. The half-life of the source is 2 days. What reading does the detector give 4 days later?
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Notes
The corrected count rate is 340 − 20 = 320 counts/min. Four days is 2 half-lives, so the corrected count rate becomes 320 ÷ 4 = 80 counts/min. The detector also records the background, which does not decay, so it reads 80 + 20 = 100 counts/min. 85 counts/min comes from halving the uncorrected reading. 65 counts/min subtracts the background after halving the uncorrected reading. 80 counts/min forgets to add the background back on.
✓ 1
Answer D
Answer: D
Exam tip: Background radiation does not halve. Subtract it before applying the half-life, then add it back if the question asks for what the detector actually reads.
111IGPC-039-01-09Radioactivity, half-life and safetyPaper CMedium1 mark
A factory produces paper sheet. A radioactive source on one side of the sheet and a detector on the other side are used to monitor its thickness continuously for many years. Which source is most suitable?
Diagram not to scale
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β-particles are partly absorbed by paper. A small change in thickness therefore causes a measurable change in the count rate. A half-life of 30 years means the count rate stays nearly constant over the working life of the equipment. α-particles would be completely stopped by the paper. γ-rays would pass through almost unaffected, so changes in thickness could not be detected. A 5-minute half-life would make the count rate fall too quickly to be useful.
✓ 1
Answer B
Answer: B
Exam tip: When choosing a source, justify both the type of radiation (how penetrating it is) and the half-life (how long it must last).
112IGPC-039-01-13Radioactivity, half-life and safetyPaper CMedium6 marks
A hospital uses a radioactive tracer to examine a patient's kidneys. The tracer emits only γ-rays and has a half-life of 6.0 hours. It is injected into the patient. A detector outside the body records the radiation.
(a)
Explain why a γ-emitter is suitable for this use.
(2)
(b)
Explain why a half-life of 6.0 hours is suitable.
(2)
(c)
The activity of the tracer at injection is 480 MBq. Calculate its activity 24 hours later.
(2)
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Part (a)
γ-rays are penetrating enough to pass out of the body and reach the detector ; they are weakly ionising, so they cause less damage to cells .
✓ 2
Part (b)
It is long enough for the tracer to reach the kidneys and for the scan to be completed ; it is short enough that the activity in the patient falls to a low level quickly, which limits the dose .
✓ 2
Part (c)
24 ÷ 6.0 = 4 half-lives
✓ 1
480 ÷ 24 = 30 MBq
✓ 1
Exam tip: When you explain a choice of half-life, give both sides: why it is not too short and why it is not too long.
113IGPC3-5-05c-03Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive tracer injected into a patient has an activity of 80 kBq and a half-life of 6.0 hours. Assuming the activity falls only because of radioactive decay, what is the activity after 24 hours?
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24 h is 4 half-lives: 80 → 40 → 20 → 10 → 5.0 kBq, so B is correct. A assumes all the nuclei have gone after 4 half-lives. C comes from counting only 3 halvings. D comes from using only 2 halvings, for example by dividing 24 by 12.
✓ 1
Answer B
Answer: B
Common error: Activity never reaches zero after a whole number of half-lives; it just halves each time.
114IGPC3-5-05c-08Radioactivity, half-life and safetyPaper CMedium1 mark
A sample contains 2000 undecayed nuclei of an isotope with a half-life of 10 s. Which statement about the decay is correct?
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Decay is random for any one nucleus, but for a large sample the half-life tells us that about half remain after 10 s: D is correct. A treats the half-life as a fixed lifetime. B forgets each half-life halves the number again (500 left after 20 s). C assumes a constant loss rather than a rate proportional to the number left.
✓ 1
Answer D
Answer: D
Common error: Half-life describes a whole sample: the number halves in each equal time interval, it does not fall by a constant amount.
115IGPC3-5-05c-13Atomic structure, isotopes, fission and fusionPaper CMedium7 marks
A nuclear power station uses uranium-235 as its fuel.
(a)
Describe what happens to a uranium-235 nucleus during nuclear fission. [3]
(3)
(b)
Explain the purpose of the control rods in the reactor. [2]
(2)
(c)
State why used fuel rods must be stored carefully for many years. [2]
(2)
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Part (a)
the nucleus absorbs a neutron
✓ 1
it splits into two smaller nuclei
✓ 1
further neutrons and energy are released
✓ 1
Part (b)
control rods absorb neutrons
✓ 1
so the rate of fission (the chain reaction) is controlled and cannot run away
✓ 1
Part (c)
they contain radioactive waste products that still emit ionising radiation
✓ 1
some of these have long half-lives so stay hazardous for a long time
✓ 1
Common error: Fission needs a neutron to start it and releases more neutrons, which is why a chain reaction is possible.
116IGPC3-5-05c-16Atomic structure, isotopes, fission and fusionPaper CMedium5 marks
In a scattering experiment a narrow beam of alpha particles is fired at a very thin sheet of gold foil. Most alpha particles pass straight through, a small number are deflected through large angles and a very few rebound almost straight back.
Diagram not to scale. Paths are not drawn to scale.
(a)
State what the fact that most alpha particles pass straight through shows about the atom. [1]
(1)
(b)
Explain why a very few alpha particles rebound almost straight back. [2]
(2)
(c)
The radius of the atom is 1.0 × 10−10 m and the radius of its nucleus is 5.0 × 10−15 m. Calculate the ratio atom radius ÷ nucleus radius. [2]
(2)
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Part (a)
most of the atom is empty space
✓ 1
Part (b)
the nucleus is positively charged and so repels the positive alpha particles
✓ 1
the nucleus is tiny and holds most of the mass, so only a head-on approach is reversed
✓ 1
Part (c)
1.0 × 10−10 ÷ 5.0 × 10−15
✓ 1
= 2.0 × 104 (20000)
✓ 1
Exam tip: Link each observation to a feature of the atom: undeflected means empty space, rebounding means a small, dense, positive centre.
117IGPC3-5-05r5-04Radioactivity, half-life and safetyPaper CMedium1 mark
A radiopharmaceutical has an activity of 240 MBq when prepared. Its half-life is 6.0 hours. What is its activity 18 hours later?
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18 h is 3 half-lives, so the activity halves three times: 240 → 120 → 60 → 30 MBq. 120 MBq is only one half-life. 60 MBq is two half-lives. 80 MBq comes from dividing by 3 instead of halving three times.
✓ 1
Answer A
Answer: A
Common error: Activity falls by a factor of 2 for each half-life, not by equal amounts per half-life.
118IGPC3-5-05r5-06Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Which statement correctly describes nuclear fission in a power station reactor?
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Fission is the splitting of a large nucleus after neutron capture, releasing energy and further neutrons that can continue the chain reaction. Joining of small nuclei is fusion. Emission of an alpha particle is radioactive decay, not splitting into two similar-sized nuclei. Electron transfer is a chemical process.
✓ 1
Answer D
Answer: D
Common error: Do not mix up fission (splitting) with fusion (joining).
119IGPC3-5-05r5-07Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Hydrogen nuclei in the Sun fuse only at extremely high temperatures. What is the main reason for this?
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Nuclei are positively charged and repel one another, so they need very high kinetic energy, which means a very high temperature, to get close enough for the strong nuclear force to act. Hydrogen nuclei are already single protons or small nuclei and need no splitting. Neutrons are not needed to be freed for attraction. Mass is not converted to kinetic energy to enable collisions.
✓ 1
Answer B
Answer: B
Exam tip: For fusion, link high temperature to high speed and then to overcoming electrostatic repulsion.
120IGPC3-5-05r5-10Radioactivity, half-life and safetyPaper CMedium6 marks
A water company adds a small amount of a gamma-emitting tracer to the water in a buried pipe. A detector held above the ground finds the leak where the count rate is highest. A sample of the tracer gives a count rate of 134 counts/min. The background count rate is 22 counts/min. After 12 hours the same sample gives 36 counts/min.
(a)
State what is meant by background radiation and give one source of it.
(1)
(b)
Suggest why a tracer with a half-life of a few hours is better for this use than one with a half-life of several years.
(1)
(c)
Calculate the corrected count rate of the sample at the start.
(1)
(d)
Determine the half-life of the tracer.
(3)
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Part (a)
radiation always present in the environment, from e.g. rocks, cosmic rays or the air
✓ 1
Part (b)
the activity falls quickly so the water supply contains little radioactivity afterwards / less hazard to people
✓ 1
Part (c)
134 − 22 = 112 counts/min
✓ 1
Part (d)
corrected count rate after 12 h = 36 − 22 = 14 counts/min
✓ 1
112 → 56 → 28 → 14 so 3 half-lives
✓ 1
half-life = 12 ÷ 3 = 4 h
✓ 1
Common error: Subtract background from every reading, not just the first.
121IGPC3-5-05r5-11Radioactivity, half-life and safetyPaper CMedium6 marks
A game designer models radioactive decay. The game starts with 320 glowing tokens. In each turn, on average, half of the tokens still in play are removed.
(a)
State what one turn in the model represents, in terms of radioactive decay.
(1)
(b)
Calculate the expected number of tokens left after 3 turns.
(2)
(c)
Calculate the expected number of tokens left after 6 turns.
(2)
(d)
In one test run the game has 43 tokens after 3 turns, not 40. Explain why.
decay is random, so the actual number only follows the expected value on average
✓ 1
Exam tip: Radioactive decay is random for each nucleus, but predictable for a large number of nuclei.
122IGPC3-5-07r5-03Radioactivity, half-life and safetyPaper CMedium1 mark
A water company wants to find a leak in a buried plastic pipe. A radioactive tracer is added to the water and a detector is carried along the ground above the pipe. Which tracer is most suitable?
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Only gamma radiation is penetrating enough to pass through the pipe wall and soil to a detector above ground, and a half-life of hours means the tracer decays away quickly, so D is correct. A is wrong because alpha particles are stopped by a few centimetres of air or a thin layer of material. B is wrong because beta particles are absorbed by the soil and pipe. C is wrong because a half-life of years leaves the water radioactive long after the test.
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Answer D
Answer: D
Common error: Choose the radiation by its penetrating power and the half-life by how long the hazard should last; both must be justified.
123IGPC3-5-07r5-04Radioactivity, half-life and safetyPaper CMedium1 mark
With no source present, a detector records 24 counts in 2.0 minutes. When a source is placed near it, the detector records 150 counts in 1.0 minute. What is the count rate due to the source alone?
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The background rate is 24 ÷ 2.0 = 12 counts per minute, so the source gives 150 − 12 = 138 counts per minute. A is correct. B subtracts 24 without converting it to a rate per minute. C ignores the background. D adds the background instead of subtracting it.
✓ 1
Answer A
Answer: A
Common error: Convert the background reading to the same time interval as the source reading before subtracting.
124IGPC3-5-07r5-12Radioactivity, half-life and safetyPaper CMedium6 marks
In a survival video game, a player's radiation meter rises to 720 units in a contaminated zone. After the player leaves the zone, the reading halves every 8.0 minutes. The designers modelled this on radioactive decay.
(a)
Calculate the reading 24 minutes after the player leaves the zone.
(2)
(b)
The player can enter a shop only when the reading is below 12 units. Determine the time after leaving the zone when this first happens. Use whole numbers of 8.0-minute intervals.
(3)
(c)
In a real radioactive sample, decay is random. State what this means about the moment at which one particular nucleus decays.
(1)
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Part (a)
24 ÷ 8.0 = 3 halvings
✓ 1
720 → 360 → 180 → 90 units
✓ 1
Part (b)
continue halving: 90 → 45 → 22.5
✓ 1
22.5 → 11.25, which is below 12 units after 6 intervals
✓ 1
6 × 8.0 = 48 minutes
✓ 1
Part (c)
it cannot be predicted
✓ 1
Exam tip: When a threshold is not reached exactly, keep halving until the value first drops below it, and say so.
125IGPC3-5-07r5-13Radioactivity, half-life and safetyPaper CMedium7 marks
A technician measures the activity of a sample of a radioactive tracer used in medical scans. Time / h: 0, 2.0, 4.0, 6.0, 8.0 Activity / kBq: 80, 57, 40, 28, 20
(a)
Use the data to determine the half-life of the tracer.
(2)
(b)
Show that the data are consistent with this half-life by using a different pair of readings.
(1)
(c)
Predict the activity at 12 h.
(2)
(d)
A patient may leave only when the activity of a sample falls to 2.5 kBq or below. Determine the minimum time needed.
(2)
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Part (a)
activity falls from 80 kBq to 40 kBq
✓ 1
this takes 4.0 h, so half-life = 4.0 h
✓ 1
Part (b)
40 kBq to 20 kBq takes 4.0 h (from 4.0 h to 8.0 h), the same time
✓ 1
Part (c)
12 h is 3 half-lives
✓ 1
80 ÷ 8 = 10 kBq
✓ 1
Part (d)
80 ÷ 2.5 = 32, which is 5 halvings
✓ 1
5 × 4.0 = 20 h
✓ 1
Exam tip: A constant half-life, checked from two different starting points, is the signature of radioactive decay.
126IGPC3-5-05r8-03Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
The carbon-fibre panels on a racing car contain atoms of carbon-12 and carbon-14. Which statement correctly describes these two types of atom?
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Isotopes of an element have the same proton number (6) and so the same number of electrons when neutral, but different neutron numbers (6 and 8). Option B reverses neutrons and protons; C is wrong because the nucleon numbers 12 and 14 differ; D is wrong because a neutral atom has 6 electrons in both cases.
✓ 1
Answer D
Answer: D
Common error: Isotopes differ in neutron number, not proton number. The chemical element is fixed by the proton number.
127IGPC3-5-05r8-04Radioactivity, half-life and safetyPaper CMedium1 mark
A hospital sample of a radioactive isotope has an activity of 640 MBq. Its half-life is 6.0 hours. What is its activity 24 hours later?
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24 h is 4 half-lives, so the activity is halved four times: 640 → 320 → 160 → 80 → 40 MBq. 160 MBq divides by 4 instead of halving 4 times. 26.7 MBq divides by 24. 107 MBq divides by 6.
✓ 1
Answer A
Answer: A
Common error: Count how many half-lives have passed, then halve that many times. Do not divide by the number of half-lives.
128IGPC3-5-05r8-10Radioactivity, half-life and safetyPaper CMedium5 marks
A honey factory checks the fill level of metal cans on a conveyor belt. A radioactive source is on one side of each can and a detector on the other side.
Diagram not to scale
(a)
Explain why a source emitting only alpha particles is unsuitable.
(2)
(b)
The background count rate is 30 counts/min. A gamma source gives 480 counts/min with an empty can and 150 counts/min with a full can. Calculate the fraction of the corrected count rate that remains when the can is full. Give your answer to 3 significant figures.
(3)
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Part (a)
Alpha particles have very short range / are stopped by thin material or a few cm of air
✓ 1
so none would reach the detector, whether the can is full or empty
✓ 1
Part (b)
Subtract background: 450 and 120 counts/min
✓ 1
Fraction = 120 ÷ 450
✓ 1
= 0.267
✓ 1
Common error: Subtract the background count from both readings before finding a ratio.
129IGPC3-5-05r8-14Atomic structure, isotopes, fission and fusionPaper CMedium4 marks
Solar panels power the sensors in a beekeeper's hive. The Sun releases energy by nuclear fusion.
(a)
State what happens to the nuclei in nuclear fusion.
(2)
(b)
State one difference between nuclear fusion and nuclear fission.
(1)
(c)
Name the element whose nuclei fuse in the Sun to form helium.
(1)
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Part (a)
Light nuclei (e.g. hydrogen) join together
✓ 1
to form a heavier nucleus, releasing energy
✓ 1
Part (b)
Fission splits a heavy nucleus whereas fusion joins light nuclei (or fusion needs very high temperature)
✓ 1
Part (c)
Hydrogen
✓ 1
Common error: Do not confuse the two: fission splits a large nucleus, fusion joins small ones.
130IGPC3-5-06r8-03Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
A violin maker dates a wooden violin using carbon-14. Carbon-12 and carbon-14 are isotopes of carbon. Which statement is correct about their nuclei?
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Both nuclei have 6 protons (carbon); the neutron numbers are 6 and 8. The others swap protons and neutrons, confuse nucleon number with proton number, or place electrons in the nucleus.
✓ 1
Answer B
Answer: B
Common error: The number in a nuclide name (carbon-14) is the nucleon number, not the proton number.
131IGPC3-5-06r8-08Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive sensor in an aircraft contains an isotope with a half-life of 2 hours. Which statement about one particular nucleus is correct?
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Radioactive decay is random: for a single nucleus the moment of decay cannot be predicted. The half-life only describes large numbers of nuclei. A and B assume a fixed time; D treats one nucleus as divisible.
✓ 1
Answer D
Answer: D
Common error: Half-life is a statistical idea for many nuclei, not a timer for one nucleus.
132IGPC3-5-06r8-11Radioactivity, half-life and safetyPaper CMedium4 marks
A student follows a radioactive tracer taken up by a crop plant. The corrected count rate from a leaf is measured at intervals.
Time / h
0
3
6
9
Corrected count rate / counts per min
960
480
240
120
(a)
Determine the half-life of the tracer.
(2)
(b)
Explain what is meant by a corrected count rate.
(1)
(c)
Predict the corrected count rate at 12 h.
(1)
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Part (a)
The count rate halves every 3 h
✓ 1
Half-life = 3 h
✓ 1
Part (b)
The measured count rate with the background count rate subtracted
✓ 1
Part (c)
120 ÷ 2 = 60 counts/min
✓ 1
Exam tip: Check that the rate halves over each equal time interval before quoting a half-life.
133IGPC3-5-06r8-14Atomic structure, isotopes, fission and fusionPaper CMedium4 marks
A nuclear power station supplies electricity to an irrigation scheme. The reactor uses nuclear fission of uranium-235.
(a)
State the particle that strikes a uranium-235 nucleus to cause fission.
(1)
(b)
Describe how a chain reaction happens.
(2)
(c)
State one problem caused by the waste from a fission reactor.
(1)
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Part (a)
A neutron
✓ 1
Part (b)
Each fission releases further neutrons
✓ 1
These neutrons cause other nuclei to split, which release more neutrons
✓ 1
Part (c)
Waste stays radioactive / hazardous for a very long time and must be stored safely
✓ 1
Common error: A chain reaction needs the neutrons released by one fission to cause further fissions.
134IGPC3-5-05r10-03Radioactivity, half-life and safetyPaper CMedium1 mark
A technician uses a sealed source of gamma rays to sterilise hospital equipment. Which action best reduces her exposure to radiation while the source is in use?
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C is correct: gamma rays are very penetrating and are absorbed significantly only by thick dense material such as lead or concrete. A, B and D would each stop alpha or at most beta particles but gamma rays pass through them almost unchanged.
✓ 1
Answer C
Answer: C
Exam tip: Match the shielding to the radiation: paper stops alpha, aluminium stops beta, thick lead or concrete is needed to reduce gamma.
135IGPC3-5-05r10-04Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive sample gives a measured count rate of 148 counts/min. The background count rate is 20 counts/min. The half-life of the sample is 3.0 hours. What is the measured count rate 6.0 hours later?
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Corrected rate now = 148 - 20 = 128 counts/min. 6.0 h is two half-lives, so the corrected rate becomes 128/4 = 32 counts/min. The measured rate adds the background back: 32 + 20 = 52 counts/min (A). B (148/4) forgets to subtract the background. C forgets to add the background back. D allows only one half-life.
✓ 1
Answer A
Answer: A
Common error: Only the count rate from the source halves; the background count rate stays the same.
136IGPC3-5-05r10-05Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
A power station generates electricity using nuclear fission. Which statement describes fission?
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D is correct. A describes fusion. B is wrong because gamma emission does not change the proton number. C describes ionisation, which is a different process and does not involve the nucleus splitting.
✓ 1
Answer D
Answer: D
Common error: Fission splits a large nucleus; fusion joins small nuclei. Do not swap them.
137IGPC3-5-05r10-10Radioactivity, half-life and safetyPaper CMedium4 marks
A hospital uses a radioactive tracer with a half-life of 6.0 hours. The tracer has an initial activity of 480 MBq.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the activity of the tracer 24 hours after it is prepared.
(2)
(c)
Suggest why a tracer with a short half-life is preferred for use inside a patient.
(1)
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Notes
Part (a)
time taken for half of the unstable nuclei (in a sample) to decay / time for the activity to halve
✓ 1
Part (b)
24 / 6.0 = 4 half-lives
✓ 1
480 / 24 = 30 MBq
✓ 1
Part (c)
activity (and so exposure of the patient) falls quickly / little radioactive material remains in the body afterwards
✓ 1
Exam tip: Work out the number of half-lives first, then halve the activity that many times.
138IGPC3-5-05r10-11Radioactivity, half-life and safetyPaper CMedium4 marks
A factory that supplies packaging to a retail chain uses a radioactive source on one side of a moving plastic film and a detector on the other side, to check that the film has a constant thickness.
Diagram not to scale
(a)
State which type of radiation, alpha, beta or gamma, is most suitable for thin plastic film and give a reason.
(2)
(b)
Explain why a source emitting alpha particles would be unsuitable.
(1)
(c)
The source has a half-life of 30 years. Suggest why a source with a half-life of a few minutes would be unsuitable.
(1)
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Part (a)
beta
✓ 1
partly absorbed by the thin film so the detected count rate changes with thickness / alpha would be stopped completely and gamma would pass through almost unchanged
✓ 1
Part (b)
alpha particles are stopped by the film (or by a few cm of air) so none reach the detector
✓ 1
Part (c)
its activity would fall quickly so the count rate would not stay constant / it would need to be replaced very often
✓ 1
Exam tip: For a thickness gauge the radiation must be partly, not completely, absorbed by the material.
139IGPC3-5-05r10-12Radioactivity, half-life and safetyPaper CMedium5 marks
A student measures the count rate from a radioactive sample using a detector. The background count rate is 30 counts/min. The measured count rates are in the table. Time / min: 0, 10, 20, 30 Measured count rate / counts/min: 270, 150, 90, 60
Measured count rate (background not subtracted)
(a)
Calculate the corrected count rate at time 0.
(1)
(b)
Determine the half-life of the sample.
(2)
(c)
Predict the measured count rate at 40 min.
(2)
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Part (a)
270 - 30 = 240 counts/min
✓ 1
Part (b)
corrected rates are 240, 120, 60, 30 counts/min
✓ 1
half-life = 10 min
✓ 1
Part (c)
corrected rate = 30 / 2 = 15 counts/min
✓ 1
measured rate = 15 + 30 = 45 counts/min
✓ 1
Common error: Subtract the background before looking for halving, and add it back when predicting a measured value.
140IGPC3-5-06r10-04Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Which row correctly describes nuclear fission and nuclear fusion?
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B is correct: fission is the splitting of a large nucleus and fusion is the joining of small nuclei, as happens in the Sun. A swaps the two processes. C wrongly says fusion splits nuclei. D confuses fusion with alpha decay.
✓ 1
Answer B
Answer: B
Common error: Do not swap the two processes: fission splits, fusion fuses.
141IGPC3-5-06r10-06Radioactivity, half-life and safetyPaper CMedium1 mark
A nucleus emits a gamma ray. What is the effect on the nucleon number and the proton number of the nucleus?
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Gamma rays are electromagnetic radiation, with no mass and no charge, so neither number changes (C). A describes alpha decay. B describes beta-minus decay. D has no basis in a gamma emission.
✓ 1
Answer C
Answer: C
Common error: Gamma emission carries away energy only; the nucleus stays the same nuclide.
142IGPC3-5-06r10-10Radioactivity, half-life and safetyPaper CMedium4 marks
A data centre smoke detector contains 0.20 µg of americium-241. The half-life of americium-241 is 432 years.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the mass of americium-241 that remains after 1296 years.
(2)
(c)
Suggest why a source with a long half-life is suitable for a smoke detector.
