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IB Physics Standard Level · Topic E.1

E.1 Structure of the atom

102 free, high-quality past-paper-style questions on Structure of the atom for IB Physics SL, each with a full mark scheme. Covers: Nuclear notation; Rutherford scattering; Energy levels & photons; and more. For when you have finished the past papers, or want to practise one topic at a time.E.1 at Higher Level →Read the E.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.486 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-01Nuclear notation Paper 1AEasy1 mark

A nucleus of iodine is represented by 12753I.

Which row gives the number of protons and the number of neutrons in this nucleus?

Number of protonsNumber of neutrons
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Step 1The lower number is the proton number Z = 53; the neutron number is A − Z = 127 − 53 = 74.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AThe two numbers have been swapped: 74 is the neutron number, and the proton number is the lower figure, Z = 53.
  • BCorrect: Z = 53 protons and N = A − Z = 127 − 53 = 74 neutrons.
  • C127 is the nucleon number A, which counts the protons and the neutrons together.
  • D180 = 127 + 53 adds Z to A; the neutron number is A − Z.
2E-1A-02Rutherford scattering Paper 1AEasy1 mark

In a Geiger–Marsden–Rutherford experiment, alpha particles are directed at a thin gold foil. The diagrams show possible paths of one alpha particle that passes close to a single gold nucleus N.

Which diagram shows a possible path?

Four possible paths A to D of an alpha particle passing close to a gold nucleus N: A curves towards and round the nucleus, B is a straight line, C curves smoothly away from the nucleus, D is two straight lines meeting at a sharp corner near the nucleus.A.NB.NC.ND.N
Diagram NOT accurately drawn
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Step 1The alpha particle and the nucleus are both positively charged, so the alpha particle is repelled by a force that grows smoothly as it approaches N and dies away as it leaves: the path curves gradually away from the nucleus.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThis path curves towards N and round it, as if the nucleus attracted the alpha particle; both are positive, so the force is repulsive.
  • BAn alpha particle passing this close to a nucleus feels a large repulsive force, so it cannot continue undeflected.
  • CCorrect: the electric repulsion increases as the particle approaches N and decreases as it moves away, so the path bends smoothly away from the nucleus.
  • DThe direction changes suddenly at one point, as if the particle bounced off a hard surface; the electric force acts at a distance, so the path is a smooth curve.
3E-1A-04Energy levels & photons Paper 1AEasy1 mark

An electron in an atom falls from an energy level E2 to a lower energy level E1.

What is the wavelength of the photon emitted? (h is the Planck constant and c is the speed of light.)

E₂E₁electron
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Step 1The photon carries the energy lost by the atom: hf = E2 − E1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With c = fλ, λ = c/f = hc/(E2 − E1).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: hf = E2 − E1 and λ = c/f, so λ = hc/(E2 − E1).
  • Bh/(E2 − E1) is 1/f, the period of the wave; the speed of light from c = fλ has been left out.
  • CThe photon energy is the difference between the two levels, not their sum.
  • DThis is 1/λ: the relation has been inverted.
4E-1A-28Emission & absorption spectra Paper 1AEasy1 mark

The diagram shows the emission spectrum of a gas. All four bright lines are produced by transitions that end on the ground state.

White light passes through a cool sample of the same gas. Which spectrum shows the light that emerges?

Top: emission spectrum of the gas, four bright lines on a dark background. Below: four spectra A to D. A: continuous spectrum with dark lines at the same four wavelengths. B: continuous spectrum with dark lines at four different wavelengths. C: bright lines on a dark background at the same four wavelengths. D: continuous spectrum with no lines.Emission spectrum of the gas400500600700wavelength / nmA.B.C.D.400500600700wavelength / nm
Diagram NOT accurately drawn
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Step 1Atoms in the cool gas absorb photons whose energies equal the gaps between their levels — the same gaps that produce the emission lines — so the continuous spectrum emerges with dark lines at exactly the emission-line wavelengths.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: the cool gas removes only the photons whose energies match its level differences, leaving dark lines at the emission wavelengths on a continuous spectrum.
  • BThe dark lines are at different wavelengths from the bright lines; absorption and emission involve the same level differences, so they occur at the same wavelengths.
  • CThis is an emission spectrum again; light passing through a cool gas gives a continuous spectrum with dark lines missing from it.
  • DThe gas does absorb light, but only at particular wavelengths, so the spectrum cannot be continuous and unbroken.
5E-1A-45Nuclear notation Paper 1AEasy1 mark

A nucleus of nuclide X is represented by AZX. A nucleus of nuclide Y contains twice as many protons as the nucleus of X and the same number of neutrons.

What is the nucleon number of Y?

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Marking pointMarkNotes
Step 1X has Z protons and A − Z neutrons, so Y has 2Z protons and A − Z neutrons.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Nucleon number of Y = 2Z + (A − Z) = A + Z.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • ADoubling the nucleon number would double the neutrons as well; only the protons are doubled.
  • BCorrect: 2Z protons plus A − Z neutrons gives A + Z nucleons.
  • CThis adds the 2Z protons of Y to A, which already contains the Z protons of X, so Z protons are counted twice.
  • D2Z is the proton number of Y; the neutrons must be added to give the nucleon number.
6E-1A-63Nuclear notation Paper 1AEasy1 mark

A neutral atom of the nuclide AZX gains two electrons and becomes a negative ion.

How many charged particles does the ion contain?

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Step 1The nucleus contains Z protons (charged) and A − Z neutrons (uncharged); the neutral atom has Z electrons, so the ion has Z + 2 electrons.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Charged particles = Z protons + (Z + 2) electrons = 2Z + 2.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AOnly the electrons have been counted; the Z protons in the nucleus are charged too.
  • BA negative ion has gained electrons, so it has Z + 2 electrons, not Z − 2.
  • CCorrect: Z protons and Z + 2 electrons are charged; the A − Z neutrons are not, so the total is 2Z + 2.
  • DThis counts every particle, including the A − Z neutrons, which carry no charge.
7E-1A-65Nuclear notation Paper 1AEasy1 mark

A nucleus of iron is represented by 5626Fe.

What is the best estimate of the mass of this nucleus? (1 u = 1.661 × 10−27 kg)

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Step 1The nucleus contains A = 56 nucleons, each of mass about 1 u.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Mass ≈ 56 × 1.661 × 10−27 = 9.3 × 10−26 kg.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A4.3 × 10−26 kg = 26 u counts only the 26 protons and leaves out the 30 neutrons.
  • B5.0 × 10−26 kg = 30 u counts only the neutrons, 56 − 26 = 30, and leaves out the protons.
  • CCorrect: the mass is about A × 1 u = 56 × 1.661 × 10−27 kg = 9.3 × 10−26 kg.
  • D1.4 × 10−25 kg = 82 u adds Z to A; the nucleon number already counts the protons.
8E-1A-66Rutherford scattering Paper 1AEasy1 mark

In a Geiger–Marsden–Rutherford experiment, alpha particles are directed at a thin gold foil in a vacuum. A detector counts the number N of alpha particles scattered through an angle θ in equal time intervals, for θ from 10° to 170°.

Which graph shows how N varies with θ?

Four sketch graphs of count N against scattering angle theta from 10 to 170 degrees. A: N falls linearly to almost zero. B: N falls very steeply at small angles and then levels off at a small non-zero value up to 170 degrees. C: N falls and reaches zero at 90 degrees, staying zero beyond. D: N falls steeply, then rises again towards 170 degrees.A.Nθ0B.Nθ0C.Nθ0D.Nθ0
Sketch graphs — not to scale
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Step 1Most alpha particles pass far from every nucleus and are deflected very little, so N is very large at small θ and falls steeply as θ increases.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2A very small number pass close to a nucleus and are scattered through large angles, even beyond 90°, so N stays small but not zero up to 170°.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AN does not fall uniformly: most alpha particles are scattered through very small angles, so the fall is very steep at small θ and then levels off.
  • BCorrect: the count is very large at small angles and falls steeply, but a small number of alpha particles are still scattered through angles greater than 90°.
  • CSome alpha particles are scattered through more than 90°: this observation is the evidence for a small, massive, positively charged nucleus.
  • DLarge-angle scattering needs a very close approach to a nucleus, which is rarer the larger the angle; N does not rise again towards 180°.
9E-1A-68Energy levels & photons Paper 1AEasy1 mark

Which graph shows how the energy E of a photon varies with its wavelength λ?

Four sketch graphs of photon energy E against wavelength lambda. A: E decreases along a curve, halving each time lambda doubles, approaching the lambda axis. B: E increases along a straight line through the origin. C: E decreases along a straight line that meets the lambda axis. D: E is constant.A.Eλ0B.Eλ0C.Eλ0D.Eλ0
Sketch graphs — not to scale
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Step 1E = hf and f = c/λ, so E = hc/λ: E is inversely proportional to λ, giving a curve that halves when λ doubles and never reaches either axis.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: E = hc/λ, so doubling the wavelength halves the photon energy; the graph is a curve approaching both axes.
  • BThis shows E proportional to λ; a longer wavelength means a lower frequency and therefore a lower photon energy.
  • CE does decrease as λ increases, but in inverse proportion: a straight line would make E zero at a finite wavelength.
  • DThe energy of a photon is fixed by its frequency, and so by its wavelength; it is not the same for all photons.
10IBP2S-035-01-01Energy levels & photons Paper 1AEasy1 mark

A bicycle light contains a blue LED that emits photons of energy 2.6 eV and a red LED that emits photons of energy 1.9 eV. What is the ratio \(\frac{\text{wavelength of red light}}{\text{wavelength of blue light}}\)?

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Photon energy is E = hc/λ, so wavelength is inversely proportional to photon energy. The lower-energy red photon therefore has the longer wavelength, and λred/λblue = Eblue/Ered = 2.6/1.9. Option A treats λ as proportional to E. Option C wrongly squares the ratio. Option D confuses equal speed with equal wavelength.✓ 1Answer B

Answer: B

Common error: A higher photon energy means a shorter wavelength. Check that the ratio you choose gives red the longer wavelength.

11IBP2S-035-01-02Energy levels & photons Paper 1AEasy1 mark

In a hydrogen discharge tube used for teaching, the applied voltage is adjusted so that electrons in the excited atoms reach the n = 3 level and no higher. All atoms then return to the ground state (n = 1). How many different wavelengths can be emitted?

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Starting from n = 3, the possible downward transitions are 3→1, 3→2 and then 2→1. Each transition has a different energy difference, so 3 different wavelengths can be emitted. Option A counts only the direct jump to the ground state. Option B counts only the jumps that start from n = 3 and forgets 2→1. Option D counts transitions involving a fourth level that is never reached.✓ 1Answer C

Answer: C

Exam tip: List every pair of levels at or below the highest occupied level. Each pair gives one line. For N levels there are N(N−1)/2 lines.

12IBP2S-035-01-08Emission & absorption spectra Paper 1AEasy1 mark

A student looks at the light from a neon advertising sign through a diffraction grating. She sees a set of bright, sharp coloured lines separated by dark gaps. What does this observation provide direct evidence for?

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Each bright line corresponds to a single photon energy. Only certain photon energies are emitted, so the electrons can only move between fixed, discrete energy levels. Option A comes from alpha-particle scattering, not from spectra. Option C would produce a continuous spectrum. Option D is wrong because the lines come from transitions between bound levels, and no ionisation is needed.✓ 1Answer B

Answer: B

Exam tip: Line spectra point to discrete energy levels. Rutherford scattering points to a small, dense nucleus. Keep the two pieces of evidence separate.

13IBP2S-035-00-01Emission & absorption spectra Paper 1AEasy1 mark

Astronomers study light from a star that has passed through the atmosphere of an orbiting exoplanet. They compare the dark absorption lines of one element in the atmosphere with the bright emission lines of the same element measured in a laboratory. Any motion of the star and planet relative to Earth can be ignored. Which statement is correct?