(1)
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Part (a)
time taken for half of the unstable nuclei (in a sample) to decay
✓ 1
Part (b)
1296 / 432 = 3 half-lives
✓ 1
0.20 / 23 = 0.025 µg
✓ 1
Part (c)
its activity stays almost constant for many years / the detector does not need replacing often
✓ 1
Exam tip: Convert the time to a number of half-lives before halving the quantity.
143IGPC3-5-06r10-12Radioactivity, half-life and safetyPaper CMedium5 marks
A student places a radioactive source in front of a detector and records the count rate with different absorbers between them. The background count rate is 30 counts/min. No absorber: 480 counts/min Sheet of paper: 478 counts/min 3 mm of aluminium: 95 counts/min 5 cm of lead: 31 counts/min
Measured count rate for each absorber. The dashed line shows the background count rate.
(a)
Calculate the corrected count rate with no absorber.
(1)
(b)
Deduce that the source does not emit alpha particles.
(2)
(c)
Deduce that the source emits both beta particles and gamma rays.
(2)
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Notes
Part (a)
480 - 30 = 450 counts/min
✓ 1
Part (b)
paper stops alpha particles
✓ 1
the count rate with paper is almost unchanged (so alpha is not present)
✓ 1
Part (c)
aluminium reduces the count rate a lot (from 450 to 65) because it stops beta particles, so beta is present
✓ 1
some radiation still passes through aluminium and is almost removed only by thick lead, so gamma is present
✓ 1
Common error: Always use the corrected count rate, after subtracting the background, when judging how much an absorber reduces the radiation.
144IGPC3-5-06r11-03Radioactivity, half-life and safetyPaper CMedium1 mark
A dock worker must move a sealed radioactive source between two storage rooms. Which action reduces the radiation dose the worker receives?
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Dose falls with greater distance and shorter exposure time, so long tongs and a short handling time help. A brings the source closer. C is wrong because paper stops only alpha particles, not beta or gamma radiation. D is wrong because thin cloth stops almost no radiation.
✓ 1
Answer B
Answer: B
Exam tip: Three ways to cut exposure are more distance, less time and more shielding.
145IGPC3-5-06r11-04Radioactivity, half-life and safetyPaper CMedium1 mark
A sealed source in a marine laboratory is tested with a Geiger–Müller tube. With no absorber the count rate is 420 counts/min. A sheet of paper reduces it to 415 counts/min, 3 mm of aluminium reduces it to 14 counts/min, and 5 cm of lead reduces it to 12 counts/min. The background count rate is 12 counts/min. Which type of radiation does the source emit?
Measured count rate for each absorber. The dashed line shows the background count rate.
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Notes
Paper barely changes the rate, so there is little alpha. Aluminium reduces the corrected rate from about 403 to about 2 counts/min, so the radiation is beta, and lead gives no further drop below background, so there is no gamma. A is wrong because alpha would be stopped by paper. C is wrong because gamma would pass through the aluminium. D is wrong for both reasons.
✓ 1
Answer B
Answer: B
Common error: Subtract the background count rate before judging how much an absorber reduces the radiation.
146IGPC3-5-06r11-05Radioactivity, half-life and safetyPaper CMedium1 mark
A logistics firm tracks pallets with a radioactive marker of half-life 6.0 hours. The initial activity of the marker is 800 Bq. What is its activity after 18 hours?
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Notes
18 hours is 3 half-lives, so the activity is 800 → 400 → 200 → 100 Bq. A uses only one half-life. B divides by the number of half-lives (800 / 3). C uses only two half-lives.
✓ 1
Answer D
Answer: D
Common error: Halve once for each half-life; do not divide by the number of half-lives.
147IGPC3-5-06r11-12Radioactivity, half-life and safetyPaper CMedium6 marks
A port uses a sealed gamma source to scan shipping containers.
(a)
State two precautions that reduce the dose to staff when the source is not in use.
(2)
(b)
Explain why gamma radiation is chosen for scanning containers rather than alpha radiation.
(2)
(c)
With no source present, a detector records 18 counts/min. State one cause of this count rate.
(1)
(d)
With the source in position, the detector records 258 counts/min. Calculate the corrected count rate due to the source alone.
(1)
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Notes
Part (a)
store the source in a lead-lined (shielded) container
✓ 1
keep staff at a distance from the source, or limit the time spent near it
✓ 1
Part (b)
alpha radiation is stopped by the container wall (or a few centimetres of air)
✓ 1
gamma radiation is highly penetrating, so it passes through the cargo and reaches the detector
✓ 1
Part (c)
background radiation, for example from rocks, cosmic rays or the air
✓ 1
Part (d)
258 − 18 = 240 counts/min
✓ 1
Exam tip: Always subtract the background count rate from a measured count rate before using it.
148IGPC3-5-06r11-13Radioactivity, half-life and safetyPaper CMedium5 marks
A radioactive isotope is extracted from a sample of reclaimed building stone. The corrected count rate from the isotope is 640 counts/min at the start. It falls to 40 counts/min after 36 minutes.
(a)
Determine the half-life of the isotope.
(2)
(b)
Predict the corrected count rate 45 minutes after the start.
(2)
(c)
Suggest why the measured count rate at 45 minutes might differ slightly from the prediction.
(1)
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Notes
Part (a)
640 to 40 is a factor of 16, which is 4 half-lives
✓ 1
36 / 4 = 9.0 min
✓ 1
Part (b)
45 min is 5 half-lives, ECF from (a)
✓ 1
640 / 25 = 20 counts/min
✓ 1
Part (c)
radioactive decay is random
✓ 1
Common error: A factor of 16 means four halvings, not sixteen half-lives.
149IGPC3-5-07r19-04Radioactivity, half-life and safetyPaper CMedium1 mark
A data centre fire alarm contains a small americium-241 source that emits alpha particles. Which statement explains why an alpha emitter is suitable for this use?
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Notes
D is correct: strong ionisation produces the current that smoke disturbs, and the low penetration makes the source safe in normal use. A is wrong because alpha particles are the least penetrating. B is wrong because the source must last for many years, so it needs a long half-life. C is wrong because alpha particles are helium nuclei, not electromagnetic waves.
✓ 1
Answer D
Answer: D
Common error: Do not confuse ionising power with penetrating power: alpha is the most ionising but the least penetrating.
150IGPC3-5-07r19-05Radioactivity, half-life and safetyPaper CMedium1 mark
A student places a rock sample in front of a Geiger-Müller counter. The counter records 380 counts in 2.0 minutes. The background count rate is 30 counts per minute. What is the corrected count rate of the rock sample?
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Notes
C is correct: 380 / 2.0 = 190 counts per minute, and 190 − 30 = 160 counts per minute. A forgets to subtract the background. B adds the background instead of subtracting it. D subtracts 30 from the total number of counts without converting to a rate.
✓ 1
Answer C
Answer: C
Exam tip: Convert to a rate first, then subtract the background rate.
151IGPC3-5-07r19-06Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Which statement about nuclear fusion in the Sun is correct?
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B is correct: nuclei repel each other, so very high temperatures are needed to give them enough kinetic energy to get close enough to fuse. A describes fission. C is wrong because the repulsion between positive nuclei must be overcome by high speeds. D is wrong because the Sun fuses light nuclei such as hydrogen.
✓ 1
Answer B
Answer: B
Common error: Fusion joins light nuclei and fission splits heavy nuclei; do not swap them.
152IGPC3-5-07r19-11Radioactivity, half-life and safetyPaper CMedium5 marks
A student monitors a sample of a radioactive isotope. The table shows the corrected count rate at different times.
Time / min
0
10
20
30
Corrected count rate / counts per minute
640
320
160
80
(a)
Define the half-life of a radioactive isotope.
(1)
(b)
Use the table to state the half-life of the isotope.
(1)
(c)
Predict the corrected count rate at 50 minutes.
(2)
(d)
Explain why the count rates in the table are described as corrected.
(1)
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Notes
Part (a)
the time taken for half of the (undecayed) nuclei in a sample to decay (accept time taken for the count rate / activity to halve)
✓ 1
Part (b)
10 minutes
✓ 1
Part (c)
50 min is 5 half-lives, so 640 / 25 (accept halving 80 → 40 → 20)
✓ 1
20 counts per minute
✓ 1
Part (d)
the background count rate has been subtracted
✓ 1
Exam tip: To predict a later count rate, count the number of half-lives and halve that many times.
153IGPC3-5-07r19-12Atomic structure, isotopes, fission and fusionPaper CMedium6 marks
A nuclear power station supplies electricity to a data centre. The reactor uses fission of uranium-235.
(a)
State what happens to a uranium-235 nucleus when it absorbs a neutron.
(2)
(b)
Explain how a chain reaction occurs.
(2)
(c)
The reactor contains control rods. State the purpose of the rods and how they work.
(2)
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Part (a)
the nucleus splits (into two smaller nuclei / fission fragments)
✓ 1
energy is released and (two or three) neutrons are emitted
✓ 1
Part (b)
the neutrons released go on to be absorbed by other uranium-235 nuclei
✓ 1
which then split and release more neutrons, so the reaction continues / grows
✓ 1
Part (c)
they control the rate of the reaction / prevent it running out of control
✓ 1
by absorbing neutrons
✓ 1
Exam tip: In a chain reaction answer, link the neutrons released to the next fission.
154IGPC3-5-07r21-03Radioactivity, half-life and safetyPaper CMedium1 mark
A freight depot stores a sealed source that emits gamma radiation. Which arrangement gives the greatest reduction in the dose received by staff?
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Notes
D is correct: thick lead absorbs much of the gamma radiation, and distance and short exposure time both reduce the dose. A is wrong because thin aluminium stops beta but not gamma. B is wrong because plastic and paper stop only alpha particles. C uses good shielding, but staying close for a long time increases the dose.
✓ 1
Answer D
Answer: D
Exam tip: Dose is reduced by shielding, increasing distance and reducing exposure time; choose a shield that suits the type of radiation.
155IGPC3-5-07r21-04Radioactivity, half-life and safetyPaper CMedium1 mark
A source is tested with different absorbers placed between it and a detector. The background count rate is 21 counts per minute. No absorber: 380 counts per minute Thick sheet of paper: 378 counts per minute 3 mm of aluminium: 24 counts per minute 5 cm of lead: 21 counts per minute Which radiation does the source emit?
Diagram not to scale
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Paper has almost no effect, so alpha particles are not being emitted. Aluminium reduces the count rate to almost the background level, so the radiation is stopped by a few mm of metal but not by paper: beta particles, so B is correct. Gamma rays would pass through 3 mm of aluminium largely unchanged. A mixture including alpha would show a fall when paper is added.
✓ 1
Answer B
Answer: B
Common error: Always subtract the background count rate before judging how much an absorber has reduced the radiation.
156IGPC3-5-07r21-10Radioactivity, half-life and safetyPaper CMedium6 marks
A researcher labels plankton with a radioactive tracer and measures the count rate of a sample. The count rates have already been corrected for background radiation. Time / h: 0, 4, 8, 12 Count rate / counts per minute: 1280, 640, 320, 160
(a)
Use the data to state the half-life of the tracer.
(1)
(b)
State what is meant by background radiation and give one source of it.
(2)
(c)
Suggest why a tracer with a half-life of 4 h is more suitable than one with a half-life of 4 years.
(1)
(d)
Calculate the count rate of the sample after 24 h.
(2)
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Part (a)
4 h
✓ 1
Part (b)
radiation that is always present in the environment
✓ 1
a valid natural source, e.g. radon gas, rocks, cosmic rays
✓ 1
Part (c)
it decays away quickly so little radiation remains in the organisms or sea after the study
✓ 1
Part (d)
24 h is 6 half-lives, so count rate = 1280 ÷ 26
✓ 1
20 counts per minute
✓ 1
Exam tip: For a half-life calculation, first find the number of half-lives by dividing the time by the half-life.
157IGPC3-5-07r21-11Radioactivity, half-life and safetyPaper CMedium5 marks
A freight depot stores a sealed source that emits gamma radiation, used to check the contents of closed steel containers.
(a)
State two precautions staff should take when handling a radioactive source.
(2)
(b)
Explain why a source that emits only alpha particles would not be suitable for checking the contents of a steel container.
(2)
(c)
State one harmful effect of ionising radiation on living cells.
(1)
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Notes
Part (a)
a valid precaution, e.g. use long tongs or keep a large distance from the source
✓ 1
a second, different valid precaution, e.g. keep handling time short, or store in a lead-lined container
✓ 1
Part (b)
alpha particles are stopped by a few centimetres of air or a sheet of paper
✓ 1
so they cannot pass through the steel wall to the detector
✓ 1
Part (c)
damages or mutates DNA, which can cause cancer or cell death
✓ 1
Exam tip: Link the penetrating power of each type of radiation to the material it must pass through.
158IGPC3-5-07r21-14Radioactivity, half-life and safetyPaper CMedium5 marks
A logistics firm tags pallets with a small radioactive marker and measures its count rate with a detector. The background count rate is 20 counts per minute. The measured count rates, which include background, are: Time / min: 0, 8, 16, 24 Measured count rate / counts per minute: 340, 180, 100, 60
(a)
Determine the half-life of the marker.
(2)
(b)
State why the measured count rate can never fall below 20 counts per minute.
(1)
(c)
Calculate the measured count rate at 40 min.
(2)
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Part (a)
corrected count rates at 0 and 8 min are 320 and 160
✓ 1
count rate halves in 8 min, so half-life is 8 min
✓ 1
Part (b)
background radiation is always detected
✓ 1
Part (c)
corrected rate = 320 ÷ 25 = 10 counts per minute
✓ 1
add background: 30 counts per minute
✓ 1
Common error: Subtract background before halving, then add it back if the question asks for the measured count rate.
159IGPC3-5-07r22-03Radioactivity, half-life and safetyPaper CMedium1 mark
A mountaineer brings back a rock sample containing a radioactive isotope. The corrected count rate from the sample falls from 800 counts per minute to 100 counts per minute in 27 hours. What is the half-life of the isotope?
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Notes
800 falls to 400, 200 and then 100, which is three half-lives. Half-life = 27 ÷ 3 = 9.0 h. 3.4 h comes from dividing 27 by the factor 8 instead of the number of halvings. 6.75 h comes from dividing by 4. 13.5 h assumes only two half-lives.
✓ 1
Answer D
Answer: D
Exam tip: Count the number of halvings (800 to 100 is three), not the overall factor, before dividing the time.
160IGPC3-5-07r22-04Radioactivity, half-life and safetyPaper CMedium1 mark
An engineer handles a gamma-emitting source while servicing equipment. Which action best reduces the engineer's exposure to radiation?
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Notes
Gamma radiation is very penetrating, so thick dense shielding and distance are effective. Paper stops only alpha particles. A thin aluminium sheet stops alpha and most beta but a large fraction of gamma passes through it. Holding the source close increases the dose because intensity falls with distance.
✓ 1
Answer A
Answer: A
Exam tip: Match the precaution to the radiation: shielding thickness and distance matter most for gamma.
161IGPC3-5-07r22-05Radioactivity, half-life and safetyPaper CMedium1 mark
In a food-science laboratory a student measures the radiation from a tracer sample. With no sample present the detector records 46 counts per minute. With the sample present it records 214 counts per minute. What is the corrected count rate due to the sample?
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Notes
The background must be subtracted: 214 − 46 = 168 counts per minute. 214 is the uncorrected reading. 260 results from adding the background instead. 46 is the background itself.
✓ 1
Answer B
Answer: B
Common error: Always subtract the background count rate from a reading taken with the source present.
162IGPC3-5-07r22-11Radioactivity, half-life and safetyPaper CMedium6 marks
A food company uses a sealed source of cobalt-60 to kill bacteria in packaged spices. Assume that the half-life of cobalt-60 is 5.0 years. When the source was installed, a detector at a fixed position recorded a corrected count rate of 2400 counts per second.
(a)
Suggest why gamma radiation, rather than alpha radiation, is used to treat sealed packets of food.
(2)
(b)
Calculate the corrected count rate at the detector 15 years after installation.
(2)
(c)
Determine the time taken for the corrected count rate to fall to 75 counts per second.
(2)
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Notes
Part (a)
gamma radiation is very penetrating and passes through the packaging and food
✓ 1
alpha particles would be absorbed by the packaging / travel only a few centimetres in air
✓ 1
Part (b)
15 years = 3 half-lives
✓ 1
2400 ÷ 8 = 300 counts per second
✓ 1
Part (c)
2400 to 1200, 600, 300, 150, 75 is 5 half-lives
✓ 1
5 × 5.0 = 25 years
✓ 1
Exam tip: For a time to reach a given rate, count halvings from the starting value, then multiply by the half-life.
163IGPC3-5-07r22-15Radioactivity, half-life and safetyPaper CMedium6 marks
A geologist at a mountain camp measures the radiation from a rock sample containing a short-lived isotope. The detector counts radiation in counts per minute. The background count rate is 24 counts per minute and stays constant. At time t = 0 the detector reading is 408 counts per minute. At t = 12 minutes the reading is 216 counts per minute.
(a)
Calculate the corrected count rate at t = 0 and at t = 12 minutes.
(2)
(b)
Show that the half-life of the isotope is 12 minutes.
(2)
(c)
Calculate the detector reading, including background, at t = 36 minutes.
(2)
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Notes
Part (a)
408 − 24 = 384 counts per minute
✓ 1
216 − 24 = 192 counts per minute
✓ 1
Part (b)
the corrected count rate halves (384 to 192) in 12 minutes
✓ 1
so the half-life is 12 minutes
✓ 1
Part (c)
36 minutes = 3 half-lives, so 384 ÷ 8 = 48 counts per minute
✓ 1
48 + 24 = 72 counts per minute
✓ 1
Exam tip: Add the background back on at the end when the question asks for the reading the detector shows.
164IGPC3-5-09r27-03Radioactivity, half-life and safetyPaper CMedium1 mark
A theatre exit sign contains tritium, which has a half-life of 12 years. What fraction of the original tritium nuclei remains after 36 years?
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Notes
36 years is 3 half-lives, and the amount halves each time: 1/2 × 1/2 × 1/2 = 1/8. 1/3 comes from dividing 36 by 12 and using the answer as the denominator directly. 1/6 comes from halving only twice and then thirding, or 3 × 2 = 6. 1/36 uses the number of years as the denominator.
✓ 1
Answer A
Answer: A
Exam tip: Work out how many half-lives have passed first, then halve that many times.
165IGPC3-5-09r27-08Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive source is placed in front of a detector. The background count rate is 15 counts/min. The detector reading is 250 counts/min with nothing between source and detector, 247 counts/min with a sheet of paper in the way, 96 counts/min with 3 mm of aluminium instead, and 15 counts/min with a thick block of lead. Which radiations does the source emit?
Diagram not to scale
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Notes
Paper changes the reading by only 3 counts/min, so no alpha is present, as alpha would be stopped by paper. Aluminium cuts the reading a great deal, so beta is present. The reading with aluminium is still well above background, so something more penetrating (gamma) is still arriving, and thick lead removes it. 'Alpha only' and 'alpha and gamma' fail because paper would have cut the count. 'Beta only' fails because the count with aluminium stays 81 counts/min above background.
✓ 1
Answer B
Answer: B
Exam tip: Alpha is stopped by paper, beta by a few mm of aluminium, and gamma is reduced only by thick lead.
166IGPC3-5-09r27-12Radioactivity, half-life and safetyPaper CMedium6 marks
Gamma radiation from a sealed source of iridium-192 is used to inspect the welds in a ship's hull. The half-life of iridium-192 is 74 days.
(a)
State two safety precautions the operator should take when using the source.
(2)
(b)
Suggest why gamma radiation is used to inspect the steel weld, rather than alpha radiation.
(2)
(c)
Calculate the fraction of the original activity that remains after 148 days.
(2)
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Notes
Part (a)
first precaution, any one from: store the source in a lead-lined container; handle it with long tongs or remote handling; stay as far from the source as possible; limit the time of exposure; wear a dosimeter (badge) to monitor the dose
✓ 1
second precaution, a different one from the same list
✓ 1
Part (b)
gamma radiation is highly penetrating and passes through the steel plate
✓ 1
alpha radiation would be stopped by the steel (or by a few centimetres of air) so would not reach the detector
✓ 1
Part (c)
148 days = 2 half-lives
✓ 1
1/4 (0.25)
✓ 1
Exam tip: Divide the total time by the half-life to find the number of half-lives, then halve that many times.
167IGPC3-5-09r27-15Radioactivity, half-life and safetyPaper CMedium5 marks
In a theatre's lighting workshop an old detector reads a background count rate of 24 counts/min. With a radioactive source placed near it, the reading is 156 counts/min.
(a)
State two sources of background radiation.
(2)
(b)
Calculate the corrected count rate due to the source.
(1)
(c)
The source has a half-life of 3.0 hours. Calculate the detector reading, including background, after 6.0 hours.
(2)
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Notes
Part (a)
first source, any one from: radon gas, rocks and soil, cosmic rays, food and drink, medical sources, building materials
Exam tip: Correct for background before you halve, and add it back if the question asks for the detector reading.
168IGPC3-5-09r28-03Radioactivity, half-life and safetyPaper CMedium1 mark
A hospital pharmacy chooses a radioactive tracer to be injected into a patient so that a detector outside the body can image an organ. Which tracer is most suitable?
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Notes
Gamma radiation passes out of the body so it can be detected, and a half-life of hours lets the activity fall quickly, limiting the dose to the patient. An alpha emitter would be absorbed within the body and could not be detected outside, and would also damage tissue. A gamma emitter with a half-life of 6 years would stay active in the patient far too long. A long-lived beta emitter is both absorbed by tissue before it can be detected and stays active for years.
✓ 1
Answer A
Answer: A
Exam tip: For a tracer inside the body, give both the radiation type (penetrating) and the half-life (short) in your answer.
169IGPC3-5-09r28-07Radioactivity, half-life and safetyPaper CMedium1 mark
A computer game models radioactive decay. A grid begins with 1000 unstable atoms. At the end of every turn, each remaining atom is removed with a probability of exactly one half, so the half-life is one turn. About how many unstable atoms are expected to remain after 3 turns?
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Notes
Each turn halves the number remaining: 1000 to 500 to 250 to 125. So about 125 atoms remain. 500 is the number left after only one turn. 250 is the number left after two turns. 333 comes from assuming a constant number is removed each turn, which is not how a half-life works: the number removed falls as fewer atoms remain.
✓ 1
Answer D
Answer: D
Common error: Radioactive decay is exponential: the amount lost each half-life gets smaller, so it is not a steady decrease.
170IGPC3-5-09r28-10Radioactivity, half-life and safetyPaper CMedium5 marks
A radioactive tracer for a medical scan has a half-life of 6.0 hours. A dose is prepared with an initial activity of 480 kBq.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the activity of the dose 24 hours after it was prepared.
(2)
(c)
Explain why a tracer with a half-life of about 6 hours is more suitable than one with a half-life of 6 seconds or one with a half-life of 6 years.
(2)
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Notes
Part (a)
the time taken for the activity (count rate or number of undecayed nuclei) to fall to half of its original value
✓ 1
Part (b)
M1 24 ÷ 6.0 = 4 half-lives, so activity is divided by 24 = 16
✓ 1
A1 480 ÷ 16 = 30 kBq
✓ 1
Part (c)
6 seconds: activity falls to almost nothing before the scan can be carried out
✓ 1
6 years: the patient would remain radioactive for a very long time / would receive a larger dose
✓ 1
Exam tip: Write the number of half-lives first, then divide by 2 raised to that number.
171IGPC3-5-09r28-11Radioactivity, half-life and safetyPaper CMedium7 marks
A water company injects a radioactive tracer into a main to find where water is escaping. A detector held above the ground records a count rate of 2430 counts per minute at the start of the test. The background count rate is 30 counts per minute. The tracer has a half-life of 15 hours.
(a)
Explain why the background count rate must be taken into account.
(1)
(b)
Calculate the count rate recorded by the detector at the same position 45 hours later. Assume the tracer has not moved and the background is unchanged.