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Absorption and emission both involve the same discrete energy differences in the atom. A photon is absorbed only if its energy hf matches a level difference ΔE. A transition between the same two levels in the opposite direction emits a photon of that same ΔE. The wavelength λ = hc/ΔE is therefore identical for both, so every absorption line matches an emission line. A cool gas may not show all the emission lines in absorption, because most of its atoms are in the lowest levels. Option B is wrong because absorbing energy does not shift the wavelength. Option C is wrong because temperature does not change the level differences. Option D is wrong because both processes use the same levels. Matching lines in this way is how elements are identified from absorption spectra.✓ 1Answer A

Answer: A

Common error: Absorption does not shift a line's wavelength. It removes photons at exactly the wavelengths the element can emit.

14IBP2S-035-00-02Ionization Paper 1AEasy1 mark

Starlight in a nebula has excited some hydrogen atoms into the n = 2 level. The energy levels of hydrogen are −13.6 eV for n = 1 and −3.40 eV for n = 2. What is the minimum energy needed to ionize a hydrogen atom that is in the n = 2 level?

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Ionization moves the electron from its current level up to E = 0. From n = 2, the energy needed is 0 − (−3.40 eV) = 3.40 eV. The value 10.2 eV is the n = 1 → n = 2 excitation energy. The value 13.6 eV applies only to an atom in the ground state.✓ 1Answer C

Answer: C

Exam tip: The ionization energy from any level is the magnitude of that level's (negative) energy.

15IBP2S-035-00-05Emission & absorption spectra Paper 1AEasy1 mark

A shop sign uses a glass tube of low-pressure neon gas through which a current passes. What is the appearance of the spectrum of the light from the tube, and why?

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Excited atoms in a low-pressure gas drop between discrete energy levels. Each drop emits a photon with a definite energy, ΔE = hf, so only certain wavelengths appear. The result is an emission line spectrum: bright lines on a dark background.✓ 1Answer B

Answer: B

Common error: A continuous spectrum comes from hot dense matter such as a filament. A low-pressure gas gives discrete lines.

16IBP2S-035-00-10Energy levels & photons Paper 1AEasy1 mark

Strontium salts give fireworks a deep red colour. What determines the wavelength of the red light emitted by the strontium atoms?

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An emitted photon carries away exactly the energy lost by the atom, so hf = hc/λ = Eupper − Elower. The colour is therefore fixed by the difference between the two levels. The upper level's energy alone does not set it. Temperature only changes how many atoms are excited, not the line wavelength.✓ 1Answer C

Answer: C

Exam tip: Always link photon energy to a difference between two levels, never to a single level.

17IBP2S-035-02-01Rutherford scattering Paper 1AEasy1 mark

In the Geiger–Marsden experiment, a beam of alpha particles was directed at a thin gold foil. The great majority of the alpha particles passed through the foil with no deflection or with only a very small deflection.
What does this observation suggest about the atom?

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If most alpha particles pass through almost undeflected, they rarely come close to anything that exerts a large force on them. So most of the atom's volume is effectively empty space. The sign of the nuclear charge comes from the rare large-angle deflections, not from this observation. Energy levels come from atomic spectra. Neutrons were not revealed by this experiment.✓ 1Answer B

Answer: B

Exam tip: Link each observation to its own conclusion. 'Most pass straight through' shows the atom is mostly empty space. 'A few rebound' shows a small, dense, charged nucleus.

18IBP2S-035-02-02Rutherford scattering Paper 1AEasy1 mark

Before the Geiger–Marsden experiment, the Thomson model described the atom as a sphere of positive charge spread evenly throughout its volume, with electrons embedded in it.
If this model were correct, what would be expected when alpha particles are fired at a thin gold foil?

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In the Thomson model the positive charge is spread out, so the electric field inside the atom is weak everywhere. Each alpha particle would feel only a small net force and be deflected through a small angle. None would be turned back. Absorption is not predicted, and large-scale reflection would need concentrated charge. Alpha particles are positive, so they are repelled by positive charge, not attracted.✓ 1Answer B

Answer: B

Common error: Alpha particles are positively charged. Near positive charge they are repelled, never pulled inwards.

19IBP2S-035-03-01Nuclear notation Paper 1AEasy1 mark

Technetium-99m is used as a tracer in hospital bone scans. Its nucleus is represented as 9943Tc. How many neutrons does this nucleus contain?

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The neutron number is N = A − Z = 99 − 43 = 56. The value 43 is the proton number and 99 is the nucleon number. 142 comes from adding A and Z instead of subtracting.✓ 1Answer B

Answer: B

Common error: The top number is the total number of nucleons, not the number of neutrons. Subtract the proton number to get the neutron number.

20IBP2S-035-03-02Nuclear notation Paper 1AEasy1 mark

Calcium in bone is present as Ca2+ ions. Consider a Ca2+ ion formed from an atom of 4020Ca. Which row gives the numbers of protons, neutrons and electrons in this ion?

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Z = 20, so there are 20 protons. N = 40 − 20 = 20 neutrons. A 2+ charge means the ion has lost two electrons, leaving 20 − 2 = 18 electrons. The 22-electron option adds electrons instead of removing them. The 18-proton option changes the nucleus, which ionisation does not do. The 40-neutron option takes the nucleon number as the neutron number.✓ 1Answer A

Answer: A

Exam tip: Ionisation changes only the number of electrons. The nuclear notation for the nucleus stays the same.

21IBP2S-035-03-06Nuclear notation Paper 1AEasy1 mark

A particle detector on a space probe identifies alpha particles in the solar wind. Compared with a neutral atom of 42He, an alpha particle

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An alpha particle is a helium-4 nucleus: 2 protons and 2 neutrons. A neutral helium atom has this same nucleus plus 2 electrons. The nucleon number, proton number and neutron number are therefore identical. The only difference is that the alpha particle has no electrons.✓ 1Answer A

Answer: A

Common error: The notation ⁴₂He describes the nucleus. It does not tell you how many electrons are present.

22E-1A-03Energy levels & photons Paper 1AMedium1 mark

An electron in an atom moves from an energy level of −3.40 eV to an energy level of −13.6 eV.

Which row describes the photon involved in this transition?

−3.40 eV−13.6 eVelectron
PhotonFrequency / Hz
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Step 1The electron moves to a lower level, so the atom loses energy and a photon of energy ΔE = −3.40 − (−13.6) = 10.2 eV = 1.63 × 10−18 J is emitted.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = ΔE/h = 1.63 × 10−18/6.63 × 10−34 = 2.5 × 1015 Hz.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe frequency is right, but the electron falls to a lower level, so energy is given out: the photon is emitted, not absorbed.
  • BCorrect: ΔE = 10.2 eV = 1.63 × 10−18 J is given out as a photon of frequency ΔE/h = 2.5 × 1015 Hz.
  • CThis uses 3.40 + 13.6 = 17.0 eV, adding the sizes of the two levels instead of finding their difference.
  • DThis divides 10.2 by h without converting the energy from electronvolts to joules.
23E-1A-05Emission & absorption spectra Paper 1AMedium1 mark

The emission spectrum of a hot gas consists of bright lines. When white light passes through the same gas when it is cool, dark lines appear on a continuous spectrum at the same wavelengths as the bright lines.

What do these observations show?

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Marking pointMarkNotes
Step 1A sharp line means photons of one definite energy hf, so an atom can gain or lose only certain amounts of energy: its energy levels are discrete.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The bright and dark lines coincide because the same pairs of levels are involved whether a photon is emitted or absorbed.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AAbsorption also happens only at particular energies — that is why the absorption spectrum shows dark lines rather than a general dimming.
  • BA continuous range of energies would give continuous spectra, not sharp lines.
  • CCorrect: only certain photon energies are emitted or absorbed, and the lines occur at the same wavelengths, so the same set of level differences is involved.
  • DThe bright and dark lines occur at exactly the same wavelengths, so they come from the same atoms.
24E-1A-07Emission & absorption spectra Paper 1AMedium1 mark

Dark lines are seen in the spectrum of sunlight at wavelengths that match the bright lines of helium measured in a laboratory.

What can be concluded?

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Step 1The continuous spectrum from the hot interior passes through the cooler gas of the Sun's outer layers, whose atoms absorb photons with energies equal to their own level differences.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each element has its own set of level differences, so dark lines at helium's wavelengths show that helium is present in that gas.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe matching wavelengths are the fingerprint of helium wherever it is found, including the Sun.
  • BOther elements have their own lines in the solar spectrum; a match shows that helium is present, not that it is the only element.
  • CThe positions of the lines depend on the energy levels of the atom, not on the temperature of the gas.
  • DCorrect: helium atoms in the Sun's outer layers absorb exactly the photon energies they would emit, removing those wavelengths from the continuous spectrum.
25E-1A-27Ionization Paper 1AMedium1 mark

The ground state of an atom is at −13.6 eV. A photon of energy 15.0 eV is incident on an atom in its ground state.

Which row describes what happens?

0ionised−13.6 eVground statephoton of energy 15.0 eV incident on the ground-state atom
What happens to the photonKinetic energy of the freed electron
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Step 1Freeing the electron from the ground state needs 0 − (−13.6) = 13.6 eV; the photon carries more, so it is absorbed and the atom is ionized (a free electron can have any energy, so no exact match is needed).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic energy of the freed electron = 15.0 − 13.6 = 1.4 eV.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AAn exact match is needed only between two bound levels; above 13.6 eV the electron is freed and can take any kinetic energy.
  • BCorrect: 13.6 eV is used to free the electron and the remaining 15.0 − 13.6 = 1.4 eV becomes its kinetic energy.
  • CThis gives the electron all of the photon's energy, forgetting the 13.6 eV needed to free it.
  • DThis calculates 15.0 − (−13.6) = 28.6 eV, adding the ionization energy instead of subtracting it.
26E-1A-29Energy levels & photons Paper 1AMedium1 mark

The diagram shows three energy levels of an atom. Photon P is emitted in the transition from −0.85 eV to −3.40 eV and photon Q in the transition from −1.51 eV to −3.40 eV.

What is (wavelength of Q)/(wavelength of P)?

Three energy levels at −0.85 eV, −1.51 eV and −3.40 eV; transition P from −0.85 eV to −3.40 eV and transition Q from −1.51 eV to −3.40 eV.Energy / eV−0.85−1.51−3.40PQ
Diagram NOT accurately drawn
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Step 1Photon energies: EP = −0.85 − (−3.40) = 2.55 eV and EQ = −1.51 − (−3.40) = 1.89 eV.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2λ = hc/E, so λQ/λP = EP/EQ = 2.55/1.89 = 1.35.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis is 0.85/1.51, the ratio of the starting levels; the photon energy is the difference between two levels.
  • BThis is EQ/EP = 1.89/2.55, which is the ratio of P's wavelength to Q's — wavelength is inversely proportional to photon energy.
  • CCorrect: λ ∝ 1/E, so λQ/λP = 2.55/1.89 = 1.35.
  • DThis is (2.55/1.89)2, as if the wavelength were inversely proportional to the square of the energy.
27E-1A-30Rutherford scattering Paper 1AMedium1 mark

In the Geiger–Marsden–Rutherford experiment, alpha particles were directed at a thin gold foil.

Which row correctly links an observation to the conclusion drawn from it?

ObservationConclusion
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Step 1Turning a fast alpha particle back needs a very large repulsive force, which is possible only if the positive charge and most of the mass are concentrated in a very small volume — the nucleus.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AUndeflected paths show that the atom is mostly empty space; on their own they say nothing about the sign of the charge.
  • BElectrons are far too light to deflect alpha particles noticeably; small deflections come from passing a nucleus at a larger distance.
  • CThe scattering depends on the charge and mass of the nucleus; it gives no information about neutrons, which were discovered later.
  • DCorrect: only a small, massive, positively charged nucleus can exert a force large enough to turn an alpha particle back.
28E-1A-46Emission & absorption spectra Paper 1AMedium1 mark

Light from the bright surface of the Sun has a continuous spectrum crossed by dark absorption lines. The dark lines are produced by a thin layer of cooler gas just above the surface. During a total eclipse the Moon hides the bright surface for a few seconds, and only light from this thin layer of gas, seen against the dark sky, enters a spectrometer.