(4)
(c)
State two sources of background radiation.
(2)
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Notes
Part (a)
B1 the detector also records radiation from other sources, so the measured rate is higher than the rate due to the tracer alone
one valid source, e.g. radon gas from rocks, cosmic rays, food and drink, building materials
✓ 1
a second, different valid source from this list
✓ 1
Common error: Subtract the background before halving, then add it back when a measured count rate is wanted.
172IGPC3-5-09r28-14Atomic structure, isotopes, fission and fusionPaper CMedium7 marks
In a strategy game, a player builds a fission power station rated at 1.2 GW. The designers base the game on a real reactor in which uranium-235 nuclei are split.
(a)
Describe what happens to a uranium-235 nucleus in nuclear fission.
(2)
(b)
Explain how a chain reaction occurs in a reactor.
(2)
(c)
State how the rate of fission is controlled in a reactor.
(1)
(d)
Calculate the energy transferred by the power station in one day (24 hours). Give your answer in joules to 3 significant figures.
(2)
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Part (a)
the nucleus absorbs a neutron and becomes unstable
✓ 1
it splits into two smaller nuclei, releasing neutrons and energy
✓ 1
Part (b)
neutrons released in one fission are absorbed by other uranium-235 nuclei
✓ 1
causing further fissions which release more neutrons, so the reaction continues
✓ 1
Part (c)
control rods absorb neutrons
✓ 1
Part (d)
M1 E = P × t = 1.2 × 109 × (24 × 3600)
✓ 1
A1 1.04 × 1014 J
✓ 1
Exam tip: Convert hours to seconds (×3600) before using E = P × t.
173IGPC3-5-09r28-15Radioactivity, half-life and safetyPaper CMedium8 marks
A hospital tests three sources P, Q and R to identify the radiation each emits. The count rate in counts per minute is measured with different absorbers placed between the source and the detector. The background count rate is 12 counts per minute.
Absorber
Source P
Source Q
Source R
none
540
800
650
paper
538
14
646
3 mm aluminium
536
12
19
50 mm lead
41
12
12
(a)
Identify the radiation emitted by each source. Give a reason for each.
(3)
(b)
Calculate the fraction of the radiation from source R, corrected for background, that is stopped by the 3 mm aluminium. Give your answer to 3 significant figures.
(3)
(c)
A factory monitors the thickness of thin aluminium foil as it is rolled. Suggest which source would be most suitable and explain why.
(2)
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Part (a)
P: gamma, because it passes through paper and aluminium almost unchanged and is only reduced by thick lead
✓ 1
Q: alpha, because it is stopped by paper
✓ 1
R: beta, because it passes through paper but is stopped by aluminium
its count rate through the foil changes noticeably as thickness changes, whereas alpha would be stopped completely and gamma would hardly change
✓ 1
Exam tip: Subtract the background from every reading before comparing or forming fractions.
174IGPC3-5-08r29-04Radioactivity, half-life and safetyPaper CMedium1 mark
A tracer is used to study drainage in a field. A sample of it gives a count rate of 960 counts/min, and 21 days later the count rate is 120 counts/min. Background radiation is negligible. What is the half-life of the tracer?
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The count rate halves 960 → 480 → 240 → 120, which is three half-lives, so the half-life is 21 / 3 = 7.0 days. 2.6 days divides 21 days by the factor 8 instead of by the number of halvings. 10.5 days uses two half-lives. 21 days is the total time elapsed.
✓ 1
Answer B
Answer: B
Common error: Count the number of halvings, not the overall factor, and divide the time by that number.
175IGPC3-5-08r29-06Radioactivity, half-life and safetyPaper CMedium1 mark
A technician must handle a sealed gamma source for a short test. Which action reduces the radiation dose to the technician most?
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Gamma radiation is penetrating and its intensity falls with distance, so using long tongs increases the distance and a short time reduces the exposure. Holding the source by hand brings it closer. Cotton gloves and polythene absorb almost no gamma radiation.
✓ 1
Answer C
Answer: C
Common error: Thin layers of cloth or plastic do not stop gamma radiation; use distance, time and dense shielding.
176IGPC3-5-08r29-10Radioactivity, half-life and safetyPaper CMedium6 marks
Phosphorus-32 is a radioactive isotope with a half-life of 14 days. A researcher adds a small amount of it to fertiliser to track uptake by wheat plants. A sample of the fertiliser gives a corrected count rate of 960 counts/min.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the corrected count rate from an identical sample after 56 days.
(3)
(c)
Suggest why an isotope with a half-life of 14 days is more suitable for this tracer than one with a half-life of a few seconds, or one lasting many years.
(2)
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Notes
Part (a)
The time taken for half the undecayed nuclei in a sample to decay (or for the count rate to halve)
✓ 1
Part (b)
M1 number of half-lives = 56 ÷ 14 = 4
✓ 1
M1 count rate halved four times (960 ÷ 24)
✓ 1
A1 60 counts/min
✓ 1
Part (c)
Long enough to be detected over the weeks of plant growth
✓ 1
Short enough that the activity falls to a low level soon after, so little radioactivity remains in the crop or soil
✓ 1
Common error: Divide the total time by the half-life to find the number of halvings before using the count rate.
177IGPC3-5-08r29-11Radioactivity, half-life and safetyPaper CMedium7 marks
An aircraft engineer tests a small sample of iridium-192, which has a half-life of 74 days, using a Geiger–Müller tube and counter. With the sample absent, the counter reads 18 counts/min. With the sample next to the tube, it reads 258 counts/min.
(a)
State what is meant by background radiation and give one source.
(2)
(b)
Calculate the corrected count rate from the sample.
(2)
(c)
Calculate the corrected count rate from the sample after 148 days.
(2)
(d)
State the reading expected on the counter at this time.
(1)
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Part (a)
Radiation that is always present in the environment
✓ 1
Any natural source: rocks, cosmic rays, radon gas, etc.
✓ 1
Part (b)
M1 subtract the background: 258 − 18
✓ 1
A1 240 counts/min
✓ 1
Part (c)
M1 148 days is 2 half-lives, so 240 ÷ 22 (ECF from (b))
✓ 1
A1 60 counts/min
✓ 1
Part (d)
A1 78 counts/min (ECF from (c): 60 + 18)
✓ 1
Common error: The counter always includes background, so subtract it to find the source count rate, and add it back to find the counter reading.
178IGPC3-5-08r29-13Atomic structure, isotopes, fission and fusionPaper CMedium7 marks
Potassium occurs naturally in soil as two isotopes, 3919K and 4019K. Potassium-40 is radioactive and decays by emitting a beta particle.
(a)
Explain why neutral atoms of the two isotopes have the same chemical properties.
(2)
(b)
State what a beta particle is and where it comes from.
(2)
(c)
For a neutral atom of potassium-40, state the number of protons, neutrons and electrons.
(3)
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Part (a)
Same number of protons, so the same number of electrons in neutral atoms
✓ 1
Chemical behaviour depends on the electrons
✓ 1
Part (b)
A fast-moving electron
✓ 1
Emitted from the nucleus
✓ 1
Part (c)
19 protons
✓ 1
21 neutrons
✓ 1
19 electrons
✓ 1
Exam tip: The number of electrons that controls chemical behaviour equals the number of protons in a neutral atom.
179IGPC3-5-09r29-03Radioactivity, half-life and safetyPaper CMedium1 mark
A smoke alarm in a shop contains a small alpha source. Smoke entering the alarm makes the current in the detector fall, which triggers the alarm. Which statement explains why an alpha source is suitable?
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Alpha particles are strongly ionising, so ionised air carries a current; smoke absorbs alpha particles, reducing ionisation and so the current. Saying alpha passes through smoke easily is wrong because alpha is the least penetrating radiation. Alpha does ionise air strongly. Smoke absorbs alpha particles rather than deflecting them to a counter.
✓ 1
Answer A
Answer: A
Common error: Do not say alpha is 'safe' because it does not ionise; it is strongly ionising and is only low-risk outside the body because it is easily absorbed.
180IGPC3-5-09r29-04Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive sample has a half-life of 3.0 hours. Its count rate, corrected for background, is 480 counts/min at 09:00. What is the corrected count rate at 18:00 on the same day?
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From 09:00 to 18:00 is 9.0 hours, which is three half-lives, so the rate is 480 ÷ 2 ÷ 2 ÷ 2 = 60 counts/min. 240 is only one half-life, 120 is two half-lives, and 160 comes from dividing 480 by the number of half-lives (3) instead of halving each time.
✓ 1
Answer C
Answer: C
Exam tip: Find the number of half-lives first, then halve that many times; the count rate is not divided by the number of half-lives.
181IGPC3-5-09r29-06Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Engineers designing a fusion power plant must heat hydrogen nuclei to extremely high temperatures. Why is such a high temperature needed for fusion?
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Nuclei repel each other because they are all positively charged, so they must collide at very high speed (high temperature) to get close enough to fuse. Fusion joins nuclei rather than splitting them. Slowing neutrons is part of fission reactors, not fusion between charged nuclei. Electrons do not join nuclei in fusion.
✓ 1
Answer A
Answer: A
Common error: Do not mix up fusion and fission: fusion joins small nuclei, fission splits large ones.
182IGPC3-5-09r29-10Radioactivity, half-life and safetyPaper CMedium6 marks
A hospital uses a radioactive tracer with a half-life of 6.0 hours. A sample of the tracer has a corrected count rate of 800 counts/s when it is prepared.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the corrected count rate 24 hours after the sample is prepared.
(3)
(c)
Calculate the time taken for the corrected count rate to fall to 100 counts/s.
(2)
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Part (a)
time taken for half of the unstable nuclei in a sample to decay (accept time for the count rate / activity to fall to half)
✓ 1
Part (b)
24 ÷ 6.0 = 4 half-lives
✓ 1
800 → 400 → 200 → 100 → 50 or 800 ÷ 24
✓ 1
50 counts/s
✓ 1
Part (c)
800 → 400 → 200 → 100 is 3 half-lives
✓ 1
3 × 6.0 = 18 hours
✓ 1
Exam tip: Give the working as a list of halvings (800, 400, 200 ...); it is clear and each step can earn credit.
183IGPC3-5-08r30-02Radioactivity, half-life and safetyPaper CMedium1 mark
Engineers check welds in a thick steel ship hull by placing a radioactive source on one side and photographic film on the other. Which type of radiation is most suitable, and why?
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Notes
Only gamma rays are penetrating enough to pass through thick steel and reach the film, with more passing through thinner or faulty regions. Alpha particles are strongly ionising but are stopped by paper or a few centimetres of air. Beta particles are less easily absorbed than alpha but are stopped by a few millimetres of aluminium or thin steel. Gamma rays are the least ionising of the three, not the most.
✓ 1
Answer C
Answer: C
Exam tip: Penetration and ionising power run in opposite directions: alpha is most ionising but least penetrating, gamma the reverse.
184IGPC3-5-08r30-04Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Four nuclides found in volcanic rock are written 3919J, 4019K, 4020L and 4221M. (J, K, L and M are not real symbols.) Which two are isotopes of the same element?
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Isotopes of the same element have the same number of protons but different numbers of neutrons. J and K both have 19 protons (20 and 21 neutrons). K and L have the same nucleon number (40) but different proton numbers, so they are different elements. K and M have the same number of neutrons (21) but different proton numbers. J and L also have the same number of neutrons (20) but different proton numbers.
✓ 1
Answer D
Answer: D
Common error: Isotopes share the same proton number (bottom number), not the same nucleon number or neutron number.
185IGPC3-5-08r30-09Atomic structure, isotopes, fission and fusionPaper CMedium7 marks
Volcanic gases contain atoms of sulfur-32 (3216S) and sulfur-34 (3416S).
(a)
State the number of protons, neutrons and electrons in a neutral atom of sulfur-34.
(3)
(b)
Explain why sulfur-32 and sulfur-34 are isotopes of the same element.
(2)
(c)
A gas sample contains 2000 sulfur atoms. 4.0% of them are sulfur-34. Calculate the total number of neutrons in the sulfur-34 atoms in the sample.
(2)
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Part (a)
16 protons
✓ 1
18 neutrons
✓ 1
16 electrons
✓ 1
Part (b)
they have the same number of protons (16)
✓ 1
but different numbers of neutrons (16 and 18)
✓ 1
Part (c)
number of sulfur-34 atoms = 0.040 × 2000 = 80
✓ 1
neutrons = 80 × 18 = 1440
✓ 1
Exam tip: Find the number of atoms of the isotope first, then multiply by the neutrons in one atom.
186IGPC3-5-08r30-10Radioactivity, half-life and safetyPaper CMedium6 marks
A theatre has smoke detectors containing a small source of americium-241, which emits alpha particles. The alpha particles ionise the air between two electrodes, producing a small steady current. Smoke entering the detector reduces the current and sounds an alarm.
Diagram not to scale
(a)
State what is meant by ionisation.
(1)
(b)
Explain why an alpha source is suitable for this detector.
(2)
(c)
Explain why a gamma source would not be suitable.
(1)
(d)
Explain why the detector is safe to use in a theatre even though it contains a radioactive source.
(2)
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Notes
Part (a)
the process in which atoms or molecules gain or lose electrons to become charged ions
✓ 1
Part (b)
alpha particles are strongly ionising, so they produce enough ions to give a measurable current
✓ 1
smoke particles absorb the alpha particles, so fewer ions form and the current falls
✓ 1
Part (c)
gamma rays are weakly ionising / pass through smoke, so the current would be tiny and would not change
✓ 1
Part (d)
alpha particles travel only a few centimetres in air
✓ 1
they are absorbed by the casing, so they cannot escape to people nearby
✓ 1
Common error: Alpha is the most ionising but the least penetrating radiation; do not describe it as the most penetrating.
187IGPC3-5-08r30-12Radioactivity, half-life and safetyPaper CMedium7 marks
At a volcano observatory a technician measures the background radiation with no source present. Three one-minute counts are 21, 18 and 24.
(a)
State two sources of background radiation.
(2)
(b)
Calculate the mean background count rate.
(2)
(c)
Suggest why the technician takes three counts and averages them.
(1)
(d)
A rock sample placed near the detector gives a count rate of 143 counts/min. Calculate the count rate due to the rock alone.
(2)
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Part (a)
any two from: radon gas in the air, cosmic rays, rocks and soil, food and drink, building materials, medical sources
✓ 2
Part (b)
total = 21 + 18 + 24 = 63
✓ 1
mean = 63 / 3 = 21 counts/min
✓ 1
Part (c)
radioactive decay is random so the counts vary; an average is more reliable
✓ 1
Part (d)
subtract the background
✓ 1
143 − 21 = 122 counts/min (ECF from (b))
✓ 1
Exam tip: Always subtract the mean background count rate from a measured count rate before using it.
188IGPC3-5-08r30-13Radioactivity, half-life and safetyPaper CMedium6 marks
Engineers inspect welds in a ship’s steel hull using a sealed source of gamma rays with a half-life of 74 days. Photographic film is placed on the opposite side of the weld.
Diagram not to scale
(a)
State two precautions that reduce the radiation dose received by the engineers.
(2)
(b)
Explain why the source is kept in a lead container.
(1)
(c)
Describe the harmful effects of ionising radiation on living cells.
(2)
(d)
Explain why a gamma source with a half-life of a few seconds would be unsuitable.
(1)
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Notes
Part (a)
any two from: keep the source in a shielded container when not in use; stay as far from the source as possible / use long tools or remote handling; limit the time of exposure; use shielding barriers; wear a dosimeter badge to monitor the dose
✓ 2
Part (b)
lead absorbs gamma rays, reducing the radiation that escapes
✓ 1
Part (c)
it can damage or mutate DNA in cells
✓ 1
this can lead to cell death or cancer
✓ 1
Part (d)
it would decay away almost immediately, so there would be no radiation left to make the image
✓ 1
Exam tip: For safety questions give the principle: more distance, less time, more shielding.
189IGPC3-5-07r38-03Radioactivity, half-life and safetyPaper CMedium1 mark
A source in a factory emits radiation that passes through a sheet of paper but is almost completely absorbed by 3 mm of aluminium. Which type of radiation does the source emit?
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Beta particles pass through paper but are stopped by a few millimetres of aluminium. Alpha particles would be stopped by the paper. Gamma rays would pass through 3 mm of aluminium with little absorption, so a mixture of alpha and gamma does not fit either observation.
✓ 1
Answer A
Answer: A
Exam tip: Remember the order of penetration: alpha is stopped by paper, beta by thin aluminium, gamma needs thick lead or concrete.
190IGPC3-5-07r38-04Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive tracer is injected into a shoal of fish. Its count rate, with background radiation ignored, is 640 counts/min. The half-life of the tracer is 3.0 hours. What is the count rate after 12 hours?
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Notes
12 hours is 4 half-lives, so the rate is halved four times: 640 → 320 → 160 → 80 → 40 counts/min. Dividing by 4 gives 160, dividing by 3 gives 210, and the count rate of a radioactive source does not fall to zero in a finite whole number of half-lives.
✓ 1
Answer C
Answer: C
Common error: Dividing by the number of half-lives is wrong; the quantity is halved once for each half-life.
191IGPC3-5-07r38-08Radioactivity, half-life and safetyPaper CMedium1 mark
Workers in a logistics depot move a sealed gamma source between containers. Which action is most effective at reducing the radiation dose they receive?
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Notes
Thick lead absorbs much of the gamma radiation and long tongs increase the distance, both of which reduce the dose. Paper and cotton gloves absorb almost no gamma radiation, and cooling does not alter radioactive emission.
✓ 1
Answer B
Answer: B
Exam tip: Dose is reduced by increasing distance, reducing time of exposure and using suitable shielding.
192IGPC3-5-07r38-10Radioactivity, half-life and safetyPaper CMedium5 marks
A marine biologist injects a radioactive tracer into a fish to follow its movements. The tracer has a half-life of 8.0 days. Immediately after injection its activity is 2400 Bq. Assume background radiation can be ignored.
(a)
State what is meant by the half-life of a radioactive isotope.
(1)
(b)
Calculate the activity of the tracer 24 days after injection.
(3)
(c)
Suggest why a tracer with a short half-life is chosen for use in a living animal.
(1)
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Notes
Part (a)
B1 time taken for the activity (or the number of undecayed nuclei) to fall to half its original value
✓ 1
Part (b)
M1 24 days is 3 half-lives
✓ 1
M1 2400 → 1200 → 600 → 300 (or 2400 ÷ 23)
✓ 1
A1 300 Bq
✓ 1
Part (c)
B1 the activity falls quickly, so little radiation remains in the animal (or less long-term damage to the animal)
✓ 1
Exam tip: Write the number of half-lives first, then halve step by step; it makes your working easy to follow.
193IGPC3-5-07r38-11Radioactivity, half-life and safetyPaper CMedium5 marks
Sealed cobalt-60 sources are delivered to a freight depot. Cobalt-60 emits gamma radiation.
(a)
State two safety precautions that depot workers should take when handling the sources.
(2)
(b)
Explain why a gamma source is more hazardous to a worker standing 1 m away than an alpha source of the same activity.
(2)
(c)
State one way in which ionising radiation can harm the human body.
(1)
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Notes
Part (a)
B1 any one from: store the source in a lead-lined container; use tongs or tools to keep it at a distance; limit time spent near the source; wear a dosimeter badge
✓ 1
B1 a second different precaution from the same list
✓ 1
Part (b)
B1 alpha particles are absorbed by a few centimetres of air (or cannot reach the worker at 1 m)
✓ 1
B1 gamma rays are highly penetrating, so they reach the worker (and pass into the body)
✓ 1
Part (c)
B1 damages or mutates DNA / can cause cancer / kills cells
✓ 1
Common error: Do not say that radiation makes a person radioactive; the harm comes from ionisation damaging cells.
194IGPC3-5-07r38-16Radioactivity, half-life and safetyPaper CMedium4 marks
A student investigates a radioactive source using a detector. With no source present, the detector gives 24 counts/min. The results with the source are shown.
Absorber: none, paper, 3 mm aluminium Count rate / counts per min: 612, 610, 25
Diagram not to scale
(a)
Calculate the count rate from the source alone with no absorber.
(1)
(b)
Deduce which type of radiation the source emits, using the data to support your answer.
(2)
(c)
Explain why the count rate with the aluminium is not zero.
(1)
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Notes
Part (a)
A1 612 − 24 = 588 counts/min
✓ 1
Part (b)
B1 beta
✓ 1
B1 paper has almost no effect (so not alpha) and 3 mm aluminium reduces the rate to about background (so not gamma)
✓ 1
Part (c)
B1 background radiation is still detected
✓ 1
Exam tip: Always take the background count rate away before comparing results for different absorbers.
195IGPC3-5-08r39-03Radioactivity, half-life and safetyPaper CMedium1 mark
A radioactive tracer with a half-life of 6.0 hours is added to the ballast water of a ship. What fraction of the original radioactive nuclei remains undecayed after 18 hours?
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Notes
18 hours is 3 half-lives, so the fraction left is (1/2)3 = 1/8. 1/3 treats the decay as linear. 1/6 multiplies the number of half-lives by 2 instead of halving each time. 1/4 allows for only two halvings.
✓ 1
Answer B
Answer: B
Common error: Each half-life halves what is left; it does not remove a fixed amount.
196IGPC3-5-08r39-11Radioactivity, half-life and safetyPaper CMedium5 marks
A radioactive gas from a volcanic vent is monitored. The count rates below have already been corrected for background radiation. 0 days: 640 counts/s 4.0 days: 320 counts/s 8.0 days: 160 counts/s 12.0 days: 80 counts/s
(a)
Determine the half-life of the gas.
(2)
(b)
Predict the corrected count rate after 24 days.
(2)
(c)
Explain why the readings were corrected for background radiation.
(1)
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Notes
Part (a)
Half-life is the time taken for the count rate to halve
✓ 1
4.0 days
✓ 1
Part (b)
24 days is 6 half-lives
✓ 1
640 / 26 = 10 counts/s
✓ 1
Part (c)
Background radiation is always present and would add to the count from the gas, so the true count from the gas would be overestimated
✓ 1
Exam tip: Find a half-life by checking that the count rate halves over equal time intervals, then count the number of half-lives.
197IGPC3-5-08r39-13Radioactivity, half-life and safetyPaper CMedium6 marks
A shipping company inspects hull welds with a sealed iridium-192 source. Take its half-life as 75 days. When new, its activity is 3.2 TBq.
(a)
State two harmful effects of ionising radiation on the human body.
(2)
(b)
State two precautions the crew should take when using the source.
(2)
(c)
Calculate the activity of the source after 225 days.
(2)
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Notes
Part (a)
Any two from: damages or mutates DNA / causes cancer / kills cells / causes burns or radiation sickness
✓ 2
Part (b)
Any two from: handle with tongs / keep in a lead-lined container when not in use / stay as far away as possible / limit time near it / wear a dosimeter badge
✓ 2
Part (c)
225 days is 3 half-lives
✓ 1
3.2 / 23 = 0.40 TBq
✓ 1
Common error: Divide the elapsed time by the half-life to count the halvings before you calculate the remaining activity.
198IGPC3-5-08r39-15Radioactivity, half-life and safetyPaper CMedium6 marks
Some old theatre exit signs glow because they contain tritium, hydrogen-3 (31H), sealed in glass tubes. Tritium emits beta particles and has a half-life of 12 years.
(a)
State the number of protons and the number of neutrons in a tritium nucleus.
(2)
(b)
Suggest why the beta radiation is not a danger to people in the theatre while the tubes are intact.
(2)
(c)
A sign is replaced when its activity has fallen to one quarter of its original value. Calculate the age of the sign when it is replaced.
(2)
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Notes
Part (a)
1 proton
✓ 1
2 neutrons
✓ 1
Part (b)
Beta particles have low penetrating power and are absorbed by the glass
✓ 1
The tritium is sealed in, so it cannot escape or be breathed in or swallowed
✓ 1
Part (c)
One quarter is 2 half-lives
✓ 1
2 × 12 = 24 years
✓ 1
Exam tip: Count how many halvings take you from the starting value to the final fraction, then multiply by the half-life.
199IGPC3-5-07r43-03Radioactivity, half-life and safetyPaper CMedium1 mark
A clinic tracing the spread of a disease uses a radioactive tracer of half-life 6.0 hours. A sample has a corrected count rate of 480 counts/min. What is the corrected count rate 24 hours later?
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Notes
24 hours is 24 ÷ 6.0 = 4 half-lives, so the count rate is halved four times: 480 → 240 → 120 → 60 → 30 counts/min. The value 20 comes from treating the fall as a steady straight-line decrease. The value 120 comes from dividing by the number of half-lives (4) instead of halving that many times. The value 60 comes from using only three half-lives.
✓ 1
Answer A
Answer: A
Common error: Halve once for each half-life; do not divide the count rate by the number of half-lives.
200IGPC3-5-07r43-05Radioactivity, half-life and safetyPaper CMedium1 mark
A technician in a warehouse handles a sealed gamma source for testing equipment. Which action best reduces the technician's exposure to the radiation?
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Notes
Gamma rays are very penetrating and only thick, dense shielding such as lead greatly reduces them, so a thick lead-lined container is best. Thin plastic does almost nothing to gamma rays. Holding the source close and for longer increases the exposure. A paper mask stops only alpha particles, not gamma rays.
✓ 1
Answer D
Answer: D
Common error: Shielding must match the radiation: paper stops alpha but only dense, thick material reduces gamma.
201IGPC3-5-07r43-08Atomic structure, isotopes, fission and fusionPaper CMedium1 mark
Which statement describes nuclear fusion, the process that releases energy in the Sun?
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Fusion is the joining of light nuclei, such as hydrogen, into a heavier nucleus such as helium, with energy released. The splitting of a heavy nucleus describes fission. Alpha emission is radioactive decay, not fusion. Absorbing a neutron to start a chain reaction describes fission in a reactor.
✓ 1
Answer C
Answer: C
Common error: Fusion joins small nuclei; fission splits large nuclei. Do not swap them.
202IGPC3-5-07r43-10Radioactivity, half-life and safetyPaper CMedium6 marks
A hospital tracking the spread of infection through a ward uses a radioactive tracer of half-life 8.0 days. A sample of the tracer has an initial corrected count rate of 800 counts/s.
(a)
Define half-life.
(1)
(b)
Calculate the corrected count rate of the sample after 24 days.
(3)
(c)
Suggest why a tracer with a half-life of a few seconds would be unsuitable and why a tracer with a half-life of many years would be unsuitable.
(2)
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Notes
Part (a)
time taken for the number of undecayed nuclei (or the activity / corrected count rate) to fall to half (its original value)
✓ 1
Part (b)
M1 24 ÷ 8.0 = 3 half-lives
✓ 1
M1 800 → 400 → 200 → 100, or 800 ÷ 23
✓ 1
A1 100 counts/s
✓ 1
Part (c)
short half-life: decays away before it can be used / detected
✓ 1
very long half-life: patient or ward remains radioactive for a very long time / higher exposure
✓ 1
Exam tip: Divide the elapsed time by the half-life to find the number of half-lives before you start halving.
203IGPC3-5-07r43-11Radioactivity, half-life and safetyPaper CMedium5 marks
A shop installs smoke detectors that contain a small source of americium-241, which emits alpha particles.
(a)
Explain why a source of alpha particles is suitable for a smoke detector. Give two reasons.
(2)
(b)
Explain why a gamma source would be unsuitable for this use.
(1)
(c)
The shop keeps boxes of spare detectors in a stockroom. State two safety precautions for storing them.
(2)
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Notes
Part (a)
first valid reason, e.g. alpha particles strongly ionise the air so a small current flows between the electrodes
✓ 1
second different valid reason, e.g. alpha has a very short range / is stopped by the casing or a few centimetres of air so it is not a hazard outside the detector
✓ 1
Part (b)
gamma is only weakly ionising (so too little current) OR very penetrating and hard to shield from users
✓ 1
Part (c)
first valid precaution, e.g. leave them in their sealed packaging / casing
✓ 1
second valid precaution, e.g. store in a labelled, locked area away from people and for as short a time as possible
✓ 1
Common error: Alpha is strongly ionising but easily stopped, so it is safe outside the body and harmful if inhaled or swallowed.
204IGPC3-5-07r43-14Atomic structure, isotopes, fission and fusionPaper CMedium5 marks
A solar farm and a wind farm both depend on the Sun's energy, which is released by a nuclear process in the Sun's core.
(a)
Name the nuclear process that releases energy in the Sun.
(1)
(b)
Describe this process.
(2)
(c)
Explain why this process happens only at very high temperatures.
(2)
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Notes
Part (a)
(nuclear) fusion
✓ 1
Part (b)
nuclei of hydrogen (small / light nuclei) join together
✓ 1
to form a nucleus of helium (a heavier nucleus) and energy is released
✓ 1
Part (c)
nuclei are positively charged and repel each other
✓ 1
at high temperature they move fast enough (have enough kinetic energy) to overcome the repulsion and get close together
✓ 1
Exam tip: Link high temperature to high kinetic energy: that is what lets positive nuclei overcome their repulsion.
205IGPC3-5-08r48-04Radioactivity, half-life and safetyPaper CMedium1 mark
A sealed beta source is used to test a gauge on a coffee packing line. The count rate from the source falls from 640 counts/min to 80 counts/min in 36 days. Background radiation is negligible. What is the half-life of the source?
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Notes
640 → 320 → 160 → 80 is three halvings, so 36 days is three half-lives and one half-life is 36 ÷ 3 = 12 days. 4.5 days comes from dividing 36 by the overall factor of 8 instead of the number of halvings. 9 days comes from counting four halvings. 18 days comes from counting only two halvings.
✓ 1
Answer B
Answer: B
Common error: Count how many times the rate halves, rather than dividing the time by the overall reduction factor.
206IGPC3-5-08r48-10Radioactivity, half-life and safetyPaper CMedium6 marks
A coffee roastery stores sacks of green beans in a store room. A technician uses a Geiger–Müller (GM) tube and counter to check the radiation levels.
(a)
State what is meant by background radiation and give one natural source of it.
(2)
(b)
With no sacks nearby, the counter records 252 counts in 10 minutes. Calculate the background count rate in counts/min.
(1)
(c)
Next to one sack, the counter records 468 counts in 10 minutes. Calculate the count rate due to the sack alone, in counts/min. Use the unrounded background rate.
(2)
(d)
Suggest why the technician counts for 10 minutes rather than for 10 seconds.
(1)
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Notes
Part (a)
radiation that is always present around us and does not come from the source being tested
✓ 1
one natural source, such as rocks, radon gas in the air, cosmic rays or food
✓ 1
Part (b)
25.2 counts/min (accept 25)
✓ 1
Part (c)
468 ÷ 10 = 46.8 counts/min
✓ 1
46.8 − 25.2 = 21.6 counts/min
✓ 1
Part (d)
radioactive decay is random, so the count varies from one short interval to the next; counting for longer gives a more reliable average
✓ 1
Exam tip: Always subtract the background count rate from the measured count rate before drawing conclusions about a source.
207IGPC3-5-08r48-11Radioactivity, half-life and safetyPaper CMedium7 marks
A gamma source is used to inspect welds in a steel bridge. The activity of the source was measured over 80 days.
Time / days
0
20
40
60
80
Activity / Bq
1600
1130
800
565
400
(a)
State what is meant by the half-life of a radioactive source.
(1)
(b)
Use the table to determine the half-life of the source.
(2)
(c)
Calculate the activity of the source 120 days after the first reading.
(2)
(d)
State two safety precautions for engineers who handle the gamma source.
(2)
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Part (a)
the time taken for the activity (or number of undecayed nuclei) of the source to fall to half of its original value
✓ 1
Part (b)
activity falls from 1600 Bq to 800 Bq between 0 and 40 days
✓ 1
half-life = 40 days
✓ 1
Part (c)
120 days is 3 half-lives
✓ 1
1600 → 800 → 400 → 200 Bq
✓ 1
Part (d)
any one from: handle with long tongs, keep as far away as possible, limit the time near the source, wear a dosimeter badge
✓ 1
a second different precaution, or store the source in a lead-lined container when not in use
✓ 1
Common error: Half-life is a time, not an amount: read off when the activity has halved from any starting point, not from zero.
208IGPC3-5-05r5-08Radioactivity, half-life and safetyPaper CHard1 mark
A student tests a sample of water for a radioactive isotope. The background count rate is 12 counts/min. The count rate measured with the sample is 76 counts/min at the start and 28 counts/min 4.0 hours later. What is the half-life of the isotope?
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Subtract the background first: 76 − 12 = 64 and 28 − 12 = 16 counts/min. The rate has fallen by a factor of 4, which is 2 half-lives in 4.0 h, so the half-life is 2.0 h. 1.0 h would be 4 half-lives. 2.8 h comes from using the raw readings, 76 to 28, without correcting. 4.0 h treats the whole time as one half-life.
✓ 1
Answer B
Answer: B
Common error: Always subtract background from every reading before comparing count rates.
209IGPC3-5-07r5-07Radioactivity, half-life and safetyPaper CHard1 mark
A radioactive source is placed near a detector. The reading is 340 counts per minute. Thirty minutes later it is 115 counts per minute. The background reading is a constant 40 counts per minute. What is the half-life of the source?
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Corrected rates are 340 − 40 = 300 and 115 − 40 = 75 counts per minute. The rate fell by a factor of 4, which is 2 half-lives, so the half-life is 30 ÷ 2 = 15 min, so B is correct. A takes the whole interval as one half-life. C ignores background (340 ÷ 115 is about 3, giving 3 half-lives). D treats a factor of 4 as 4 half-lives.
✓ 1
Answer B
Answer: B
Exam tip: Subtract background from every reading before comparing readings to find a half-life.
210IGPC3-5-07r5-16Radioactivity, half-life and safetyPaper CHard8 marks
A student investigates a radioactive source with a detector. With the source removed, the detector records a constant 20 counts per second. The table shows the count rate with the source present. Time / min: 0, 20, 40, 60 Count rate / counts per s: 340, 180, 100, 60
(a)
State what is meant by background radiation and give one source of it.
(2)
(b)
Determine the half-life of the source.
(3)
(c)
Predict the count rate at 100 min.
(2)
(d)
Explain why the measured count rate never falls to zero, however long the student waits.
(1)
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Notes
Part (a)
radiation that is always present in the environment, not from the source being tested
✓ 1
one source, for example radon gas, rocks, cosmic rays or food
✓ 1
Part (b)
subtract background to get 320, 160, 80 and 40 counts per second
✓ 1
the corrected rate falls from 320 to 160 in 20 min
✓ 1
half-life = 20 min
✓ 1
Part (c)
100 min is 5 half-lives, so 320 ÷ 32 = 10 counts per second from the source
✓ 1
add background: 10 + 20 = 30 counts per second
✓ 1
Part (d)
background radiation is still detected
✓ 1
Common error: Taking the half-life from the raw readings (340 to 180) gives a wrong value because background was not removed first.
211IGPC3-5-06r8-16Radioactivity, half-life and safetyPaper CHard6 marks
A packaging factory monitors the thickness of thin plastic film by passing it between a source and a detector. The sources available are: X (alpha emitter, half-life 430 years), Y (beta emitter, half-life 5.3 years), Z (gamma emitter, half-life 2.6 years).
Diagram not to scale
(a)
Identify the most suitable source and give two reasons for your choice.
(3)
(b)
Explain why a long half-life is an advantage in this use.
(1)
(c)
Calculate the fraction of the activity of Y remaining after 10.6 years.
(2)
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Part (a)
Source Y
✓ 1
Beta passes through thin plastic but is partly absorbed, so the count rate changes with thickness
✓ 1
Alpha (X) is stopped completely and gamma (Z) passes almost unchanged, so neither shows changes in thickness
✓ 1
Part (b)
Count rate stays almost constant over time, so little recalibration is needed
✓ 1
Part (c)
10.6 ÷ 5.3 = 2 half-lives
✓ 1
(1/2)2 = 0.25
✓ 1
Exam tip: For source choice, match radiation type to the material it must pass through and the half-life to the length of use.
212IGPC3-5-05r10-08Radioactivity, half-life and safetyPaper CHard1 mark
A radioactive source has an activity of 640 Bq. Fifteen days later its activity is 20 Bq. What is the half-life of the source?
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640/20 = 32 = 25, so the activity halved 5 times in 15 days. Half-life = 15/5 = 3.0 days (D). A assumes one half-life is half the time. B assumes the full time is one half-life. C divides the time by 32 rather than by the number of halvings.
✓ 1
Answer D
Answer: D
Common error: Count how many times the activity halves, not the overall ratio, before finding the half-life.
213IGPC3-5-06r10-08Radioactivity, half-life and safetyPaper CHard1 mark
A radioactive source gives a measured count rate of 340 counts/min. The background count rate is 20 counts/min. After 5.0 hours the measured count rate is 100 counts/min. What is the half-life of the source?
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Notes
Corrected rates: 340 - 20 = 320 and 100 - 20 = 80 counts/min. 320 to 80 is two halvings, so half-life = 5.0/2 = 2.5 hours (B). A comes from using the uncorrected rates (340/100 = 3.4 halvings). C treats the fall to a quarter as one half-life. D uses four halvings.
✓ 1
Answer B
Answer: B
Exam tip: Subtract the background from both readings before comparing them.
214IGPC3-5-07r21-16Radioactivity, half-life and safetyPaper CHard6 marks
Four radioactive sources are available to an engineering firm. P: alpha emitter, half-life 430 years Q: beta emitter, half-life 28 years R: gamma emitter, half-life 6 hours S: beta emitter, half-life 12 minutes For each use below, choose the most suitable source and give one reason for your choice.
(a)
Tracing the flow of water through a buried pipe network using detectors at ground level.
(2)
(b)
Monitoring the thickness of aluminium foil as it is made, with the same source in use for many years.
(2)
(c)
A smoke detector fitted to the ceiling of a building.
(2)
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Part (a)
R
✓ 1
gamma rays pass through soil and pipe walls, and the short half-life means little radiation remains afterwards
✓ 1
Part (b)
Q
✓ 1
beta particles are partly absorbed by thin foil so the reading changes with thickness, and the long half-life keeps the output almost constant
✓ 1
Part (c)
P
✓ 1
alpha particles ionise the air in the small gap and are stopped before leaving the unit, and the long half-life means the source lasts for years
✓ 1
Exam tip: Match the type of radiation to what it must pass through, and the half-life to how long the source must last.
215IGPC3-5-07r22-08Radioactivity, half-life and safetyPaper CHard1 mark
A source is monitored by a detector in a laboratory. The constant background count rate is 20 counts per second. At the start the detector reads 340 counts per second. After 40 minutes it reads 100 counts per second. What is the half-life of the source?
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Notes
Corrected rates are 340 − 20 = 320 and 100 − 20 = 80 counts per second. The rate has fallen by a factor of 4, which is two half-lives, so the half-life is 40 ÷ 2 = 20 min. 23 min comes from using the uncorrected readings. 13 min assumes three half-lives. 10 min assumes four half-lives.
✓ 1
Answer A
Answer: A
Common error: Subtract the background from both readings before comparing them; otherwise the half-life comes out wrong.
216IGPC3-5-08r39-07Radioactivity, half-life and safetyPaper CHard1 mark
A detector near a volcanic vent records 270 counts per minute with a radioactive gas sample present. With the sample removed it records 30 counts per minute. Twelve hours later the count rate with the sample present is 60 counts per minute. What is the half-life of the gas?
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Notes
After subtracting the 30 counts per minute of background, the sample's count rate falls from 240 to 30 counts per minute. That is a fall by a factor of 8, which is 3 half-lives, so the half-life is 12 / 3 = 4.0 hours. 6.0 hours comes from not subtracting the background from the final reading, giving 240 to 60, which is only 2 half-lives. 1.5 hours comes from dividing 12 by the factor 8. 12 hours takes the elapsed time as the half-life.
✓ 1
Answer D
Answer: D
Common error: Subtract the background count rate from every reading before comparing count rates.
217IGPC3-5-08r39-16Radioactivity, half-life and safetyPaper CHard7 marks
A volcanology team monitors a radioactive gas sample. The background count rate is 25 counts/min. The total count rates measured with the sample present are: 0 min: 345 counts/min 30 min: 185 counts/min 60 min: 105 counts/min 90 min: 65 counts/min
(a)
Calculate the corrected count rate from the sample at 0 min and at 90 min.
(2)
(b)
Determine the half-life of the sample.
(2)
(c)
Calculate the total count rate that would be measured at 150 min.
(3)
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Notes
Part (a)
345 − 25 = 320 counts/min
✓ 1
65 − 25 = 40 counts/min
✓ 1
Part (b)
Corrected rates 320, 160, 80, 40 halve every 30 min
✓ 1
Half-life = 30 min
✓ 1
Part (c)
150 min is 5 half-lives
✓ 1
Corrected rate = 320 / 25 = 10 counts/min
✓ 1
Total = 10 + 25 = 35 counts/min
✓ 1
Common error: Add the background count rate back on when asked for the rate the detector would actually show.
218IGPC3-5-08r48-08Radioactivity, half-life and safetyPaper CHard1 mark
A technician in a coffee roastery measures a radioactive sample. The count rate is 440 counts/min at the start and 90 counts/min exactly 9.0 hours later. The background count rate is 40 counts/min. What is the half-life of the sample?
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Notes
The corrected rates are 400 counts/min and 50 counts/min, a reduction by a factor of 8, which is three halvings. So 9.0 hours is three half-lives and the half-life is 3.0 hours. 3.9 hours comes from using the uncorrected rates without subtracting background. 4.5 hours comes from assuming two halvings. 9.0 hours takes the whole time as one half-life.
✓ 1
Answer C
Answer: C
Exam tip: Subtract the background count rate from every reading before comparing count rates to find a half-life.
219IGPC3-5-08r48-15Radioactivity, half-life and safetyPaper CHard8 marks
A technician in a coffee roastery measures the count rate from a radioactive sample at 2-hour intervals. The background count rate is 30 counts/min.
Time / h
0
2
4
6
8
Measured count rate / counts/min
270
150
90
60
45
(a)
Calculate the count rate due to the sample alone at each time.
(2)
(b)
Determine the half-life of the sample.
(2)
(c)
Predict the count rate that the technician would measure 14 hours after the start.
(3)
(d)
Suggest why the half-life is hard to determine from readings taken a long time after the start.
(1)
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Notes
Part (a)
subtract the background rate of 30 counts/min from each reading
✓ 1
240, 120, 60, 30, 15 counts/min
✓ 1
Part (b)
time for the corrected count rate to fall to half, e.g. 240 to 120
✓ 1
2 hours
✓ 1
Part (c)
14 hours is 7 half-lives
✓ 1
240 ÷ 27 = 1.875 counts/min from the sample
✓ 1
add background: 1.875 + 30 = 31.9 counts/min
✓ 1
Part (d)
the count rate from the sample becomes very small compared with the background, so random variations make the corrected value unreliable
✓ 1
Exam tip: The counter always measures sample plus background, so add the background back on when predicting what a counter will read.
220IGPC3-5-07r21-07Atomic structure, isotopes, fission and fusionPaper EEasy1 mark
A research ship has a nuclear reactor that uses uranium-235 as fuel. Which statement describes how the reactor releases energy?
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Notes
Fission of uranium-235 occurs when a nucleus absorbs a neutron and splits into smaller nuclei, releasing energy and more neutrons; A is correct. B describes fusion, as in the Sun. C describes burning, a chemical change. D describes ionisation, which releases far less energy.
✓ 1
Answer A
Answer: A
Common error: Fission splits a large nucleus; fusion joins small nuclei. Do not mix the two up.
221IGPC-039-00-04Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
An iron nucleus is represented as 5626Fe. What is the relative charge on this nucleus?
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Only protons in the nucleus are charged, and each has a relative charge of +1. Neutrons are uncharged. The nucleus contains 26 protons, so its relative charge is +26. The value +30 counts the neutrons instead, +56 counts every nucleon, and 0 is the charge of the whole neutral atom, not of the nucleus.
✓ 1
Answer A
Answer: A
Exam tip: The relative charge of a nucleus equals the proton number. The relative mass is approximately equal to the nucleon number.
222IGPC-039-00-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
In an alpha-particle scattering experiment, a beam of alpha particles is directed at a very thin gold foil. Almost all of the alpha particles pass straight through the foil with no deflection. What does this observation show?
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Notes
If almost all alpha particles meet nothing that affects them, the atom must be mostly empty space. The positive charge of the nucleus is shown by the few large deflections, not by the particles that pass straight through. The experiment gives no evidence about neutrons or electron shells.
✓ 1
Answer B
Answer: B
Exam tip: Link each observation to its own conclusion. Straight through means mostly empty space. Large deflection means a small, dense, positive nucleus.
223IGPC-039-00-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
A uranium-235 nucleus absorbs a neutron and undergoes fission: 23592U + 10n → 14054Xe + 94ZSr + 210n What is the value of Z?
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Notes
Proton numbers must balance on both sides of the equation. The left side has 92 + 0 = 92. Neutrons have a proton number of 0, so 92 = 54 + Z, which gives Z = 38. The value 40 wrongly counts the 2 neutrons as charged. The value 54 is xenon's proton number, and 94 is the nucleon number of strontium.
✓ 1
Answer A
Answer: A
Common error: Giving the neutron a proton number of 1. A neutron has proton number 0, so it adds nothing to the bottom-row total.
224IGPC-039-00-14Atomic structure, isotopes, fission and fusionPaper EMedium7 marks
In a nuclear power station, one possible fission reaction is: 23592U + 10n → 14456Ba + 8936Kr + x10n
(a)
State what is meant by nuclear fission.
(1)
(b)
Calculate the value of x.
(2)
(c)
Explain how this reaction can lead to a chain reaction.
(2)
(d)
State how the total mass of the products compares with the total mass of the reactants, and state the main form of the energy released.
(2)
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Notes
Part (a)
the splitting of a large (heavy) nucleus into two smaller nuclei
✓ 1
Part (b)
nucleon numbers balance: 235 + 1 = 144 + 89 + x ; x = 3
✓ 2
Part (c)
each fission releases more than one neutron / releases 3 neutrons ; these neutrons are absorbed by other uranium-235 nuclei, causing further fissions
✓ 2
Part (d)
the total mass of the products is (slightly) less than that of the reactants ; energy is released mainly as kinetic energy of the products
✓ 2
Common error: Forgetting the incoming neutron on the left-hand side. The total nucleon number on the left is 236, not 235.
225IGPC-039-01-05Radioactivity, half-life and safetyPaper EMedium1 mark
The nuclide polonium-210, \({}^{210}_{84}\text{Po}\), decays by emitting an α-particle to form an isotope of lead (Pb). Which nuclide is formed?
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Notes
An α-particle is \({}^{4}_{2}\text{He}\). It carries away 4 nucleons and 2 protons. The nucleon number becomes 210 − 4 = 206 and the proton number becomes 84 − 2 = 82, giving \({}^{206}_{82}\text{Pb}\). \({}^{210}_{82}\text{Pb}\) changes only the proton number. \({}^{206}_{84}\text{Pb}\) changes only the nucleon number. \({}^{214}_{86}\text{Pb}\) adds the α-particle instead of removing it.
✓ 1
Answer A
Answer: A
Exam tip: Check that the top numbers balance and the bottom numbers balance on both sides of a decay equation.
226IGPC-039-01-06Radioactivity, half-life and safetyPaper EMedium1 mark
A nucleus emits a β−-particle. What happens to the proton number and to the nucleon number of the nucleus?