What spectrum is observed during these few seconds?

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Marking pointMarkNotes
Step 1Atoms in the thin layer absorb photons whose energies equal the differences between their energy levels; the excited atoms then return to lower levels and emit photons of these same energies in all directions.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With the bright surface hidden there is no continuous light behind the gas, so only the photons emitted by the gas are seen: bright lines on a dark background at the wavelengths of the normal dark lines.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the gas emits photons of the same energies that it absorbs; with nothing bright behind it, these appear as bright lines at the wavelengths of the dark lines.
  • BEmission and absorption between the same pair of energy levels involve the same energy difference, so the bright lines must be at the same wavelengths as the dark lines.
  • CDark lines appear only when a source of continuous light lies behind the absorbing gas; during the eclipse the bright surface is hidden, so there is no light for the gas to remove.
  • DA thin gas emits only photons whose energies equal the differences between its discrete energy levels, so its spectrum is a line spectrum, not a continuous one.
29E-1A-48Energy levels & photons Paper 1AMedium1 mark

A laser projector contains a red laser of wavelength 650 nm and a blue laser of wavelength 450 nm. Both lasers emit light with the same power.

What is (number of photons emitted per second by the red laser)/(number of photons emitted per second by the blue laser)?

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Marking pointMarkNotes
Step 1Each photon has energy E = hc/λ, so the number of photons per second is P/E = Pλ/hc, which is proportional to λ at the same power.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = 650/450 = 1.44.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A450/650 = 0.69 treats the photon energy as proportional to the wavelength; red photons carry less energy, so more of them are needed.
  • BEqual power means equal energy per second, not an equal number of photons, because the photons carry different energies.
  • CCorrect: a red photon has less energy than a blue photon, so the red laser emits 650/450 = 1.44 times as many photons each second.
  • D(650/450)2 = 2.09 squares the ratio; the number of photons per second is proportional to λ, not to λ2.
30E-1A-64Nuclear notation Paper 1AMedium1 mark

The nucleus of a nuclide has a charge of +1.28 × 10−18 C and contains 18 nucleons.

Which is the nuclear notation for this nuclide? (X represents the chemical symbol.)

Show mark scheme
Marking pointMarkNotes
Step 1Proton number Z = charge/e = 1.28 × 10−18/1.60 × 10−19 = 8.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The nucleon number A = 18 is written above and Z = 8 below: 188X.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: Z = 1.28 × 10−18/1.60 × 10−19 = 8 protons and A = 18 nucleons, so the nuclide is 188X.
  • BThe two numbers have been swapped: the nucleon number goes at the top and the proton number at the bottom.
  • C10 is the neutron number, 18 − 8; the lower number must be the proton number Z.
  • DThis treats the 18 nucleons as 18 neutrons and adds the 8 protons; the nucleon number already includes the protons.
31E-1A-67Rutherford scattering Paper 1AMedium1 mark

The diagram shows the path of an alpha particle that passes close to a single nucleus N in a thin metal foil. Q is the point on the path closest to N; P and R are points far from N.

Which row gives the point at which the speed of the alpha particle is least and the direction of the force on the alpha particle at that point?

Path of an alpha particle passing a nucleus N. The particle enters from the left moving horizontally, curves away from N and leaves deflected upwards. P is on the incoming part of the path, Q is the point of the path nearest to N, and R is on the outgoing part.NPQRalpha particle
Diagram NOT accurately drawn
Point of least speedDirection of force at that point
Show mark scheme
Marking pointMarkNotes
Step 1Both N and the alpha particle are positive, so the electric force on the alpha particle acts along the line from N to the particle, directly away from N.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2While the particle approaches N the repulsive force does negative work on it, so its kinetic energy and speed are least at Q, the point nearest N; it speeds up again as it moves away.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe speed is least at Q, but the nucleus and the alpha particle are both positive, so the force is repulsive.
  • BNo energy is lost: the force does positive work as the particle moves away from N, so at R its speed is again as large as at P.
  • CAt Q the particle moves at right angles to the line from N; the electric force acts along that line, not along the velocity.
  • DCorrect: the speed is least at Q, the point of the path nearest N, where the repulsive force acts directly away from N.
32E-1A-69Rutherford scattering Paper 1AMedium1 mark

In a Geiger–Marsden–Rutherford experiment, a narrow beam of alpha particles is directed at a thin metal foil in a vacuum, and the fraction of the alpha particles that are scattered through more than 90° is measured.

Which change, with everything else unchanged, increases this fraction?

Show mark scheme
Marking pointMarkNotes
Step 1An alpha particle is turned through more than 90° only if it passes so close to a nucleus that the repulsive electric force on it is very large.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2A slower alpha particle, with less kinetic energy and momentum, is turned back by a smaller force, so it need not pass as close to a nucleus: a larger fraction of the beam is scattered through more than 90°.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AA faster alpha particle must pass even closer to a nucleus to be turned back, so the fraction decreases; the electric force does not grow with the speed of the particle.
  • BCorrect: a slower alpha particle is turned back by a smaller repulsive force, so more of the alpha particles that pass near a nucleus are scattered through more than 90°.
  • CMore alpha particles per second increases the number scattered through more than 90° in a given time, but not the fraction of the beam that is scattered through these angles.
  • DA thinner foil has fewer nuclei in the path of the beam, so fewer alpha particles pass close to a nucleus and the fraction decreases.
33E-1A-70Emission & absorption spectra Paper 1AMedium1 mark

Two discharge tubes contain chlorine atoms. One contains only 3517Cl and the other only 3717Cl. The two tubes produce line spectra that are almost identical.

What is the reason for this?

Show mark scheme
Marking pointMarkNotes
Step 1Spectral lines come from transitions of the electrons between atomic energy levels; these levels are set by the charge of the nucleus, Ze, and the number of electrons, both fixed by Z = 17 for both isotopes.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThe isotopes have different numbers of neutrons, 18 and 20; neutrons are uncharged and do not affect the electric force on the electrons.
  • BVisible line spectra come from electron transitions between atomic energy levels, not from changes inside the nucleus.
  • CThe masses differ by about 6 %, and the speed of the atoms does not decide which photon energies are emitted.
  • DCorrect: the same proton number gives the same nuclear charge and the same 17 electrons, so the atomic energy levels, and therefore the lines, are the same.
34E-1A-72Energy levels & photons Paper 1AMedium1 mark

The diagram shows three energy levels of an atom, measured from the ground state. Atoms of a gas are excited to the 3.54 eV level and then return to the ground state.

What is the longest wavelength of the photons emitted?

Three energy levels drawn to scale: ground state 0, then 2.48 eV and 3.54 eV above the ground state.0 (ground state)2.48 eV3.54 eV
Energy levels drawn to scale
Show mark scheme
Marking pointMarkNotes
Step 1The longest wavelength comes from the smallest energy difference: 3.54 − 2.48 = 1.06 eV = 1.70 × 10−19 J.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2λ = hc/ΔE = 6.63 × 10−34 × 3.00 × 108/1.70 × 10−19 = 1.17 × 10−6 m ≈ 1170 nm.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • A206 nm uses 3.54 + 2.48 = 6.02 eV; a photon's energy is the difference between two levels, not their sum.
  • B351 nm comes from the largest gap, 3.54 eV: this is the shortest wavelength emitted, not the longest.
  • C501 nm comes from the 2.48 eV → ground-state transition; the 3.54 eV → 2.48 eV transition has a smaller energy and so a longer wavelength.
  • DCorrect: the smallest gap, 1.06 eV, gives the longest wavelength, hc/ΔE = 1170 nm (in the infrared).
35IBP2S-035-01-03Emission & absorption spectra Paper 1AMedium1 mark

A space telescope records the spectrum of a star. The spectrum is continuous but crossed by dark lines at specific wavelengths. Which statement best explains the dark lines?

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Marking pointMarkNotes
The hot interior emits a continuous spectrum. Atoms in the cooler outer layers absorb only photons whose energy equals a difference between their energy levels. The excited atoms then de-excite and emit photons in all directions, so the intensity along the line of sight towards the observer drops at those wavelengths. Option B is wrong because excited atoms do de-excite. Option C is wrong because the interior spectrum is continuous. Option D is wrong because the process is absorption and re-emission, not reflection.✓ 1Answer A

Answer: A

Common error: Absorbed energy is not 'lost forever'. The dark lines appear because re-emission is in random directions, not because the atoms never emit again.

36IBP2S-035-01-04Energy levels & photons Paper 1AMedium1 mark

Water purifiers use mercury lamps to produce ultraviolet light. In a mercury atom, the ground state has energy −10.4 eV and an excited state has energy −5.5 eV. What is the wavelength of the photon emitted when an electron falls from this excited state to the ground state?
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

−5.5 eVexcited state−10.4 eVground statephoton, λ = ?
Show mark scheme
Marking pointMarkNotes
The photon energy equals the difference between the levels: ΔE = −5.5 − (−10.4) = 4.9 eV = 4.9 × 1.60×10−19 = 7.84×10−19 J. Then λ = hc/ΔE = (6.63×10−34 × 3.00×108)/7.84×10−19 = 2.5×10−7 m. Option A uses 10.4 eV, the energy of the ground level alone. Option B uses 5.5 eV, the energy of the excited level alone. Option D comes from leaving the energy in eV instead of converting it to joules.✓ 1Answer C

Answer: C

Exam tip: Photon energy is always the difference between two levels, never the value of a single level. Convert eV to J before using λ = hc/E.

37IBP2S-035-01-05Ionization Paper 1AMedium1 mark

An atom in a gas has a ground state of −5.0 eV, a first excited state of −2.6 eV and a second excited state of −1.4 eV. Two experiments are carried out on separate ground-state atoms: one is struck by a photon of energy 3.0 eV, and the other is struck by a free electron of kinetic energy 3.0 eV. Which statement is correct?

−1.4 eVsecond excited state−2.6 eVfirst excited state−5.0 eVground state
Show mark scheme
Marking pointMarkNotes
From the ground state, the possible excitation energies are 2.4 eV (to −2.6 eV) and 3.6 eV (to −1.4 eV). A photon is absorbed only if its whole energy exactly matches a level difference, and 3.0 eV matches neither, so the photon is not absorbed. A colliding electron can give up part of its kinetic energy. It can transfer 2.4 eV to raise the atom to the first excited state and continue with 0.6 eV. Options A and B wrongly assume the photon can be partly absorbed. Option D wrongly applies the exact-match rule to electrons.✓ 1Answer C

Answer: C

Common error: Photons are absorbed whole or not at all. Colliding electrons can transfer any amount of energy up to their kinetic energy.

38IBP2S-035-01-10Emission & absorption spectra Paper 1AMedium1 mark

Visible light from a hot star passes through a cold cloud of hydrogen gas in interstellar space at about 20 K. The visible Balmer absorption lines of hydrogen, which involve transitions starting from n = 2, are almost absent from the spectrum. What is the best explanation?

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Marking pointMarkNotes
A Balmer absorption line needs an atom that starts in n = 2. At 20 K, collisions do not have enough energy to excite atoms to n = 2 (about 10 eV above the ground state), so almost every atom is in n = 1. Ground-state atoms can only absorb the higher-energy ultraviolet (Lyman) photons. Option A is wrong because Balmer photons have less energy than Lyman photons. Option C is wrong because cold atoms are not excited and so cannot emit. Option D is wrong because ionised hydrogen has no bound electron to make the transition.✓ 1Answer B

Answer: B

Exam tip: An absorption line needs atoms already in the lower level of that transition. Ask which levels are actually occupied at the temperature given.

39IBP2S-035-00-03Ionization Paper 1AMedium1 mark

Radiation with a continuous range of photon energies, from visible to far ultraviolet, passes through cold mercury vapour. Every atom in the vapour is in its ground state. In a simplified model, the only bound energy levels of mercury are −10.4 eV (ground state), −5.5 eV, −3.7 eV and −1.6 eV. Which photon energy will not be absorbed by the vapour?