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Notes
In β− decay, a neutron in the nucleus changes into a proton and a fast-moving electron, and the electron is emitted. The nucleus gains one proton and loses one neutron, so the proton number rises by 1. The total number of nucleons stays the same. The first option treats the emitted electron as if it removes a proton. The other options wrongly assume the nucleon number changes.
✓ 1
Answer B
Answer: B
Common error: Losing a negative particle does not lower the proton number. A neutron turns into a proton, so the proton number goes up.
227IGPC-039-01-07Radioactivity, half-life and safetyPaper EMedium1 mark
A narrow beam containing α-particles, β-particles and γ-rays passes between two parallel charged metal plates in a vacuum. Which statement describes what happens?
Diagram not to scale. Vacuum between the plates.
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α-particles are positive and β-particles are negative, so the electric field deflects them in opposite directions. A β-particle has a much smaller mass than an α-particle, so it is deflected much more, even though its charge is only half as large. γ-rays are uncharged and pass straight through. The last option assumes the larger charge of the α-particle gives the larger deflection, ignoring its much greater mass.
✓ 1
Answer C
Answer: C
Exam tip: Deflection depends on both charge and mass. β-particles are thousands of times lighter than α-particles, so they bend far more.
228IGPC-039-01-14Radioactivity, half-life and safetyPaper EMedium6 marks
Strontium-90, \({}^{90}_{38}\text{Sr}\), decays by β− emission to form yttrium (Y). The half-life of strontium-90 is 29 years.
(a)
Write the nuclide notation for the isotope of yttrium formed.
(2)
(b)
Describe the change that occurs inside the nucleus during β− decay.
(1)
(c)
The yttrium nucleus also decays by β− emission to form zirconium (Zr). State the nucleon number and the proton number of the zirconium nucleus.
(1)
(d)
Calculate the time taken for the activity of a sample of strontium-90 to fall to \(\frac{1}{8}\) of its initial value.
(2)
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Part (a)
nucleon number 90
✓ 1
proton number 39, giving \({}^{90}_{39}\text{Y}\)
✓ 1
Part (b)
A neutron changes into a proton and an electron, and the electron is emitted .
✓ 1
Part (c)
nucleon number 90 and proton number 40
✓ 1
Part (d)
\(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), so 3 half-lives
✓ 1
3 × 29 = 87 years
✓ 1
Common error: In β⁻ decay the nucleon number stays the same. Do not subtract 1 from it because an electron leaves.
229IGPC-039-00-13Atomic structure, isotopes, fission and fusionPaper EMedium7 marks
In a vacuum, a narrow beam of alpha particles is directed at a gold foil that is only a few hundred atoms thick. Detectors placed around the foil record the directions in which the alpha particles travel after they hit the foil.
Diagram not to scale. The apparatus is in a vacuum.
(a)
Describe what is observed about the paths of the alpha particles.
(3)
(b)
Explain what these observations show about the structure of the atom.
(4)
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Notes
Part (a)
most alpha particles pass straight through the foil undeflected ; a small number are deflected through small angles ; a very small number are deflected through large angles / bounce back towards the source
✓ 3
Part (b)
most of the atom is empty space, because most alpha particles pass straight through ; the nucleus is positively charged, because it repels / deflects the positive alpha particles ; the nucleus is very small compared with the atom, because very few alpha particles are deflected through large angles ; the nucleus contains most of the mass of the atom / is very dense, because some alpha particles bounce back
✓ 4
Exam tip: In part (b), state each conclusion separately. Link each one to the specific observation from part (a) that provides the evidence for it.
230IGPC-039-00-10Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Two hydrogen nuclei fuse: 21H + 31H → 42He + X What is particle X?
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Notes
Nucleon numbers: 2 + 3 = 4 + A, so A = 1. Proton numbers: 1 + 1 = 2 + Z, so Z = 0. A particle with nucleon number 1 and proton number 0 is a neutron. A proton would need Z = 1. An electron has A = 0 and Z = −1. An alpha particle has A = 4 and Z = 2.
✓ 1
Answer B
Answer: B
Exam tip: Balance the top and bottom rows separately, then match the leftover numbers to a particle.
231IGPC3-5-05c-04Radioactivity, half-life and safetyPaper EMedium1 mark
Strontium-90, 9038Sr, released in a nuclear accident, decays by beta-minus (β−) emission. Which nuclide is formed?
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Notes
In β− decay a neutron changes into a proton and an electron is emitted, so the proton number rises by 1 (38 → 39) while the nucleon number stays 90: B is correct. A has the proton number falling. C loses a nucleon, which does not happen in beta decay. D is the product of alpha decay (nucleon number −4, proton number −2).
✓ 1
Answer B
Answer: B
Common error: In beta-minus decay the nucleon number stays the same and the proton number goes up by one.
232IGPC3-5-05c-10Radioactivity, half-life and safetyPaper EMedium6 marks
A smoke alarm contains a tiny sample of americium-241, 24195Am, which decays by emitting alpha particles. Its half-life is 432 years.
(a)
Complete the equation for the decay: 24195Am → ……Np + 42He [2]
(2)
(b)
Explain why an alpha-emitting source is suitable for use in a smoke alarm. [2]
(2)
(c)
Calculate the fraction of the americium-241 that remains undecayed after 1296 years. [2]
(2)
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Part (a)
nucleon number of neptunium = 237
✓ 1
proton number of neptunium = 93
✓ 1
Part (b)
alpha particles ionise the air strongly, producing a small steady current between electrodes
✓ 1
they travel only a few centimetres in air, so very little radiation escapes from the alarm
✓ 1
Part (c)
1296 ÷ 432 = 3 half-lives
✓ 1
(1/2)3 = 1/8 (0.125)
✓ 1
Exam tip: In a decay equation, the nucleon numbers on each side add up to the same total and so do the proton numbers.
233IGPC3-5-05r5-05Radioactivity, half-life and safetyPaper EMedium1 mark
A carbon-14 nucleus has nucleon number 14 and proton number 6. It decays by beta-minus emission. What are the nucleon number and proton number of the nucleus formed?
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Notes
In beta-minus decay a neutron becomes a proton and an electron is emitted, so the nucleon number stays at 14 and the proton number rises by 1 to 7. A proton number of 5 treats the emitted electron as taking away a proton. 10 and 4 describe alpha decay. 13 and 6 would mean a neutron was lost.
✓ 1
Answer C
Answer: C
Exam tip: In beta-minus decay the nucleon number is unchanged and the proton number goes up by one.
234IGPC3-5-05r5-12Radioactivity, half-life and safetyPaper EMedium5 marks
Phosphorus-32 emits beta-minus particles and is used in pharmacology to treat some blood disorders. A nucleus of phosphorus-32 has nucleon number 32 and proton number 15. It decays into an isotope of sulfur (proton number 16).
(a)
State the nucleon number and the proton number of the sulfur nucleus formed.
(2)
(b)
Describe what happens inside the nucleus when it emits a beta-minus particle.
(1)
(c)
A dose of phosphorus-32 has an activity of 80 MBq. Its half-life is taken as 14 days. Calculate the activity after 42 days.
(2)
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Notes
Part (a)
nucleon number 32
✓ 1
proton number 16
✓ 1
Part (b)
a neutron changes into a proton and an electron, which is emitted
✓ 1
Part (c)
42 days is 3 half-lives
✓ 1
80 ÷ 23 = 10 MBq
✓ 1
Exam tip: Check your decay equation by confirming that the nucleon numbers and the proton (charge) numbers each balance on both sides.
235IGPC3-5-05r5-13Radioactivity, half-life and safetyPaper EMedium6 marks
Radium-226 (nucleon number 226, proton number 88) decays by alpha emission to radon-222, a radioactive gas that can dissolve in well water. For this question take the half-life of radon-222 as 3.5 days.
(a)
State the nucleon number and proton number of radon-222, and of an alpha particle.
(2)
(b)
Explain why an alpha emitter is more dangerous if swallowed or breathed in than if it is held outside the body.
(2)
(c)
A water sample contains radon-222 with an activity of 560 Bq. Calculate the activity after 14 days.
(2)
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Notes
Part (a)
radon-222: nucleon number 222 and proton number 86
✓ 1
alpha particle: nucleon number 4 and proton number 2
✓ 1
Part (b)
alpha particles are strongly ionising
✓ 1
outside the body they are stopped by air or skin, but inside they are absorbed by living cells, damaging them
✓ 1
Part (c)
14 days is 4 half-lives
✓ 1
560 ÷ 24 = 35 Bq
✓ 1
Common error: Alpha radiation is harmless outside the body only because it is easily stopped. Inside the body, that same stopping is what deposits all its energy in tissue.
236IGPC3-5-07r5-05Radioactivity, half-life and safetyPaper EMedium1 mark
A nucleus of radium-226 has 88 protons. It decays by emitting an alpha particle. What are the nucleon number and the proton number of the new nucleus?
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Notes
An alpha particle is a helium nucleus (4 nucleons, 2 protons). The nucleon number falls by 4 to 222 and the proton number falls by 2 to 86, so D is correct. A forgets that the proton number changes. B changes only the proton number. C adds the alpha particle's numbers instead of removing them.
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Answer D
Answer: D
Exam tip: In a nuclear equation, the nucleon numbers on each side must add to the same total and so must the proton numbers.
237IGPC3-5-07r5-06Radioactivity, half-life and safetyPaper EMedium1 mark
Carbon-14 decays by beta-minus emission. Which statement describes what happens in the nucleus?
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Notes
In beta-minus decay a neutron becomes a proton and an electron is created and ejected, so the proton number rises by 1 and the nucleon number is unchanged. A is correct. B reverses the change, which would lower the proton number. C describes ionisation of an atom, not a change in the nucleus. D describes alpha decay.
✓ 1
Answer A
Answer: A
Common error: The beta particle does not come from the electron shells; it comes from the nucleus.
238IGPC3-5-07r5-08Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Fusion of hydrogen nuclei in a star needs a temperature of millions of degrees. What is the reason for this?
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Notes
Hydrogen nuclei are positively charged and repel each other. They must move fast enough, which means a very high temperature, to come close enough for the strong nuclear force to join them, so C is correct. A describes fission, not fusion. B confuses fusion with a chain reaction that releases neutrons. D is wrong because temperature does not make nuclei more massive.
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Answer C
Answer: C
Common error: Do not mix up fusion (joining small nuclei) with fission (splitting large nuclei); the conditions needed are very different.
239IGPC3-5-07r5-15Atomic structure, isotopes, fission and fusionPaper EMedium6 marks
A coastal desalination plant that supplies drinking water is powered by a nuclear fission reactor. Each fission of a uranium-235 nucleus releases 3.2 × 10−11 J. The reactor produces thermal power of 80 MW.
(a)
Describe how a chain reaction occurs in the reactor.
(3)
(b)
State the purpose of the control rods.
(1)
(c)
Calculate the number of fissions each second needed to produce 80 MW.
(2)
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Notes
Part (a)
a neutron is absorbed by a uranium-235 nucleus, which splits into two smaller nuclei
✓ 1
the fission releases more neutrons
✓ 1
these neutrons cause further fissions in other nuclei
✓ 1
Part (b)
they absorb neutrons to control the rate of fission (the chain reaction)
✓ 1
Part (c)
energy each second = 80 × 106 J
✓ 1
number = 80 × 106 ÷ 3.2 × 10−11 = 2.5 × 1018 per second
✓ 1
Exam tip: 1 MW is 10⁶ J every second, so power in watts gives the energy per second directly.
240IGPC3-5-05r8-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Hydrogen nuclei in a star must be at an extremely high temperature for fusion to happen. What is the reason for this?
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Nuclei are all positively charged and repel each other; only very fast-moving nuclei (high temperature) get close enough to fuse. B is wrong because melting is irrelevant to nuclear forces. C describes fission, not fusion. D is wrong because removing electrons leaves the nucleus positive, not neutral.
✓ 1
Answer B
Answer: B
Exam tip: Link high temperature to high speed of the nuclei, then to overcoming repulsion between like charges.
241IGPC3-5-05r8-07Radioactivity, half-life and safetyPaper EMedium1 mark
A smoke detector in a honey-processing room contains americium-241, 24195Am, which decays by emitting an alpha particle. Which nucleus is formed?
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Notes
An alpha particle is 42He, so the nucleon number falls by 4 (241 → 237) and the proton number falls by 2 (95 → 93): neptunium-237. Cm-241 shows the proton number rising by 1, as in beta decay. Am-237 changes only the nucleon number. Bk-245 adds the alpha particle instead of removing it.
✓ 1
Answer D
Answer: D
Common error: In alpha decay both numbers go down: nucleon number by 4 and proton number by 2.
242IGPC3-5-05r8-11Radioactivity, half-life and safetyPaper EMedium5 marks
A racing team adds a radioactive tracer to a piston ring to study wear. The tracer has a half-life of 12 days. The initial activity is 3200 Bq.
(a)
State what is meant by half-life.
(1)
(b)
Calculate the activity after 48 days.
(2)
(c)
Calculate the time taken for the activity to fall to 25 Bq.
(2)
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Notes
Part (a)
Time taken for the activity (or number of undecayed nuclei) to fall to half of its value
✓ 1
Part (b)
48 days = 4 half-lives
✓ 1
3200 ÷ 16 = 200 Bq
✓ 1
Part (c)
3200 ÷ 25 = 128 = 27, so 7 half-lives
✓ 1
7 × 12 = 84 days
✓ 1
Exam tip: Write the number of half-lives as a power of 2 to avoid counting mistakes.
243IGPC3-5-05r8-15Radioactivity, half-life and safetyPaper EMedium4 marks
Cobalt-60 is used in a hospital to treat tumours. A cobalt nucleus 6027Co decays by beta emission to form a nickel nucleus.
(a)
Describe what a beta particle is.
(1)
(b)
State the nucleon number and the proton number of the nickel nucleus formed.
(2)
(c)
Calculate the number of neutrons in the nickel nucleus.
(1)
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Notes
Part (a)
A fast-moving electron emitted from the nucleus
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Part (b)
Nucleon number 60
✓ 1
Proton number 28
✓ 1
Part (c)
60 − 28 = 32
✓ 1
Common error: In beta decay the nucleon number stays the same and the proton number rises by 1.
244IGPC3-5-06r8-04Radioactivity, half-life and safetyPaper EMedium1 mark
A wooden flute contains carbon-14, which has a half-life of 5700 years. What fraction of the original carbon-14 remains after 17 100 years?
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Notes
17 100 ÷ 5700 = 3 half-lives, so the fraction is (1/2)3 = 0.125. 0.333 divides by 3. 0.167 is 1/(3×2). 0.250 uses only 2 half-lives.
✓ 1
Answer D
Answer: D
Common error: Halve once per half-life; do not divide by the number of half-lives.
245IGPC3-5-06r8-07Radioactivity, half-life and safetyPaper EMedium1 mark
Radon-222, 22286Rn, seeps into a farm cellar and decays by alpha emission. Which nucleus is formed?
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Notes
Alpha emission removes 4 nucleons and 2 protons: 222 − 4 = 218 and 86 − 2 = 84, polonium-218. Ra-218 raises the proton number; Fr-222 is beta decay; Ra-226 adds the alpha particle.
✓ 1
Answer B
Answer: B
Common error: Alpha decay lowers both numbers (nucleon by 4, proton by 2).
246IGPC3-5-06r8-13Atomic structure, isotopes, fission and fusionPaper EMedium4 marks
A smoke detector in a concert hall contains americium-241, 24195Am, which emits alpha particles.
(a)
Describe what an alpha particle is.
(1)
(b)
State the nucleon number and the proton number of the nucleus formed after one alpha emission.
(2)
(c)
Calculate the number of neutrons in the nucleus formed.
(1)
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Notes
Part (a)
A helium nucleus: 2 protons and 2 neutrons
✓ 1
Part (b)
Nucleon number 237
✓ 1
Proton number 93
✓ 1
Part (c)
237 − 93 = 144
✓ 1
Common error: Alpha emission reduces the proton number by 2, not by 4.
247IGPC3-5-06r8-15Atomic structure, isotopes, fission and fusionPaper EMedium6 marks
A reactor in a power station supplies a food-processing region. It releases 3.2×10−11 J of energy in each fission and has a thermal power output of 1.6×108 W.
(a)
State the main energy transfer in the fission process.
(2)
(b)
Calculate the number of fissions per second.
(2)
(c)
Calculate the number of fissions in one hour. Use the exact value from (b).
(2)
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Part (a)
Nuclear energy transfers to kinetic energy of the fission fragments
✓ 1
which is then transferred as thermal energy
✓ 1
Part (b)
Number per second = power ÷ energy per fission = 1.6×108 ÷ 3.2×10−11
✓ 1
= 5.0×1018 per second
✓ 1
Part (c)
1 h = 3600 s: 5.0×1018 × 3600
✓ 1
= 1.8×1022
✓ 1
Exam tip: Check powers of ten: divide by 10⁻¹¹ adds 11 to the power.
248IGPC3-5-05r10-06Radioactivity, half-life and safetyPaper EMedium1 mark
Smoke detectors often contain americium-241, 24195Am, which decays by emitting an alpha particle. Which nuclide is formed?
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Notes
An alpha particle is 42He. The nucleon number falls by 4 (241 to 237) and the proton number falls by 2 (95 to 93), giving 23793Np (D). A keeps the nucleon number unchanged. B has the proton number rising. C lowers only the nucleon number.
✓ 1
Answer D
Answer: D
Common error: In alpha decay both numbers change: nucleon number down 4, proton number down 2.
249IGPC3-5-05r10-13Radioactivity, half-life and safetyPaper EMedium5 marks
Radon-222 is a radioactive gas that can collect in the basements of buildings. It is formed when radium-226, 22688Ra, emits an alpha particle.
(a)
Determine the nucleon number and the proton number of the nucleus formed from radium-226.
(2)
(b)
Explain why radon gas that has been breathed in is considered more hazardous than an alpha source held outside the body.
(2)
(c)
Calculate the number of neutrons in a nucleus of radon-222.
(1)
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Notes
Part (a)
nucleon number = 226 - 4 = 222
✓ 1
proton number = 88 - 2 = 86
✓ 1
Part (b)
alpha particles are strongly ionising
✓ 1
outside the body they are absorbed by air / skin, but inside the lungs they irradiate living tissue directly
✓ 1
Part (c)
222 - 86 = 136
✓ 1
Exam tip: Alpha emission lowers the nucleon number by 4 and the proton number by 2.
250IGPC3-5-06r10-05Radioactivity, half-life and safetyPaper EMedium1 mark
Polonium-210, 21084Po, decays by emitting an alpha particle. Which nuclide is formed?
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Notes
Alpha particle = 42He: nucleon number 210 - 4 = 206, proton number 84 - 2 = 82, giving 20682Pb (D). A has its proton number raised by 1 (beta decay). B only changes the nucleon number. C has numbers that increase.
✓ 1
Answer D
Answer: D
Exam tip: Subtract 4 from the nucleon number and 2 from the proton number for alpha decay.
251IGPC3-5-06r10-13Radioactivity, half-life and safetyPaper EMedium5 marks
Carbon-14 is used to date archaeological finds. A nucleus of carbon-14, 146C, decays by beta-minus emission to a nucleus of nitrogen.
(a)
Determine the proton number of the nitrogen nucleus formed.
(2)
(b)
Explain, in terms of a change inside the nucleus, why the proton number increases.
(2)
(c)
Calculate the number of neutrons in the nitrogen nucleus.
(1)
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Notes
Part (a)
nucleon number stays at 14
✓ 1
proton number = 6 + 1 = 7
✓ 1
Part (b)
a neutron changes into a proton
✓ 1
and an electron (the beta particle) is emitted
✓ 1
Part (c)
14 - 7 = 7
✓ 1
Exam tip: In beta-minus decay, the nucleon number is unchanged and the proton number rises by 1.
252IGPC3-5-06r11-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Smoke detectors in a building contain americium-241, which has 95 protons and decays by emitting an alpha particle. What are the proton number and nucleon number of the nucleus formed?
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Notes
An alpha particle has 2 protons and 2 neutrons, so the proton number falls by 2 to 93 and the nucleon number falls by 4 to 237. A shows the change for beta-minus decay. B forgets that the nucleon number falls. D adds the alpha particle instead of removing it.
✓ 1
Answer C
Answer: C
Exam tip: Alpha decay lowers the proton number by 2 and the nucleon number by 4.
253IGPC3-5-06r11-08Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Energy from the Sun supports photosynthesis in marine algae. The Sun's energy comes from nuclear fusion. Why must the temperature be extremely high for fusion to occur?
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Notes
Nuclei are positively charged and repel one another, so they need very high speeds, and hence temperatures, to get close enough to join. A describes fission in a reactor. B confuses fusion with splitting nuclei. C is wrong because the nuclei do not need to be radioactive.
✓ 1
Answer D
Answer: D
Common error: Fusion joins small nuclei and needs high temperature; fission splits large nuclei and uses neutrons.
254IGPC3-5-06r11-14Radioactivity, half-life and safetyPaper EMedium7 marks
Smoke detectors in the atrium of a building contain americium-241, which has proton number 95. It decays by emitting an alpha particle. A different radioactive isotope in the building, bismuth-214, has proton number 83 and decays by beta-minus emission.
(a)
State the composition of an alpha particle.
(1)
(b)
Determine the proton number and nucleon number of the nucleus formed when americium-241 decays.
(2)
(c)
Determine the proton number and nucleon number of the nucleus formed when bismuth-214 decays.
(2)
(d)
Explain how a beta particle is produced in a nucleus.
(2)
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Notes
Part (a)
2 protons and 2 neutrons
✓ 1
Part (b)
proton number 95 − 2 = 93
✓ 1
nucleon number 241 − 4 = 237
✓ 1
Part (c)
proton number 83 + 1 = 84
✓ 1
nucleon number unchanged at 214
✓ 1
Part (d)
a neutron changes into a proton
✓ 1
an electron (the beta particle) is emitted from the nucleus
✓ 1
Common error: Beta-minus decay changes the proton number by +1 but leaves the nucleon number unchanged.
255IGPC3-5-07r19-07Radioactivity, half-life and safetyPaper EMedium1 mark
Radioactive phosphorus-32, 3215P, is used as a tracer to study how plants take up fertiliser. It decays by beta-minus emission. Which nuclide is formed?
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Notes
B is correct: in beta-minus decay a neutron becomes a proton, so the nucleon number stays 32 and the proton number rises from 15 to 16, giving sulfur. A has the proton number decreasing. C is the result of alpha emission, where the nucleon number falls by 4 and proton number by 2. D has lost a neutron, which does not happen in beta decay.
✓ 1
Answer B
Answer: B
Exam tip: In beta-minus decay the nucleon number is unchanged and the proton number increases by one.
256IGPC3-5-07r19-13Radioactivity, half-life and safetyPaper EMedium6 marks
Polonium-210 is an alpha emitter. Its decay can be written 21084Po → AZPb + 42He. Lead-214, 21482Pb, is a beta-minus emitter.
(a)
Determine the values of A and Z for the lead nucleus formed by the decay of polonium-210.
(2)
(b)
Describe what happens inside a nucleus during beta-minus decay.
(1)
(c)
State the nucleon number and the proton number of the nucleus formed when lead-214 decays by beta-minus emission.
(2)
(d)
State the change in nucleon number and proton number when a nucleus emits a gamma ray.
(1)
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Notes
Part (a)
A = 210 − 4 = 206
✓ 1
Z = 84 − 2 = 82
✓ 1
Part (b)
a neutron changes into a proton (and an electron is emitted)
✓ 1
Part (c)
nucleon number 214
✓ 1
proton number 82 + 1 = 83
✓ 1
Part (d)
no change in either number
✓ 1
Common error: In beta-minus decay the proton number rises by one; students often reduce it by mistake.
257IGPC3-5-07r21-05Radioactivity, half-life and safetyPaper EMedium1 mark
A marine biologist tracks a radioactive tracer in plankton. The count rate of a sample, already corrected for background, is 640 counts per minute at the start and 80 counts per minute after 3.0 hours. What is the half-life of the tracer?