−1.6 eV−3.7 eV−5.5 eV−10.4 eVground state
Show mark scheme
Marking pointMarkNotes
Atoms in the ground state can only absorb photons whose energy lifts the electron from the ground state. The allowed energies are 4.9 eV (to −5.5 eV), 6.7 eV (to −3.7 eV) and 8.8 eV (to −1.6 eV). A photon of 10.4 eV or more ionizes the atom, so the 11.0 eV photon is also absorbed. The energy 1.8 eV equals the gap between −5.5 eV and −3.7 eV. No atoms are in the −5.5 eV level, and no level lies 1.8 eV above the ground state. So 1.8 eV is not absorbed.✓ 1Answer B

Answer: B

Common error: Photons with more than the ionization energy do not need to match a level exactly. The excess energy becomes kinetic energy of the freed electron.

40IBP2S-035-00-04Emission & absorption spectra Paper 1AMedium1 mark

Light from a hot tungsten filament passes through a glass cell containing cool sodium vapour. Observer P looks along the beam, straight through the cell. Observer Q looks at the cell from the side, at 90° to the beam, against a dark background. What do P and Q see when the light is passed through a spectrometer?

hot tungstenfilamentglass cell of cool sodium vapourPQlooks along the beamlooks from the side
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The sodium atoms absorb photons at their characteristic wavelengths, so P sees a continuous spectrum with dark lines. The excited atoms then de-excite and re-emit photons of these wavelengths in random directions. Some of this light reaches Q, who therefore sees bright emission lines at the same wavelengths against a dark background.✓ 1Answer C

Answer: C

Exam tip: An absorption line is dark because the re-emitted light is spread in all directions, so very little continues along the original beam.

41IBP2S-035-00-06Energy levels & photons Paper 1AMedium1 mark

A forensic lab uses a gas-discharge lamp. Two of its spectral lines come from transitions with photon energies E1 = 2.0 eV and E2 = 3.0 eV. What is the ratio \(\frac{\lambda_1}{\lambda_2}\) of the wavelengths of these two lines?

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Marking pointMarkNotes
Since E = hc/λ, wavelength is inversely proportional to photon energy. So λ1/λ2 = E2/E1 = 3.0/2.0 = 1.5. The lower-energy photon has the longer wavelength.✓ 1Answer B

Answer: B

Common error: 0.67 comes from treating λ as proportional to E. Lower photon energy means longer wavelength.

42IBP2S-035-00-09Ionization Paper 1AMedium1 mark

Ultraviolet radiation of wavelength 60 nm from a hot young star ionizes ground-state hydrogen atoms in a nearby gas cloud. The ionization energy of hydrogen is 13.6 eV. What is the maximum kinetic energy of a freed electron?
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

Show mark scheme
Marking pointMarkNotes
Photon energy E = hc/λ = (6.63×10−34 × 3.00×108)/(60×10−9) = 3.32×10−18 J = 20.7 eV.
Removing the electron uses 13.6 eV, leaving 20.7 − 13.6 = 7.1 eV as the electron's kinetic energy.
The distractors come from common slips:
  • 20.7 eV ignores the ionization energy.
  • 34.3 eV adds the two energies instead of subtracting.
✓ 1Answer A

Answer: A

Common error: Energy is conserved: kinetic energy of the freed electron = photon energy − ionization energy.

43IBP2S-035-02-03Rutherford scattering Paper 1AMedium1 mark

Why was a very thin metal foil used in the Geiger–Marsden scattering experiment?

Show mark scheme
Marking pointMarkNotes
A thin foil makes it unlikely that an alpha particle passes close to more than one nucleus. A large deflection can then be interpreted as a single encounter with one nucleus, rather than the sum of many small deflections. Electrons are far too light to turn an alpha particle back. The scintillations occur on a separate screen, not the foil. Thickness does not change the spacing of nuclei.✓ 1Answer A

Answer: A

Exam tip: 'Single scattering' is the key phrase when explaining why the foil must be thin.

44IBP2S-035-02-04Rutherford scattering Paper 1AMedium1 mark

A semiconductor engineer uses alpha-particle backscattering to check silicon wafers for gold contamination. Per atom, gold impurity atoms send far more alpha particles back towards the detector than silicon atoms do.
Which property of gold is mainly responsible?

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Marking pointMarkNotes
Large-angle scattering is caused by the Coulomb repulsion between the alpha particle and a nucleus. The force is proportional to the nuclear charge Ze: Z = 79 for gold compared with Z = 14 for silicon. A gold nucleus can therefore turn an alpha particle back from a greater distance. Electrons are too light to cause backscattering. Atomic sizes are similar. Bulk density is irrelevant when comparing individual atoms.✓ 1Answer B

Answer: B

Common error: Electrons have negligible effect on alpha-particle paths. Only the nucleus is massive and charged enough to produce large deflections.

45IBP2S-035-02-10Rutherford scattering Paper 1AMedium1 mark

What is the main reason the alpha source, foil and detector in a Rutherford scattering experiment are placed inside an evacuated chamber?

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Marking pointMarkNotes
Alpha particles are strongly ionising, so they lose energy quickly in air. Their range in air is only a few centimetres. Without a vacuum, most would never reach the foil or the detector with their original energy. Air nuclei (Z ≈ 7–8) are light and rarely cause large deflections. Gold does not oxidise. Air does not neutralise a moving alpha particle.✓ 1Answer A

Answer: A

Exam tip: Remember the typical range of alpha particles in air: a few centimetres. It explains both the vacuum and why the source sits close to the foil.

46IBP2S-035-03-03Nuclear notation Paper 1AMedium1 mark

Nuclear fuel is enriched by increasing the proportion of 23592U relative to 23892U. Chemical methods cannot separate the two. Enrichment in gas centrifuges is possible because atoms of the two isotopes have different

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Marking pointMarkNotes
Isotopes have the same proton number, here 92. They therefore have the same number of electrons in the neutral atom and the same chemistry. They differ only in neutron number (143 and 146), so their atoms have different masses. A centrifuge separates particles by mass.✓ 1Answer C

Answer: C

Common error: Different neutron numbers do not give different chemistry. Chemical behaviour is set by the electrons, and the number of electrons equals Z for a neutral atom.