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Notes
640 → 320 → 160 → 80 is three halvings in 3.0 h, so the half-life is 3.0 ÷ 3 = 1.0 h; A is correct. B divides 3.0 h by the ratio 8 instead of the number of halvings. C assumes only two halvings. D is the total time.
✓ 1
Answer A
Answer: A
Exam tip: Count how many times the quantity halves, then divide the total time by that number.
258IGPC3-5-07r21-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Americium-241, written 24195Am, decays by emitting an alpha particle. What are the nucleon number and the proton number of the nucleus formed?
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Notes
An alpha particle is a helium nucleus with 4 nucleons and 2 protons, so the nucleon number falls by 4 to 237 and the proton number falls by 2 to 93; D is correct. A describes beta decay. B adds 2 to the proton number instead of subtracting it. C leaves the nucleon number unchanged.
✓ 1
Answer D
Answer: D
Common error: In alpha decay both numbers decrease: nucleon number by 4 and proton number by 2.
259IGPC3-5-07r21-13Atomic structure, isotopes, fission and fusionPaper EMedium6 marks
A research submarine is powered by a nuclear reactor that uses uranium-235.
(a)
Describe what happens to a uranium-235 nucleus when it absorbs a neutron in the reactor.
(2)
(b)
Explain how a chain reaction can occur in the reactor.
(2)
(c)
State one way in which nuclear fusion differs from nuclear fission.
(1)
(d)
State where nuclear fusion occurs naturally.
(1)
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Notes
Part (a)
the nucleus splits into two smaller nuclei
✓ 1
releasing energy and further neutrons
✓ 1
Part (b)
neutrons released are absorbed by other uranium-235 nuclei
✓ 1
which then split and release more neutrons, so the process repeats and sustains itself
✓ 1
Part (c)
fusion joins light nuclei to form a heavier nucleus, whereas fission splits a heavy nucleus
✓ 1
Part (d)
in the Sun or other stars
✓ 1
Common error: Fission splits heavy nuclei; fusion joins light nuclei. Name the process correctly in every answer.
260IGPC3-5-07r22-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Granite in a mountain hut contains radium-226, which has 88 protons and 138 neutrons. It decays by emitting an alpha particle. How many protons and neutrons are in the nucleus formed?
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Notes
An alpha particle is 2 protons and 2 neutrons, so the new nucleus has 88 − 2 = 86 protons and 138 − 2 = 136 neutrons. 88 protons and 136 neutrons removes only neutrons. 86 protons and 140 neutrons gains neutrons. 84 protons and 138 neutrons removes 4 protons instead of 2.
✓ 1
Answer C
Answer: C
Common error: An alpha particle carries away both 2 protons and 2 neutrons; check that both numbers fall.
261IGPC3-5-07r22-07Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Caesium-137 has 55 protons and nucleon number 137. A sample found in mushrooms decays by beta-minus emission. What are the proton number and nucleon number of the nucleus formed?
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Notes
In beta-minus decay a neutron changes into a proton, so the proton number rises by 1 to 56 while the nucleon number stays at 137. 54 and 137 would be a decrease in proton number, which does not occur in beta-minus decay. 55 and 136 would lose a nucleon. 53 and 133 is what happens in alpha decay.
✓ 1
Answer D
Answer: D
Exam tip: In beta-minus decay the nucleon number is unchanged and the proton number increases by one.
262IGPC3-5-07r22-12Radioactivity, half-life and safetyPaper EMedium6 marks
Radium-226 occurs naturally in granite. In a poorly ventilated mountain hut built from granite, radon gas can build up. Radium-226 (proton number 88) decays by alpha emission to radon: 22688Ra → AZRn + 42He
(a)
Determine the values of A and Z.
(2)
(b)
State the composition of an alpha particle.
(1)
(c)
Radon-222 is also an alpha emitter. Explain why breathing in radon gas is hazardous.
(2)
(d)
Calculate the number of neutrons in a nucleus of radon-222.
(1)
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Notes
Part (a)
A = 222
✓ 1
Z = 86
✓ 1
Part (b)
2 protons and 2 neutrons
✓ 1
Part (c)
alpha particles are strongly ionising
✓ 1
inside the lungs there is no skin or air to absorb them, so they damage nearby cells
✓ 1
Part (d)
222 − 86 = 136
✓ 1
ECF from (a)
—
Common error: In a nuclear equation the nucleon numbers must add to the same total on both sides, and so must the proton numbers.
263IGPC3-5-09r27-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Cobalt-60, 6027Co, decays by emitting a β− particle. Which nuclide is formed?
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Notes
In β− decay a neutron changes into a proton, so the proton number rises by 1 to 28 while the nucleon number stays at 60. The nuclide with proton number 26 has the proton number going down instead of up. The nuclide with nucleon number 59 has lost a neutron but kept the proton number, which is not what happens. The last option shows what an alpha decay would give (nucleon number down 4, proton number down 2).
✓ 1
Answer C
Answer: C
Common error: In β⁻ decay the nucleon number does not change; only the proton number goes up by 1.
264IGPC3-5-09r27-06Radioactivity, half-life and safetyPaper EMedium1 mark
A nucleus of uranium-238, 23892U, emits one α particle and then two β− particles. What is the nuclide that results?
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Notes
The α particle takes away 4 nucleons and 2 protons, giving 23490Th. Each β− decay leaves the nucleon number alone and adds 1 to the proton number, so two of them give proton number 92 and nucleon number 234. The radium option subtracts 1 for each beta particle instead of adding 1. The thorium-230 option wrongly reduces the nucleon number during beta emission and does not change the proton number. The plutonium option ignores the alpha emission altogether.
✓ 1
Answer A
Answer: A
Exam tip: Deal with each emission one at a time: alpha is (−4, −2), beta-minus is (0, +1) for (nucleon, proton) number.
265IGPC3-5-09r27-14Radioactivity, half-life and safetyPaper EMedium6 marks
An old film projector in a theatre museum has an anti-static brush containing polonium-210, an alpha emitter. In a laboratory test, a source emitting alpha, beta and gamma radiation sends a narrow beam into the gap between two parallel plates in a vacuum. One plate is positively charged and the other is negatively charged.
Diagram not to scale
(a)
State towards which plate the beta particles are deflected.
(1)
(b)
Explain why the alpha particles are deflected much less than the beta particles, even though an alpha particle has a larger charge.
(2)
(c)
State what happens to the gamma rays.
(1)
(d)
Polonium-210, 21084Po, decays to an isotope of lead by emitting an alpha particle. Complete the nuclear equation: 21084Po → lead nuclide + alpha particle.
(2)
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Notes
Part (a)
the positive plate
✓ 1
Part (b)
an alpha particle has a much greater mass than a beta particle
✓ 1
so the acceleration (and hence deflection) of the alpha particle is much smaller
✓ 1
Part (c)
they pass straight through, undeflected, because they are uncharged
✓ 1
Part (d)
lead nuclide is 20682Pb
✓ 1
alpha particle is 42He
✓ 1
Exam tip: In a nuclear equation the nucleon numbers must add up to the same total on each side, and so must the proton numbers.
266IGPC3-5-09r28-04Radioactivity, half-life and safetyPaper EMedium1 mark
A radioactive tracer is added to the water in an underground pipe to locate a leak. Its measured activity is 800 counts per minute at the start and 100 counts per minute after 9.0 days (background has already been subtracted). What is the half-life of the tracer?
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Notes
The activity falls from 800 to 100, which is a factor of 8, so three half-lives have passed. The half-life is 9.0 / 3 = 3.0 days. 1.1 days comes from dividing 9.0 days by the factor of 8 instead of by the number of halvings. 2.3 days comes from counting four halvings. 4.5 days comes from assuming the activity only halved once.
✓ 1
Answer C
Answer: C
Common error: Count how many times the activity halves (800 to 400 to 200 to 100), not the overall factor.
267IGPC3-5-09r28-12Radioactivity, half-life and safetyPaper EMedium6 marks
Radon-222 is a radioactive gas that can dissolve in groundwater. It forms when radium-226, 22688Ra, decays by alpha emission. An alpha particle is a helium nucleus, 42He.
(a)
Complete the nuclide equation for the decay of radium-226: 22688Ra → AZRn + 42He. State the values of A and Z.
(2)
(b)
Radon-222 itself decays by alpha emission to form an isotope of polonium. State the nucleon number of this polonium nucleus and its proton number.
(2)
(c)
Explain why an alpha emitter is more hazardous inside the body than outside it.
(2)
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Part (a)
A1 A = 222
✓ 1
A1 Z = 86
✓ 1
Part (b)
A1 nucleon number 218
✓ 1
A1 proton number 84
✓ 1
Part (c)
alpha particles are strongly ionising
✓ 1
outside, they are absorbed by air or skin, but inside all their energy is absorbed by living cells, damaging DNA and possibly causing cancer
✓ 1
Exam tip: In a nuclide equation, the top numbers must add up to the same total on both sides, and so must the bottom numbers.
268IGPC3-5-08r29-07Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Which statement about nuclear fusion in the Sun is correct?
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In fusion, light nuclei such as hydrogen join to form a heavier nucleus and release energy; very high temperature is needed because the positively charged nuclei repel each other. Splitting a heavy nucleus is fission, nuclei repel rather than attract at long range, and the Sun contains no oxygen-based chemical burning.
✓ 1
Answer B
Answer: B
Common error: Fission splits heavy nuclei; fusion joins light nuclei. Do not swap them.
269IGPC3-5-08r29-14Atomic structure, isotopes, fission and fusionPaper EMedium5 marks
A solar-powered aircraft flies using electrical energy from sunlight. The energy released by the Sun comes from nuclear fusion.
(a)
Describe the process of nuclear fusion.
(2)
(b)
Explain why fusion needs a very high temperature.
(1)
(c)
In a fission reactor on the ground, a chain reaction is sustained. Explain how.
(2)
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Part (a)
Two light nuclei join together
✓ 1
to form a heavier nucleus, releasing energy
✓ 1
Part (b)
Nuclei are positively charged and repel, so they need high kinetic energy to get close enough to fuse
✓ 1
Part (c)
Each fission of uranium-235 releases neutrons
✓ 1
These neutrons are absorbed by other uranium nuclei and cause further fissions
✓ 1
Common error: Fusion joins light nuclei; fission splits heavy nuclei. State which nuclei are involved in each.
270IGPC3-5-09r29-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
A hospital stores a sealed source of caesium-137, 13755Cs, which decays by beta-minus emission. Which nuclide is formed?
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Notes
In beta-minus decay a neutron changes into a proton, so the proton number rises by 1 to 56 while the nucleon number stays at 137, giving barium-137. Proton number 54 would mean it fell. Nucleon number 136 would mean the nucleus lost a nucleon, which does not happen in beta decay. Iodine-133 is the nuclide that alpha emission would produce (loss of 4 nucleons and 2 protons), not beta-minus emission.
✓ 1
Answer C
Answer: C
Common error: In beta-minus decay the nucleon number stays the same and the proton number goes up by one.
271IGPC3-5-09r29-13Radioactivity, half-life and safetyPaper EMedium6 marks
Americium-241 is used in smoke alarms. A nucleus of americium-241 has 95 protons and 241 nucleons and decays by alpha emission to neptunium. Iodine-131 (53 protons, 131 nucleons) is used in a hospital to treat the thyroid gland and decays by beta-minus emission.
(a)
Determine the nucleon number and the proton number of the neptunium nucleus formed.
(2)
(b)
Determine the proton number of the nucleus formed when iodine-131 decays, and state its nucleon number.
(2)
(c)
Explain why the proton number increases in beta-minus decay.
(2)
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Part (a)
nucleon number = 241 − 4 = 237
✓ 1
proton number = 95 − 2 = 93
✓ 1
Part (b)
proton number = 53 + 1 = 54
✓ 1
nucleon number = 131
✓ 1
Part (c)
a neutron in the nucleus changes into a proton
✓ 1
an electron (beta particle) is emitted / the nucleon number is unchanged because a neutron is replaced by a proton
✓ 1
Exam tip: In a decay equation, both the nucleon numbers and the proton numbers must balance on each side.
272IGPC3-5-08r30-03Radioactivity, half-life and safetyPaper EMedium1 mark
A volcanologist measures a sample of radioactive gas. After correcting for background, the count rate falls from 640 counts/min to 80 counts/min in 24 days. What is the half-life of the isotope?
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640 → 320 → 160 → 80 is three halvings, so 3 half-lives take 24 days and the half-life is 24 / 3 = 8.0 days. 3.0 days divides the time by the ratio of count rates (8) instead of by the number of halvings. 12 days assumes only two half-lives. 6.0 days assumes four half-lives.
✓ 1
Answer B
Answer: B
Exam tip: Count the number of halvings, not the overall ratio, then divide the total time by that number.
273IGPC3-5-08r30-14Atomic structure, isotopes, fission and fusionPaper EMedium6 marks
Some icebreaker ships are powered by nuclear fission reactors using uranium-235 (23592U).
(a)
Describe how a chain reaction occurs in the reactor.
(3)
(b)
State the purpose of the control rods in the reactor.
(1)
(c)
In the Sun, hydrogen nuclei undergo fusion. Explain why fusion requires a very high temperature.
(2)
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Notes
Part (a)
a neutron is absorbed by a uranium-235 nucleus, which becomes unstable and splits into two smaller nuclei
✓ 1
energy is released together with two or three more neutrons
✓ 1
these neutrons are absorbed by other uranium-235 nuclei, causing further fission
✓ 1
Part (b)
they absorb neutrons to control the rate of fission / keep the chain reaction steady
✓ 1
Part (c)
nuclei are positively charged, so they repel each other
✓ 1
at high temperature they have enough kinetic energy to overcome the repulsion and get close enough to fuse
✓ 1
Common error: Fission is the splitting of a large nucleus and fusion is the joining of small nuclei; do not swap them.
274IGPC3-5-07r38-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Steel beams in a building are checked using gamma radiography. The source is cobalt-60, whose nucleus has 27 protons and a nucleon number of 60, and it decays by beta emission. What does the daughter nucleus contain?
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Notes
In beta decay a neutron changes into a proton and an electron, so the proton number rises by 1 to 28 while the nucleon number stays at 60. A fall to 26 protons has the direction of change wrong, and a nucleon number of 59 would mean a nucleon had been lost, which does not happen in beta decay.
✓ 1
Answer A
Answer: A
Exam tip: In a beta-minus decay the nucleon number is unchanged and the proton number goes up by one.
275IGPC3-5-07r38-12Atomic structure, isotopes, fission and fusionPaper EMedium5 marks
Caesium-137, with proton number 55 and nucleon number 137, is a beta emitter found in marine sediment samples. It decays to an isotope of barium.
(a)
State what a beta particle is.
(1)
(b)
The decay can be written 13755Cs → AZBa + 0−1e. Determine the proton number Z and the nucleon number A of the barium nucleus.
(2)
(c)
A caesium nucleus contains no electrons. Explain how it can emit a beta particle.
(2)
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Part (a)
B1 a fast-moving electron emitted from the nucleus
✓ 1
Part (b)
B1 Z = 56 (since 55 = Z − 1)
✓ 1
B1 A = 137
✓ 1
Part (c)
B1 a neutron in the nucleus changes into a proton
✓ 1
B1 an electron (the beta particle) is created and emitted at the same time
✓ 1
Exam tip: Check a nuclear equation by confirming that the nucleon numbers and the proton numbers add up to the same totals on both sides.
276IGPC3-5-08r39-04Radioactivity, half-life and safetyPaper EMedium1 mark
A nucleus in volcanic rock decays by beta-minus (β−) emission. Which row describes the change to the nucleus?
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Notes
In β− decay a neutron changes into a proton and an electron, and the electron is emitted. The total number of nucleons stays the same and the proton number rises by 1. The first alternative describes alpha decay. Decreasing the proton number by 1 wrongly assumes losing an electron removes a positive charge. The last alternative has a nucleon lost with no change in protons, which does not happen in any of the common decays.
✓ 1
Answer D
Answer: D
Exam tip: Beta-minus decay changes a neutron into a proton, so A stays the same and Z goes up by one.
277IGPC3-5-08r39-05Radioactivity, half-life and safetyPaper EMedium1 mark
Radium-226 (proton number 88) is found in some volcanic spring water. A nucleus of radium-226 emits one alpha particle and then one beta-minus particle. What are the nucleon number and proton number of the nucleus formed?
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Notes
The alpha particle removes 4 nucleons and 2 protons, giving 222 and 86. The beta-minus decay then raises the proton number by 1 with no change in nucleon number, giving 222 and 87. The option with 85 makes the beta decay lower the proton number. The option with 226 and 89 shows only the beta decay. The option with 230 and 91 adds the alpha particle instead of removing it.
✓ 1
Answer C
Answer: C
Common error: Alpha decay lowers A by 4 and Z by 2; beta-minus raises Z by 1; do the two changes one after the other.
278IGPC3-5-08r39-08Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
Why does nuclear fusion in the Sun require an extremely high temperature?
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Notes
Light nuclei repel each other because they are positively charged. They must have a very high speed, which means a very high temperature, to get close enough to join. Splitting large nuclei describes fission. A neutron chain reaction is also a feature of fission. Hydrogen in the Sun is already a plasma, so a change of state is not the reason.
✓ 1
Answer B
Answer: B
Exam tip: Link fusion to the charge on the nuclei: high temperature gives the high speeds needed to overcome repulsion.
279IGPC3-5-08r39-12Atomic structure, isotopes, fission and fusionPaper EMedium6 marks
A nuclear-powered ship uses fission of uranium-235 to heat water for its turbines.
(a)
Describe what happens in a fission reaction of uranium-235.
(3)
(b)
Explain the purpose of the control rods in the reactor.
(2)
(c)
State why the reactor is surrounded by thick concrete and steel.
(1)
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Part (a)
A uranium-235 nucleus absorbs a neutron
✓ 1
The nucleus splits into two smaller nuclei
✓ 1
Further neutrons and energy are released
✓ 1
Part (b)
The rods absorb neutrons
✓ 1
This keeps the chain reaction steady, with on average one neutron from each fission causing another fission
✓ 1
Part (c)
To absorb gamma rays and neutrons so the crew are not exposed to radiation
✓ 1
Exam tip: For a chain reaction, each fission must release neutrons that cause further fissions.
280IGPC3-5-07r43-04Radioactivity, half-life and safetyPaper EMedium1 mark
A nucleus of thorium-234 (nucleon number 234, proton number 90) decays by beta-minus emission. Which row describes the nucleus formed?
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Notes
In beta-minus decay a neutron changes into a proton and an electron is emitted. The nucleon number stays the same at 234 and the proton number rises by one to 91. A proton number of 89 has the change the wrong way round. Nucleon number 230 with proton number 88 is what alpha decay would give. Nucleon number 233 with proton number 90 would mean a neutron had simply been lost, which does not happen in beta decay.
✓ 1
Answer C
Answer: C
Exam tip: For beta-minus decay the nucleon number is unchanged and the proton number goes up by 1.
281IGPC3-5-07r43-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
An americium-241 nucleus (nucleon number 241, proton number 95) in a shop's smoke detector decays by emitting an alpha particle. How many neutrons are in the nucleus that is formed?
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Notes
An alpha particle carries away 2 protons and 2 neutrons, so the new nucleus has nucleon number 237 and proton number 93. Neutrons = 237 − 93 = 144. The value 146 is the neutron number of the original nucleus. The value 142 comes from using 237 − 95 and forgetting the proton number also falls. The value 148 comes from adding the two neutrons instead of removing them.
✓ 1
Answer B
Answer: B
Exam tip: Write the new nucleon number and proton number first, then subtract to find neutrons.
282IGPC3-5-07r43-12Radioactivity, half-life and safetyPaper EMedium6 marks
A packaging company that supplies shops monitors the thickness of plastic film using a strontium-90 source. A nucleus of strontium-90 (nucleon number 90, proton number 38) decays by beta-minus emission.
Diagram not to scale
(a)
Describe what happens inside the nucleus when a beta-minus particle is emitted.
(2)
(b)
Explain why beta radiation is suitable for monitoring the thickness of thin plastic film, but alpha and gamma radiation are not.
(2)
(c)
State the nucleon number and the proton number of the nucleus formed.
(2)
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Part (a)
a neutron changes into a proton
✓ 1
an electron (the beta particle) is emitted at high speed
✓ 1
Part (b)
beta is partly absorbed by the film, so the count rate at the detector changes with thickness
✓ 1
alpha is completely stopped by the film OR gamma passes through almost unchanged, so no useful change is detected
✓ 1
Part (c)
A1 nucleon number 90
✓ 1
A1 proton number 39
✓ 1
Exam tip: Beta-minus decay raises the proton number by 1 and leaves the nucleon number unchanged.
283IGPC3-5-08r48-05Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
A nucleus of plutonium-238 (nucleon number 238, proton number 94) in a space probe's power unit decays by emitting an alpha particle. What are the nucleon number and the proton number of the nucleus that is formed?
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Notes
An alpha particle is a helium nucleus with 4 nucleons and 2 protons, so the nucleon number falls by 4 to 234 and the proton number falls by 2 to 92. A proton number of 90 comes from subtracting 4 from both numbers. Nucleon 238 and proton 95 is what happens in beta decay, not alpha decay. Nucleon 236 comes from subtracting only 2 from the nucleon number.
✓ 1
Answer B
Answer: B
Common error: In alpha decay the nucleon number falls by 4 but the proton number falls by only 2.
284IGPC3-5-08r48-06Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
A cobalt-60 nucleus (nucleon number 60, proton number 27) in a bridge radiography source decays by emitting a beta particle (an electron). What are the nucleon number and proton number of the nucleus formed?
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Notes
In beta decay a neutron changes into a proton and an electron, so the nucleon number stays at 60 and the proton number rises by 1 to 28. Nucleon 56 and proton 25 is the result of alpha decay. Proton number 26 has the proton number falling instead of rising. Nucleon 59 with proton 27 treats the beta particle as removing a nucleon, which it does not.
✓ 1
Answer D
Answer: D
Exam tip: Beta decay leaves the nucleon number unchanged and raises the proton number by one.
285IGPC3-5-08r48-07Atomic structure, isotopes, fission and fusionPaper EMedium1 mark
A space probe observes the Sun, where hydrogen nuclei fuse to form helium nuclei. Which statement explains why fusion needs a very high temperature?
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Notes
Hydrogen nuclei are all positively charged and repel each other, so they must travel very fast (a high temperature) to get close enough for the strong nuclear force to join them. Splitting nuclei is fission, not fusion. Neutrons do not need to be separated from protons for fusion to happen. At these temperatures the matter is a plasma and is not liquefied.
✓ 1
Answer A
Answer: A
Common error: Fusion joins small nuclei together; do not confuse it with fission, in which a large nucleus splits.
286IGPC3-5-08r48-12Radioactivity, half-life and safetyPaper EMedium6 marks
The power unit of a deep-space probe contains plutonium-238, which decays by emitting alpha particles.
(a)
Complete the nuclear equation for the decay: plutonium-238 (nucleon number 238, proton number 94) → X + alpha particle. State the nucleon number and the proton number of the nucleus X.
(2)
(b)
Explain why an alpha emitter sealed in a metal case is suitable as a heat source.
(2)
(c)
The power output of the unit is 300 W when launched and is proportional to the activity of the plutonium. The half-life is 88 years. Calculate the power output after 264 years.
(2)
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Part (a)
nucleon number of X = 234
✓ 1
proton number of X = 92
✓ 1
Part (b)
alpha particles are stopped (absorbed) within a very short distance, so they cannot escape the case
✓ 1
their kinetic energy is transferred to thermal energy in the material, which heats the unit
✓ 1
Part (c)
264 years is 3 half-lives
✓ 1
300 → 150 → 75 → 37.5 W
✓ 1
Exam tip: In a nuclear equation, the nucleon numbers must add up to the same total on both sides, and so must the proton numbers.
287IGPC-039-00-09Atomic structure, isotopes, fission and fusionPaper EHard1 mark
Three nuclei are described below. Nucleus P: proton number 17, nucleon number 35 Nucleus Q: 17 protons and 20 neutrons Nucleus R: 18 protons and 17 neutrons Which statement is correct?
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Notes
P has 17 protons, and Q also has 17 protons, so they are the same element. Q has nucleon number 37 and P has 35, so P and Q are isotopes with the same relative charge (+17) but different relative masses. R has 18 protons, so it is a different element. R has the same nucleon number as P (35), but a different relative charge (+18).
✓ 1
Answer A
Answer: A
Exam tip: Work out the proton number and nucleon number of every nucleus before comparing them. Isotopes share the proton number, not the nucleon number.
288IGPC-039-00-16Atomic structure, isotopes, fission and fusionPaper EHard7 marks
An iron nucleus is represented as 5626Fe. The charge on a proton is 1.6×10−19 C. The mass of a single proton or neutron is about 1.67×10−27 kg.
(a)
State the number of neutrons in this nucleus.
(1)
(b)
Calculate the charge on the nucleus.
(2)
(c)
Calculate the approximate mass of the nucleus.
(2)
(d)
Another isotope of iron is 5426Fe. State how the charge and the mass of its nucleus compare with those of 5626Fe.
(2)
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Notes
Part (a)
56 − 26 = 30
✓ 1
Part (b)
charge = 26 × 1.6×10−19 ; = 4.2×10−18 C
✓ 2
Part (c)
mass = 56 × 1.67×10−27 ; = 9.4×10−26 kg
✓ 2
Part (d)
the charge is the same, because it has the same number of protons (26) ; the mass is smaller, because it has 2 fewer neutrons
✓ 2
Common error: Using the nucleon number to work out the charge. Only protons are charged, so multiply the proton number by the proton charge.
289IGPC-039-01-10Radioactivity, half-life and safetyPaper EHard1 mark
A detector near a radioactive sample reads 430 counts/min. Twenty-four minutes later it reads 80 counts/min. The background count rate throughout is 30 counts/min. What is the half-life of the sample?
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Notes
Subtract the background from both readings. The corrected count rates are 430 − 30 = 400 counts/min and 80 − 30 = 50 counts/min. 400 → 200 → 100 → 50 is 3 half-lives. These take 24 min, so the half-life is 24 ÷ 3 = 8 min. Using the uncorrected readings (430 → 80) gives about 2.4 halvings and a half-life of about 10 min. 12 min assumes only two half-lives. 6 min comes from miscounting four halvings.
✓ 1
Answer B
Answer: B
Common error: Subtract the background from every reading before counting halvings. Uncorrected readings never halve cleanly.
290IGPC-039-01-15Radioactivity, half-life and safetyPaper EHard8 marks
A student measures the count rate near a radioactive sample every 2.0 minutes. The background count rate in the room is 40 counts/min. Her measured count rates are: t = 0 min: 1240 counts/min t = 2.0 min: 796 counts/min t = 4.0 min: 516 counts/min t = 6.0 min: 340 counts/min t = 8.0 min: 229 counts/min
Measured count rate (background not subtracted)
(a)
Explain why the readings must be corrected before the half-life is found.
(1)
(b)
Calculate the corrected count rates at t = 0 and at t = 6.0 min.
(2)
(c)
Use your corrected count rates at t = 0 and t = 6.0 min to determine the half-life of the sample.
(2)
(d)
Predict the reading on the detector at t = 12.0 min.
(3)
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Part (a)
The detector also records background radiation, which does not come from the sample (and does not decay), so it must be subtracted to give the count rate from the sample alone .
✓ 1
Part (b)
1240 − 40 = 1200 counts/min
✓ 1
340 − 40 = 300 counts/min
✓ 1
Part (c)
1200 → 600 → 300 is 2 half-lives
✓ 1
half-life = 6.0 ÷ 2 = 3.0 min
✓ 1
Part (d)
12.0 min is 4 half-lives
✓ 1
corrected count rate = 1200 ÷ 16 = 75 counts/min
✓ 1
reading = 75 + 40 = 115 counts/min
✓ 1
Exam tip: If the question asks what the detector reads, add the background back on after applying the half-life.
291IGPC-039-00-15Atomic structure, isotopes, fission and fusionPaper EHard6 marks
In some fusion reactions, deuterium nuclei join together: 21H + 21H → 32He + X
(a)
Identify particle X. Give its nucleon number and proton number.
(2)
(b)
Fusion happens only at extremely high temperatures, such as inside stars. Explain why.
(3)
(c)
State one difference between fusion and fission, other than the temperature needed.
(1)
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Notes
Part (a)
a neutron ; nucleon number 1 and proton number 0
✓ 2
Part (b)
the nuclei are both positively charged, so there is electrostatic repulsion between them ; at high temperature the nuclei have very high kinetic energy / speed ; this lets them overcome the repulsion and get close enough to join / fuse, held by the strong nuclear force
✓ 3
Part (c)
any one valid difference, e.g. fusion joins light nuclei whereas fission splits a heavy nucleus / fission is started by a nucleus absorbing a neutron but fusion is not / fission products are highly radioactive but fusion products are mainly not / fusion releases more energy per unit mass of fuel
✓ 1
Exam tip: For 'explain' questions about fusion, name the force involved: electrostatic repulsion between positive nuclei.
292IGPC3-5-05c-06Radioactivity, half-life and safetyPaper EHard1 mark
In a laboratory a Geiger–Müller counter records a background count rate of 20 counts/min. A source of half-life 30 minutes is placed beside it and the counter reads 340 counts/min. What count rate will the counter read 90 minutes later with everything in the same position?
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Notes
Count rate from the source alone = 340 − 20 = 320 counts/min. 90 min is 3 half-lives, so 320 ÷ 8 = 40 counts/min. The counter still detects the background, so it reads 40 + 20 = 60 counts/min: C is correct. A forgets to add the background back. B divides the total reading (including background) by 8. D uses only 2 half-lives: 320 ÷ 4 + 20 = 100.
✓ 1
Answer C
Answer: C
Common error: Background radiation does not decay with the source: subtract it before halving and add it back for the reading.
293IGPC3-5-05c-15Radioactivity, half-life and safetyPaper EHard7 marks
A paper mill measures the thickness of paper as it is made, using a radioactive source on one side and a detector on the other. The source has a half-life of 30 years.
Diagram not to scale
(a)
Explain why a beta-emitting source is chosen rather than an alpha-emitting or gamma-emitting source. [3]
(3)
(b)
Suggest why a source with a long half-life is suitable. [1]
(1)
(c)
When new, the detector reads 840 counts/s with no paper present. Ignoring background radiation, estimate the count rate 45 years later. [3]
(3)
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Part (a)
alpha particles would be completely absorbed by the paper, so nothing would be detected
✓ 1
gamma rays pass through almost unchanged, so the count rate would hardly change with thickness
✓ 1
beta particles are partly absorbed, so the count rate changes noticeably with thickness
✓ 1
Part (b)
the activity stays almost constant for years, so the readings stay reliable and the source rarely needs replacing
✓ 1
Part (c)
45 ÷ 30 = 1.5 half-lives
✓ 1
remaining fraction = (1/2)1.5 = 0.354
✓ 1
840 × 0.354 = 297 counts/s
✓ 1
Common error: A thickness gauge needs radiation that is partly, not completely or hardly, absorbed.
294IGPC3-5-05r5-14Radioactivity, half-life and safetyPaper EHard6 marks
A student has three radioactive sources P, Q and R. The count rate is measured with each source with different absorbers placed between the source and the detector. The background count rate is 21 counts/min. Source P: no absorber 412; paper 410; 3 mm aluminium 408; 10 cm lead 24 Source Q: no absorber 530; paper 524; 3 mm aluminium 23; 10 cm lead 21 Source R: no absorber 800; paper 24; 3 mm aluminium 21; 10 cm lead 21 (all values in counts/min)
Diagram not to scale
(a)
Identify the radiation emitted by each of P, Q and R. Give a reason for each.
(3)
(b)
The radiation from Q passes between two charged metal plates, the upper plate positive and the lower plate negative. State the direction in which the radiation is deflected and explain why.
(2)
(c)
Calculate the corrected count rate from Q with no absorber.
(1)
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Notes
Part (a)
P is gamma, since only thick lead reduces the count rate to near background
✓ 1
Q is beta, since it passes through paper but is stopped by 3 mm aluminium
✓ 1
R is alpha, since it is stopped by paper
✓ 1
Part (b)
towards the upper (positive) plate
✓ 1
beta particles are negatively charged (electrons), so they are attracted to the positive plate
✓ 1
Part (c)
530 − 21 = 509 counts/min
✓ 1
Model sketch
Model answer: β⁻ particles are negatively charged, so they are deflected towards the positive plate
Exam tip: Identify radiation by which absorber first reduces the count rate to background level, after allowing for background.
295IGPC3-5-05r5-15Radioactivity, half-life and safetyPaper EHard6 marks
A batch of radiopharmaceutical is contaminated with a second isotope. At time zero the batch contains isotope X with a half-life of 2.0 hours and an activity of 640 MBq, and isotope Y with a half-life of 8.0 hours and an activity of 640 MBq.
(a)
Calculate the activity of X after 8.0 hours.
(2)
(b)
Calculate the activity of Y after 8.0 hours.
(1)
(c)
Calculate the percentage of the total activity at 8.0 hours that is due to Y.
(2)
(d)
Suggest why contamination by Y is more of a problem for a patient than contamination by X.
(1)
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Part (a)
8.0 h is 4 half-lives of X
✓ 1
640 ÷ 24 = 40 MBq
✓ 1
Part (b)
8.0 h is 1 half-life of Y, so 640 ÷ 2 = 320 MBq
✓ 1
Part (c)
total activity = 40 + 320 = 360 MBq (ECF from (a) and (b))
✓ 1
320 ÷ 360 × 100 = 88.9%
✓ 1
Part (d)
Y has a longer half-life, so it stays active in the patient's body for longer and keeps irradiating tissue
✓ 1
Common error: Total activity is the sum of the separate activities. Never add half-lives or average them.
296IGPC3-5-07r5-14Radioactivity, half-life and safetyPaper EHard8 marks
Radon-222, 22286Rn, is a radioactive gas that can dissolve in groundwater used for water supply. It decays by emitting an alpha particle and has a half-life of 3.8 days.
(a)
The decay is 22286Rn → AZPo + 42He. Determine A and Z.
(2)
(b)
Explain why radon in drinking water is more hazardous when swallowed than alpha radiation from a source held outside the body.
(2)
(c)
A water sample has an activity of 1200 Bq. Calculate its activity 11.4 days later.
(2)
(d)
Determine the time taken for the activity of the sample to fall to 18.75 Bq.
(2)
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Part (a)
A = 218
✓ 1
Z = 84
✓ 1
Part (b)
outside the body alpha particles are absorbed by air or the outer layer of skin
✓ 1
inside the body they are absorbed by nearby tissue, where strong ionisation damages cells
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Part (c)
11.4 ÷ 3.8 = 3 half-lives
✓ 1
1200 → 600 → 300 → 150 Bq
✓ 1
Part (d)
1200 ÷ 18.75 = 64, which is 6 halvings
✓ 1
6 × 3.8 = 22.8 days
✓ 1
Common error: Do not forget that the polonium nucleus has a lower proton number as well as a lower nucleon number than radon.
297IGPC3-5-05r8-08Radioactivity, half-life and safetyPaper EHard1 mark
A tracer is added to a racing-engine piston ring. The background count rate is 36 counts/min. The count rate from the ring is 156 counts/min at the start and 66 counts/min 15 minutes later. What is the half-life of the tracer?
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Corrected rates: 156 − 36 = 120 and 66 − 36 = 30 counts/min. 120 → 60 → 30 is two half-lives in 15 min, so the half-life is 7.5 min. 15 min treats the fall as one half-life. 12 min comes from using the uncorrected rates (ratio 2.36, about 1.24 half-lives). 5.0 min assumes three half-lives.
✓ 1
Answer B
Answer: B
Common error: Subtract background from every reading before comparing count rates for half-life.
298IGPC3-5-05r8-16Radioactivity, half-life and safetyPaper EHard6 marks
A beekeeping researcher tags pollen with a radioactive tracer. The count rate from a sample is measured over time. The background count rate is 4 counts/min.
Time / min
0
20
40
60
Measured count rate / counts per min
244
124
64
34
Measured count rate (background not subtracted)
(a)
Explain why the background count rate is subtracted.
(1)
(b)
Determine the half-life of the tracer.
(3)
(c)
Predict the measured count rate at 100 min. Give your answer in counts/min to 1 decimal place.
(2)
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Part (a)
The detector also counts radiation from the surroundings, which does not come from the tracer
✓ 1
Part (b)
Corrected rates: 240, 120, 60, 30 counts/min
✓ 1
The rate halves every 20 min
✓ 1
Half-life = 20 min
✓ 1
Part (c)
100 min = 5 half-lives, so corrected rate = 240 ÷ 32 = 7.5
✓ 1
Measured rate = 7.5 + 4 = 11.5 counts/min
✓ 1
Exam tip: Add the background back on when asked for a measured count rate.
299IGPC3-5-06r8-05Radioactivity, half-life and safetyPaper EHard1 mark
A nuclear reactor supplies electricity to a farming cooperative. Boron control rods are lowered into the core. What is their purpose?
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Notes
Control rods absorb neutrons, keeping the chain reaction steady. Slowing neutrons is the job of the moderator; carrying energy away is the coolant; shielding is the thick concrete wall.
✓ 1
Answer A
Answer: A
Common error: Control rods absorb neutrons; the moderator slows them. Do not mix these roles.
300IGPC3-5-06r8-12Radioactivity, half-life and safetyPaper EHard6 marks
An aircraft exit sign contains tritium, a beta emitter with a half-life of 12 years. When new, the activity is 1.6 GBq (1600 MBq).
(a)
Explain why a low-energy beta source is suitable inside a sealed glass sign.
(2)
(b)
Calculate the activity after 36 years. Give your answer in MBq.
(2)
(c)
The sign is replaced when its activity falls below 25 MBq. Calculate the age of the sign when this happens.
(2)
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Part (a)
Beta particles of low energy are stopped by the glass wall, so little radiation reaches passengers
✓ 1
The beta particles can still make the coating inside glow
✓ 1
Part (b)
36 years = 3 half-lives
✓ 1
1600 ÷ 8 = 200 MBq
✓ 1
Part (c)
1600 ÷ 25 = 64 = 26, so 6 half-lives
✓ 1
6 × 12 = 72 years
✓ 1
Exam tip: Convert the ratio of activities into a power of 2 to find the number of half-lives.
301IGPC3-5-05r10-07Atomic structure, isotopes, fission and fusionPaper EHard1 mark
A nucleus of cobalt-60 has 27 protons and 33 neutrons. It decays by beta-minus emission. How many protons and neutrons are in the nucleus formed?
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In beta-minus decay a neutron changes into a proton and an electron is emitted. Protons: 27 + 1 = 28; neutrons: 33 - 1 = 32; nucleon number stays 60 (B). A has no gain of a proton. C has the change the wrong way round. D is what alpha emission would give (2 protons and 2 neutrons lost).
✓ 1
Answer B
Answer: B
Exam tip: In beta-minus decay the nucleon number stays the same and the proton number goes up by 1.
302IGPC3-5-05r10-14Atomic structure, isotopes, fission and fusionPaper EHard5 marks
A nuclear power station uses fission of uranium-235 to provide low-carbon electricity. One fission reaction is 23592U + 10n → 9236Kr + 14156Ba + x10n
(a)
Calculate x, the number of neutrons released in this reaction.
(2)
(b)
Explain how a chain reaction can continue in the reactor.
(2)
(c)
State why the nuclei in a fusion reactor must be at a very high temperature.
(1)
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Part (a)
235 + 1 = 236 and 92 + 141 = 233
✓ 1
x = 236 - 233 = 3
✓ 1
Part (b)
neutrons released go on to be absorbed by other uranium-235 nuclei
✓ 1
causing further fissions, each releasing more neutrons
✓ 1
Part (c)
nuclei are positively charged and repel / they need high speeds (kinetic energy) to get close enough to join
✓ 1
Exam tip: Check both the nucleon numbers and the proton numbers balance on each side of a nuclear equation.
303IGPC3-5-05r10-15Radioactivity, half-life and safetyPaper EHard4 marks
A hospital stores waste containing iodine-131, which has a half-life of 8.0 days. The initial activity of the waste is 350 Bq. The waste may be thrown away only when its activity is below 4.0 Bq.
(a)
Calculate the activity of the waste after 48 days. Give your answer to 2 significant figures.
(2)
(b)
Determine the shortest time for which the waste must be stored, assuming the activity halves in whole numbers of half-lives only.
(2)
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Part (a)
48 / 8.0 = 6 half-lives
✓ 1
350 / 26 = 5.5 Bq
✓ 1
Part (b)
after 6 half-lives the activity is 5.5 Bq (still above 4.0 Bq), after 7 half-lives it is 2.7 Bq (below 4.0 Bq) (ECF from (a))
✓ 1
7 × 8.0 = 56 days
✓ 1
Exam tip: Test whole numbers of half-lives one by one until the activity drops below the limit.
304IGPC3-5-06r10-07Radioactivity, half-life and safetyPaper EHard1 mark
Strontium-90, 9038Sr, decays by beta-minus emission. Which equation is correct?
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Beta-minus emission gives an electron (0-1e). To balance, the proton number rises by 1 (38 to 39) and the nucleon number is unchanged at 90, giving yttrium-90 (B). A lowers the proton number, which would not balance the charge. C shows alpha decay. D shows gamma emission.
✓ 1
Answer B
Answer: B
Common error: The beta particle has charge -1, so the daughter nucleus must gain +1 to keep the total charge the same.
305IGPC3-5-06r10-14Radioactivity, half-life and safetyPaper EHard7 marks
The activity of a radioactive source used in a data centre sensor is measured over several days. Time / days: 0, 9, 18, 27 Activity / Bq: 720, 360, 180, 90 The source must be replaced when its activity falls below 10 Bq.
(a)
Determine the half-life of the source.
(2)
(b)
Calculate the activity after 45 days.
(2)
(c)
Determine the shortest time, in a whole number of half-lives, after which the source must be replaced.
(3)
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Part (a)
activity halves from 720 to 360 Bq in 9 days
✓ 1
half-life = 9 days
✓ 1
Part (b)
45 / 9 = 5 half-lives
✓ 1
720 / 25 = 22.5 Bq
✓ 1
Part (c)
after 6 half-lives the activity is 11.25 Bq (still above 10 Bq)
✓ 1
after 7 half-lives the activity is 5.6 Bq (below 10 Bq)
✓ 1
7 × 9 = 63 days
✓ 1
Exam tip: Check each whole number of half-lives in turn against the limit.
306IGPC3-5-06r11-07Radioactivity, half-life and safetyPaper EHard1 mark
A detector on a buoy measures a radioactive isotope in seawater. It reads 220 counts/min at the start and 70 counts/min exactly 3.0 hours later. The background count rate is 20 counts/min. What is the half-life of the isotope?
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Notes
The corrected rates are 200 and 50 counts/min, a factor of 4, which is 2 half-lives in 3.0 h, so the half-life is 1.5 h. C ignores the background (220 / 70 is about 3.1). A assumes four half-lives. D takes the elapsed time as the half-life.
✓ 1
Answer B
Answer: B
Common error: Subtract the background count rate from every reading before comparing them.
307IGPC3-5-06r11-15Radioactivity, half-life and safetyPaper EHard6 marks
A sealed source with a half-life of 5.0 days is kept in a warehouse before shipping. A detector near the store reads 148 counts/min and the background count rate is 20 counts/min. The store may be re-opened when the detector reading, including background, has fallen to 28 counts/min.
(a)
Calculate the corrected count rate due to the source at present.
(1)
(b)
Calculate the time that must pass before the store may be re-opened.
(3)
(c)
Describe a sketch of the corrected count rate against time for the first 20 days.
a curve falling from 128 counts/min at t = 0, getting less steep with time
✓ 1
passing through 64 counts/min at 5.0 days, 32 at 10 days and 8 at 20 days
✓ 1
Model sketch
Model answer: the count rate halves every 5.0 days
Common error: The re-opening limit includes background, so subtract it before counting half-lives.
308IGPC3-5-06r11-16Atomic structure, isotopes, fission and fusionPaper EHard8 marks
A research ship has a nuclear reactor. Each fission of a uranium-235 nucleus releases 3.2 × 10−11 J of energy. The reactor releases energy at a rate of 6.4 × 107 W.
(a)
Describe how a chain reaction is sustained in the reactor.
(3)
(b)
Calculate the number of fissions each second.
(2)
(c)
Calculate the number of fissions in 1.0 hour.
(2)
(d)
State one safety problem caused by the waste produced in a reactor.
(1)
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Part (a)
a neutron strikes a uranium-235 nucleus, which splits into smaller nuclei
✓ 1
the fission releases further neutrons
✓ 1
these neutrons strike other uranium-235 nuclei and cause more fissions
✓ 1
Part (b)
number per second = power / energy per fission = 6.4 × 107 / 3.2 × 10−11
✓ 1
= 2.0 × 1018
✓ 1
Part (c)
1.0 h = 3600 s
✓ 1
2.0 × 1018 × 3600 = 7.2 × 1021, ECF from (b)
✓ 1
Part (d)
the waste stays radioactive for a very long time, so it must be stored securely
✓ 1
Exam tip: Energy per second is power, so the number of events per second is power divided by energy per event.
309IGPC3-5-07r19-08Radioactivity, half-life and safetyPaper EHard1 mark
A radioactive sample is monitored with a counter. The background count rate is 20 counts per minute. At the start, the measured count rate is 340 counts per minute. After 3.0 hours the measured count rate is 30 counts per minute. What is the half-life of the sample?
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Notes
D is correct: the corrected rates are 320 and 10 counts per minute, a factor of 32 = 25, so there are 5 half-lives in 3.0 h and the half-life is 0.60 h. A comes from not subtracting the background (340/30 is about 11, about 3.5 half-lives). B assumes 4 half-lives. C assumes 2 half-lives.
✓ 1
Answer D
Answer: D
Common error: Subtract the background from every reading before comparing count rates to find the number of half-lives.
310IGPC3-5-07r19-15Radioactivity, half-life and safetyPaper EHard7 marks
A hospital receives a radioactive tracer that emits gamma rays. At the time of delivery, its activity is 480 MBq. The half-life is 6.0 hours. A scan needs an activity of at least 30 MBq.
(a)
Calculate the activity of the tracer 18 hours after delivery.
(2)
(b)
Calculate the longest time after delivery at which the scan can begin.
(2)
(c)
Explain why a gamma emitter is suitable as a tracer inside the body but an alpha emitter is not.
(2)
(d)
Suggest why an isotope with a half-life of 6.0 hours is chosen rather than one with a half-life of 6 days.
(1)
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Notes
Part (a)
18 h is 3 half-lives
✓ 1
480 / 8 = 60 MBq
✓ 1
Part (b)
480 → 240 → 120 → 60 → 30 is 4 half-lives
✓ 1
4 × 6.0 = 24 hours
✓ 1
Part (c)
gamma rays are penetrating enough to pass out of the body to a detector
✓ 1
alpha particles are absorbed by tissue and are strongly ionising, so would cause damage inside the body
✓ 1
Part (d)
the activity (and so the patient's exposure) falls quickly after the scan / does not stay in the body long
✓ 1
Exam tip: Write the successive halved values to show each half-life; it earns the method mark safely.
311IGPC3-5-07r19-16Radioactivity, half-life and safetyPaper EHard7 marks
A technician in a cycling-component laboratory uses a radioactive tracer in a brake pad to study wear. The background count rate in the laboratory is 24 counts per minute. A counter near the brake pad records 408 counts per minute at the start. After 7.5 hours it records 72 counts per minute.
(a)
State two sources of background radiation.
(2)
(b)
Calculate the corrected count rate at the start and after 7.5 hours.
(2)
(c)
Determine the half-life of the tracer.
(3)
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Part (a)
any two from: cosmic rays; radon gas / radioactive gases in the air; rocks and soil; buildings; food and drink; medical sources
✓ 1
second different correct source
✓ 1
Part (b)
408 − 24 = 384 counts per minute
✓ 1
72 − 24 = 48 counts per minute
✓ 1
Part (c)
384 / 48 = 8 = 23, so 3 half-lives (ECF from (b))
✓ 1
7.5 / 3
✓ 1
2.5 hours
✓ 1
Common error: Using the uncorrected readings (408 and 72) gives a wrong number of half-lives.
312IGPC3-5-07r21-08Radioactivity, half-life and safetyPaper EHard1 mark
A radioactive sample is used to tag crates in a warehouse. At time zero a detector reads 212 counts per minute. Exactly 6.0 hours later it reads 44 counts per minute. The background count rate is 20 counts per minute. What is the half-life of the sample?
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Notes
Corrected count rates are 212 − 20 = 192 and 44 − 20 = 24. 192 → 96 → 48 → 24 is three halvings in 6.0 h, so the half-life is 2.0 h; C is correct. A comes from not subtracting background and treating 212 → 44 as about two halvings. D divides by 8, the ratio of the count rates, instead of by the number of halvings. B is the total time.
✓ 1
Answer C
Answer: C
Exam tip: Subtract background from every reading before comparing count rates.
313IGPC3-5-07r21-12Radioactivity, half-life and safetyPaper EHard7 marks
Radon-222, written 22286Rn, is a radioactive gas that can build up in the basements of buildings. It decays by emitting an alpha particle. The half-life of radon-222 is 3.8 days.
(a)
Determine the nucleon number and the proton number of the nucleus formed in the decay.
(2)
(b)
Alpha particles are stopped by a sheet of paper, yet radon is hazardous if breathed in. Explain why.
(2)
(c)
A sealed basement sample of radon has an activity of 3200 Bq. Calculate its activity after 19 days. Assume no more radon enters the sample.
(3)
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Notes
Part (a)
nucleon number 218
✓ 1
proton number 84
✓ 1
Part (b)
alpha particles are strongly ionising
✓ 1
inside the lungs there is no skin or air to absorb them, so all their energy is absorbed by sensitive tissue
✓ 1
Part (c)
19 ÷ 3.8 = 5 half-lives
✓ 1
3200 halved five times, or 3200 ÷ 25
✓ 1
100 Bq
✓ 1
Exam tip: Activity halves in each half-life, so after n half-lives it is the original activity divided by 2ⁿ.
314IGPC3-5-07r22-13Radioactivity, half-life and safetyPaper EHard8 marks
A telecoms cable factory checks the thickness of plastic insulation using a strontium-90 source. Strontium-90 (proton number 38, nucleon number 90) decays by beta-minus emission to yttrium. The half-life of strontium-90 is 29 years.
Diagram not to scale
(a)
Complete the equation: 9038Sr → AZY + 0−1e. State A and Z.
(2)
(b)
Describe the change that happens inside the nucleus when a beta-minus particle is emitted.
(2)
(c)
Beta particles travel between two horizontal charged plates. The top plate is positive and the bottom plate is negative. State the direction in which the beta particles are deflected and give a reason.
(2)
(d)
After a number of years the activity of the source is 6.25% of its original value. Determine this number of years.
(2)
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Part (a)
A = 90
✓ 1
Z = 39
✓ 1
Part (b)
a neutron changes into a proton
✓ 1
and an electron (the beta particle) is emitted
✓ 1
Part (c)
upwards / towards the positive plate
✓ 1
beta particles are negatively charged
✓ 1
Part (d)
6.25% = 1/16, so 4 half-lives
✓ 1
4 × 29 = 116 years
✓ 1
Exam tip: Convert a percentage remaining into a fraction such as 1/16, then count the powers of two.
315IGPC3-5-07r22-16Atomic structure, isotopes, fission and fusionPaper EHard6 marks
In a nuclear reactor a uranium-235 nucleus (proton number 92) absorbs a neutron and splits into barium-141 (proton number 56), krypton-92 (proton number 36) and x neutrons: 23592U + 10n → 14156Ba + 9236Kr + x 10n
(a)
Calculate x.
(2)
(b)
Explain how the neutrons released can lead to a chain reaction.
(2)
(c)
Nuclear fusion requires very high temperatures. Explain why.
(2)
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Notes
Part (a)
nucleon number on left = 235 + 1 = 236, so 141 + 92 + x = 236
✓ 1
x = 3
✓ 1
Part (b)
the released neutrons are absorbed by other uranium-235 nuclei
✓ 1
causing further fissions that release more neutrons, and so on
✓ 1
Part (c)
nuclei are positively charged and repel each other
✓ 1
high temperature gives nuclei enough kinetic energy to get close enough to join
✓ 1
Exam tip: Check the proton numbers also balance (92 = 56 + 36) to confirm your value of x.
316IGPC3-5-09r27-07Radioactivity, half-life and safetyPaper EHard1 mark
The count rate from a sample of a radioactive isotope, measured with a detector, is 340 counts/min at the start. The background count rate is 20 counts/min. After 6.0 hours the detector reads 60 counts/min. What is the half-life of the isotope?
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Notes
Subtract background first: the count rate from the isotope falls from 320 to 40 counts/min. 320 → 160 → 80 → 40 is 3 half-lives, so the half-life is 6.0 / 3 = 2.0 hours. 2.4 hours comes from using 340 and 60 without correcting for background (about 2.5 half-lives). 1.5 hours assumes 4 half-lives. 3.0 hours assumes only 2 half-lives.
✓ 1
Answer D
Answer: D
Common error: Always subtract the background count rate from every reading before looking for a halving.
317IGPC3-5-09r27-11Atomic structure, isotopes, fission and fusionPaper EHard7 marks
A nuclear-powered ship has a reactor in which uranium-235 undergoes fission.
(a)
Describe what happens when a nucleus of uranium-235 undergoes fission.
(3)
(b)
Explain how control rods are used to keep the reaction in the reactor steady.
(2)
(c)
The thermal power of the reactor is 150 MW. Each fission releases 3.2 × 10−11 J. Calculate the number of fissions per second. Give your answer to 2 significant figures.
(2)
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Part (a)
a neutron is absorbed by the uranium-235 nucleus
✓ 1
the nucleus splits into two smaller nuclei
✓ 1
further neutrons (two or three) and energy are released
✓ 1
Part (b)
control rods absorb neutrons
✓ 1
lowering the rods further into the reactor reduces the number of neutrons causing fission (raising them increases it), so the rate stays steady
✓ 1
Part (c)
number per second = power / energy per fission = 150 × 106 / 3.2 × 10−11
✓ 1
4.7 × 1018 per second
✓ 1
Exam tip: Convert MW to W before dividing: 1 MW = 1 × 10⁶ W.
318IGPC3-5-09r27-16Atomic structure, isotopes, fission and fusionPaper EHard7 marks
A ship's reactor produces iodine-131, a fission product with a half-life of 8.0 days. Iodine-131, 13153I, decays by β− emission to xenon (Xe). In the Sun, fusion reactions release energy.
(a)
Explain why a very high temperature is needed for nuclear fusion to take place.
(2)
(b)
Write the nuclear equation for the decay of iodine-131 to xenon.
(2)
(c)
A sample of iodine-131 contains 4.8 × 1020 nuclei at the start. Calculate the number of nuclei that have decayed after 24 days.
(3)
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Notes
Part (a)
nuclei are positively charged and repel each other
✓ 1
very high temperatures give the nuclei enough kinetic energy (speed) to overcome the repulsion and get close enough to fuse
✓ 1
Part (b)
xenon nuclide is 13154Xe
✓ 1
beta particle written as 0−1e
✓ 1
Part (c)
24 days = 3 half-lives
✓ 1
remaining = 4.8 × 1020 / 8 = 6.0 × 1019
✓ 1
decayed = 4.8 × 1020 − 6.0 × 1019 = 4.2 × 1020
✓ 1
Common error: The question asks for the number that have decayed, not the number remaining; subtract from the starting number.
319IGPC3-5-09r28-05Atomic structure, isotopes, fission and fusionPaper EHard1 mark
A nucleus of uranium-238 (proton number 92) emits one alpha particle and then two beta-minus particles. What are the nucleon number and the proton number of the final nucleus?
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Notes
The alpha particle removes 4 nucleons and 2 protons, giving 234 and 90. Each beta-minus particle raises the proton number by 1 and leaves the nucleon number unchanged, so the final nucleus has 234 nucleons and 92 protons. Nucleon number 234 with proton number 90 ignores the effect of the two beta particles. Nucleon number 236 treats the alpha particle as removing only 2 nucleons. Proton number 88 subtracts one proton for each beta particle instead of adding one.
✓ 1
Answer C
Answer: C
Exam tip: Track nucleon number and proton number separately through each step of a decay chain.
320IGPC3-5-09r28-13Radioactivity, half-life and safetyPaper EHard7 marks
A radiopharmacy prepares a sample that contains two radioactive isotopes. Isotope A has a half-life of 2.0 hours and isotope B has a half-life of 8.0 hours. At time zero, the activity of A is 800 Bq and the activity of B is 800 Bq.
(a)
Calculate the activity of A after 8.0 hours.
(2)
(b)
Calculate the activity of B after 8.0 hours.
(1)
(c)
Determine the total activity of the sample after 8.0 hours.
(1)
(d)
Show that after 24 hours isotope B is responsible for more than 99% of the total activity of the sample. Use unrounded activities.
A1 fraction = 100 ÷ (100 + 0.195) = 0.998, which is 99.8% and is greater than 99%
✓ 1
Common error: A longer half-life does not mean a bigger starting activity: the isotope that survives longer takes over the total later.
321IGPC3-5-09r28-16Radioactivity, half-life and safetyPaper EHard8 marks
A radioactive tracer is used to check the flow in a buried water pipe. The corrected count rate from a sample of the tracer falls from 1280 counts per minute to 160 counts per minute over 36 hours.
(a)
Determine the half-life of the tracer.
(3)
(b)
Calculate the corrected count rate a further 18 hours later. Give your answer to the nearest whole number.
(3)
(c)
Explain why the tracer must emit gamma radiation rather than alpha radiation if it is to be detected from above ground.
(2)
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Notes
Part (a)
M1 1280 ÷ 160 = 8 = 23, so 3 half-lives
✓ 1
M1 36 ÷ 3
✓ 1
A1 12 hours
✓ 1
Part (b)
M1 18 ÷ 12 = 1.5 half-lives
✓ 1
M1 160 ÷ 21.5 (accept 80 ÷ √2)
✓ 1
A1 57 counts per minute
✓ 1
Part (c)
alpha radiation is absorbed by water, the pipe wall and soil, so it cannot reach the detector
✓ 1
gamma radiation is very penetrating and can pass through to be detected at the surface
✓ 1
Exam tip: A non-whole number of half-lives still works: use 2 raised to the number of half-lives.
322IGPC3-5-08r29-08Radioactivity, half-life and safetyPaper EHard1 mark
A soil sample contains a radioactive tracer of half-life 12 hours. The background count rate is 20 counts/min. The detector reads 580 counts/min with the sample present. What does the detector read 36 hours later?
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Notes
The corrected count rate is 580 − 20 = 560 counts/min. In 36 hours there are 3 half-lives, so the corrected rate is 560 / 8 = 70 counts/min, and the detector also records background, so it reads 70 + 20 = 90 counts/min. 70 forgets to add the background back. 72.5 divides the uncorrected reading by 8. 140 uses only two half-lives.
✓ 1
Answer D
Answer: D
Exam tip: Subtract background first, apply the half-lives, then add background back if the question asks what the detector reads.
323IGPC3-5-08r29-12Radioactivity, half-life and safetyPaper EHard7 marks
Old aircraft instrument dials were painted with luminous paint containing radium-226, 22688Ra, which decays by emitting an alpha particle to form radon.
(a)
Determine the proton number and the nucleon number of the radon nucleus formed.
(2)
(b)
Explain why the paint is a small hazard while the dial is intact, but a serious hazard if radium is swallowed.
(2)
(c)
The half-life of radium-226 is 1600 years. A dial contains 3.2 mg of radium-226. Calculate the mass of radium-226 left after 4800 years.
(3)
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Part (a)
Proton number 86
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Nucleon number 222
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Part (b)
Alpha particles have very short range and are stopped by the glass cover, skin or a few centimetres of air
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Alpha is strongly ionising, so inside the body it damages cells and can cause cancer
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Part (c)
M1 number of half-lives = 4800 ÷ 1600 = 3
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M1 mass halved 3 times: 3.2 ÷ 23
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A1 0.40 mg
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Exam tip: Alpha particles are least penetrating but most ionising, which is why the risk depends on whether the source is outside or inside the body.
324IGPC3-5-09r29-08Radioactivity, half-life and safetyPaper EHard1 mark
In an epidemiology study, a gamma-emitting tracer has a half-life of 20 minutes. A detector near a sample records 85 counts/min at 10:00. The background count rate is a constant 5 counts/min. What count rate is recorded at 11:00?
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The source contributes 85 − 5 = 80 counts/min at 10:00. In 60 minutes there are three half-lives, so the source falls to 80 ÷ 8 = 10 counts/min, and the detector records 10 + 5 = 15 counts/min. 11 comes from halving the uncorrected 85 three times. 5 ignores the source completely. 20 comes from using only two half-lives.
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Answer B
Answer: B
Common error: Background radiation does not decay: subtract it before halving, then add it back to find what the detector reads.
325IGPC3-5-09r29-14Radioactivity, half-life and safetyPaper EHard6 marks
A sample of a radioactive gas collected from a geothermal borehole has a corrected count rate of 960 counts/s. After 45 minutes the corrected count rate is 120 counts/s.
(a)
Determine the half-life of the gas.
(2)
(b)
Calculate the corrected count rate 90 minutes after the count rate was 960 counts/s.
(2)
(c)
The count rate shown by the detector varies slightly from one second to the next. Explain why.
(2)
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Part (a)
960 → 480 → 240 → 120 is 3 half-lives
✓ 1
45 ÷ 3 = 15 minutes
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Part (b)
90 ÷ 15 = 6 half-lives (ECF from (a))
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960 ÷ 26 = 15 counts/s
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Part (c)
radioactive decay is random
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it is impossible to predict when a particular nucleus will decay, so the number of decays in each second varies
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Exam tip: Check your half-life by halving back from the start; the halvings should land exactly on the later reading.
326IGPC3-5-08r30-05Atomic structure, isotopes, fission and fusionPaper EHard1 mark
A nucleus of uranium-238 (23892U) in volcanic rock emits two alpha particles and then one beta-minus particle. What are the nucleon number and proton number of the resulting nucleus?
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Each alpha emission lowers the nucleon number by 4 and the proton number by 2: 238/92 becomes 234/90 and then 230/88. Beta-minus emission leaves the nucleon number unchanged and raises the proton number by 1, giving 230/89. Nucleon number 230 with proton number 87 has beta emission lowering the proton number. Nucleon number 234 with proton number 91 includes only one alpha emission. Nucleon number 234 with proton number 89 takes alpha emission as lowering the nucleon number by 2.
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Answer A
Answer: A
Exam tip: Alpha: A falls by 4, Z falls by 2. Beta-minus: A unchanged, Z rises by 1. Track both numbers after each emission.
327IGPC3-5-08r30-08Radioactivity, half-life and safetyPaper EHard1 mark
A detector reads 24 counts/min when no source is present. A rock sample with a radioactive isotope of half-life 5.0 days makes the detector read 120 counts/min. What will the detector read 10 days later?
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The count rate due to the sample is 120 − 24 = 96 counts/min. In 10 days (two half-lives) this falls to 96 / 4 = 24 counts/min, and the detector still records the 24 counts/min of background, so it reads 48 counts/min. 30 counts/min divides the total reading by 4 and ignores the background. 24 counts/min gives only the corrected count rate. 72 counts/min allows for only one half-life.
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Answer C
Answer: C
Common error: Background radiation does not decay: subtract it before halving, then add it back to find the reading.
328IGPC3-5-08r30-11Radioactivity, half-life and safetyPaper EHard7 marks
A volcanologist follows a radioactive gas escaping from a vent. With no source present, the detector reads a background of 20 counts/min. When the gas sample is first measured, the detector reads 980 counts/min. After 12 hours it reads 140 counts/min.
(a)
Define half-life.
(1)
(b)
Calculate the corrected count rates at the start and after 12 hours.
(1)
(c)
Determine the half-life of the gas.
(2)
(d)
Calculate the count rate the detector will show 20 hours after the first measurement.
(3)
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Part (a)
the time taken for half of the nuclei of a radioactive isotope in a sample to decay
✓ 1
Part (b)
960 counts/min and 120 counts/min
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Part (c)
960 → 480 → 240 → 120 is three halvings
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half-life = 12 / 3 = 4.0 hours
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Part (d)
20 hours = 5 half-lives
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corrected count rate = 960 / 25 = 30 counts/min
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add background: 30 + 20 = 50 counts/min
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Common error: Subtract the background before using half-lives, and add it back if the question asks what the detector reads.
329IGPC3-5-07r38-07Radioactivity, half-life and safetyPaper EHard1 mark
A cargo scanner contains a source of half-life 20 minutes. A detector with background radiation of 10 counts/min measures a total count rate of 90 counts/min. What is the total count rate measured 40 minutes later?
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The count rate from the source is 90 − 10 = 80 counts/min. Two half-lives pass, so it falls to 20 counts/min, and the background of 10 counts/min is still detected, giving a total of 30 counts/min. Forgetting to add the background gives 20, halving the uncorrected 90 twice gives 22.5, and halving only once gives 45.
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Answer D
Answer: D
Common error: Subtract background before applying half-lives, then add it back if the question asks for the measured count rate.
330IGPC3-5-07r38-15Radioactivity, half-life and safetyPaper EHard6 marks
Waste from a research laboratory contains a radioactive isotope of half-life 30 years. It is to be stored in a vault under a museum until its activity has fallen to a safe level of 100 Bq. The initial activity of the waste is 6400 Bq.
(a)
Calculate the activity of the waste after 60 years.
(2)
(b)
Calculate the time that the waste must be stored before its activity falls to 100 Bq.
(3)
(c)
Suggest why an isotope with a long half-life needs to be stored securely for a very long time.
(1)
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Part (a)
M1 60 years is 2 half-lives
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A1 1600 Bq
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Part (b)
M1 6400 ÷ 100 = 64 (the activity must fall by a factor of 64)
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M1 64 = 26, so 6 half-lives are needed
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A1 6 × 30 = 180 years
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Part (c)
B1 its activity stays high for a long time so it remains a hazard (it continues to emit ionising radiation)
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Exam tip: If a quantity falls by a factor that is a power of two, the power is the number of half-lives.
331IGPC3-5-08r39-14Radioactivity, half-life and safetyPaper EHard7 marks
Radon-222 is a radioactive gas released from volcanic soil.
(a)
A nucleus of radon-222, 22286Rn, decays by emitting an alpha particle to form polonium (Po). Write the nuclear equation for this decay.
(3)
(b)
A nucleus of lead-214, 21482Pb, decays by beta-minus emission. Determine the nucleon number and the proton number of the nucleus formed.
(2)
(c)
A sealed sample holds 4.0×106 radon-222 nuclei. The half-life of radon-222 is 3.8 days. Calculate the number of radon-222 nuclei left after 11.4 days.
(2)
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Part (a)
22286Rn → 21884Po + 42He
—
Nucleon number of polonium 218
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Proton number of polonium 84
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Alpha particle written as helium with 4 and 2
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Part (b)
Nucleon number 214
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Proton number 83
✓ 1
Part (c)
11.4 days is 3 half-lives
✓ 1
4.0×106 / 8 = 5.0×105
✓ 1
Common error: In a nuclear equation the nucleon numbers and the proton numbers must each add up to the same total on both sides.
332IGPC3-5-07r43-07Radioactivity, half-life and safetyPaper EHard1 mark
A detector near a sample of radioactive waste records 340 counts/min. The background count rate is 20 counts/min. After 12 hours the detector records 100 counts/min. What is the half-life of the waste?
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Subtract the background: the corrected rate falls from 320 to 80 counts/min, a factor of 4, which is two half-lives in 12 hours, so the half-life is 6.0 hours. The value 6.8 hours comes from using the uncorrected readings (340 to 100). The value 4.0 hours comes from assuming three half-lives. The value 12 hours treats the whole elapsed time as one half-life, although the count rate has fallen by much more than half.
✓ 1
Answer A
Answer: A
Common error: Always subtract background before comparing count rates; only the source's counts decay.
333IGPC3-5-07r43-13Radioactivity, half-life and safetyPaper EHard7 marks
A tracer containing a radioactive isotope with a half-life of 6.0 hours is used to follow a pathogen through hospital drains. When prepared, the sample has an activity of 240 MBq.
(a)
Calculate the time taken for the activity of the sample to fall to 7.5 MBq.
(3)
(b)
A detector beside a drain records 100 counts/min with the sample present, of which 12 counts/min is background radiation. Calculate the reading the detector would give 12 hours later. Assume the count rate from the sample is proportional to its activity.
(3)
(c)
Explain why the background count rate must be removed before using half-lives to predict the reading.
M1 12 hours is 2 half-lives, so 88 ÷ 4 = 22 counts/min
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A1 22 + 12 = 34 counts/min
✓ 1
Part (c)
background does not come from the sample, so it does not decay with the sample's half-life / stays constant
✓ 1
Common error: Halving the whole reading is wrong: only the corrected count rate from the source halves each half-life; background stays the same.
334IGPC3-5-07r43-15Atomic structure, isotopes, fission and fusionPaper EHard6 marks
In a nuclear power station, uranium-235 nuclei undergo fission. Each fission releases 3.2 × 10−11 J of energy. The reactor has a thermal power output of 9.6 × 108 W.
(a)
Describe how a chain reaction occurs in the reactor.
(2)
(b)
Calculate the number of fission events per second in the reactor.
(2)
(c)
State the purpose of the control rods in the reactor.
(2)
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Part (a)
a neutron is absorbed by a uranium-235 nucleus, which splits and releases energy and more neutrons
✓ 1
these neutrons cause further nuclei to split, so the process repeats / continues
✓ 1
Part (b)
M1 number per second = 9.6 × 108 ÷ 3.2 × 10−11
✓ 1
A1 3.0 × 1019 per second
✓ 1
Part (c)
they absorb neutrons
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to control / reduce the rate of fission so the chain reaction does not run out of control
✓ 1
Exam tip: Power is energy per second, so divide the power by the energy per event to find the events per second.
335IGPC3-5-08r48-14Atomic structure, isotopes, fission and fusionPaper EHard7 marks
Uranium-235 is used in some nuclear power systems. Scientists also study fusion as a future energy source for space travel.
(a)
Describe what happens to a uranium-235 nucleus in nuclear fission.
(3)
(b)
Explain how a chain reaction can occur.
(2)
(c)
Explain why nuclear fusion needs a very high temperature.
(2)
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Part (a)
the nucleus absorbs a neutron
✓ 1
it splits into two smaller nuclei
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two or three neutrons are released, together with a large amount of energy
✓ 1
Part (b)
the neutrons released go on to strike other uranium-235 nuclei
✓ 1
these nuclei split and release more neutrons, so the process continues and grows
✓ 1
Part (c)
nuclei are positively charged, so they repel each other
✓ 1
at high temperature they have enough kinetic energy to overcome the repulsion and get close enough to join
✓ 1
Common error: Fission splits one large nucleus into smaller ones; fusion joins small nuclei into a larger one. Do not swap the two.
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