47IBP2S-035-03-04Nuclear notation Paper 1AMedium1 mark

Caesium-137, 13755Cs, is a contaminant monitored after nuclear accidents. Which nuclide has the same number of neutrons as 13755Cs?

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Marking pointMarkNotes
Cs-137 has N = 137 − 55 = 82. For Ba-138, N = 138 − 56 = 82, which matches. Ba-137 has the same nucleon number but N = 81. Cs-135 is an isotope of caesium with N = 80. Xe-137 has N = 83.✓ 1Answer D

Answer: D

Exam tip: Work out N = A − Z for every option. Equal nucleon numbers or equal proton numbers do not mean equal neutron numbers.

48IBP2S-035-03-05Nuclear notation Paper 1AMedium1 mark

A study of corrosion in steel pipelines involves the iron isotope 5626Fe. Nucleus X has one fewer proton and three more neutrons than a 5626Fe nucleus. Which is the correct notation for X?

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Marking pointMarkNotes
For X, Z = 26 − 1 = 25. Fe-56 has N = 30, so X has N = 33. The nucleon number is A = Z + N = 25 + 33 = 58, giving 5825X. The value 59 adds the neutrons but ignores the lost proton. Z = 27 adds a proton instead of removing one. The value 55 ignores the extra neutrons.✓ 1Answer B

Answer: B

Exam tip: Find Z and N separately, then add them to get A. Do not adjust A by one change only.

49IBP2S-035-03-07Nuclear notation Paper 1AMedium1 mark

A nuclear submarine reactor uses fuel containing 23592U. The unified atomic mass unit is 1.66 × 10−27 kg. What is the approximate mass of one 23592U nucleus?

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Marking pointMarkNotes
Each nucleon has a mass of about 1 u, so the nucleus has a mass of about 235 u. This gives 235 × 1.66 × 10−27 ≈ 3.9 × 10−25 kg. The value 1.5 × 10−25 kg counts only the 92 protons. The value 2.4 × 10−25 kg counts only the 143 neutrons. The value 3.9 × 10−22 is the mass in grams (235 g ÷ 6.02 × 1023), wrongly labelled as kilograms.✓ 1Answer C

Answer: C

Exam tip: The nucleon number A gives the mass of a nucleus in u to a good approximation. Convert to kg using 1 u = 1.66 × 10⁻²⁷ kg.

50IBP2S-035-03-08Rutherford scattering Paper 1AMedium1 mark

A cyclotron used to produce medical isotopes can accelerate deuterons (21H nuclei) or alpha particles (42He nuclei). What is the approximate value of \(\frac{\text{charge-to-mass ratio of a deuteron}}{\text{charge-to-mass ratio of an alpha particle}}\)?

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Marking pointMarkNotes
For the deuteron, q/m ≈ e/(2 u). For the alpha particle, q/m ≈ 2e/(4 u) = e/(2 u). The ratio is therefore approximately 1. The value 1/2 compares masses only. The value 2 compares charges only. The value 1/4 multiplies the two ratios instead of dividing them.✓ 1Answer D

Answer: D

Common error: Take the charge from Z and the mass from A. Both are needed, because the alpha particle has twice the charge and twice the mass of the deuteron.

51IBP2S-035-03-09Nuclear notation Paper 1AMedium1 mark

Some nuclear reactors use heavy water, D2O, as a moderator. D is the isotope 21H and the oxygen is 168O. How many neutrons are there in one molecule of heavy water?

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Marking pointMarkNotes
Each 21H nucleus has 2 − 1 = 1 neutron, and 168O has 16 − 8 = 8 neutrons. The total is 2 × 1 + 8 = 10. The answer 8 is correct for ordinary water, where the hydrogen nuclei have no neutrons. 18 is the nucleon number of ordinary water, and 20 is the nucleon number of heavy water.✓ 1Answer B

Answer: B

Exam tip: In a molecule, count the particles in each nucleus separately. Multiply by the number of atoms of each kind, then add.

52E-1A-06Emission & absorption spectra Paper 1AHard1 mark

The diagram shows the four lowest energy levels of an atom. A cool gas of these atoms, all in the ground state, is illuminated with light containing a wide range of photon energies.

Which photon energy can be absorbed by the gas?

Four energy levels of an atom at −1.6 eV, −3.7 eV, −5.5 eV and −10.4 eV (ground state), drawn to scale.Energy / eV−10.4ground state−5.5−3.7−1.6
Diagram drawn to scale
Show mark scheme
Marking pointMarkNotes
Step 1In a cool gas the atoms are in the ground state, so a photon is absorbed only if its energy equals the difference between the ground state and a higher level: 4.9 eV, 6.7 eV or 8.8 eV.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Only 4.9 eV = −5.5 − (−10.4) is among the options; 1.8 eV and 2.1 eV are gaps between excited levels, which are empty.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A1.8 eV = −3.7 − (−5.5) would need atoms already in the −5.5 eV level; in a cool gas that level is empty.
  • B2.1 eV = −1.6 − (−3.7) would need atoms already in the −3.7 eV level; in a cool gas that level is empty.
  • CCorrect: 4.9 eV = −5.5 − (−10.4) exactly matches the gap from the ground state to the first excited level.
  • D5.5 eV is the size of the −5.5 eV level read as a photon energy; it matches no gap from the ground state (the photon energy must be a difference between two levels).
53E-1A-26Energy levels & photons Paper 1AHard1 mark

The diagram shows energy levels of an atom. An atom in the ground state absorbs a photon of energy 12.09 eV.

Which row gives the level to which the electron is raised and the number of different photon energies that can then be emitted as it returns to the ground state?

Energy levels of an atom at −13.6 eV (ground state), −3.40 eV, −1.51 eV and −0.85 eV, with the ionization level at 0.Energy / eV0ionization−0.85−1.51−3.40−13.6ground state
Diagram NOT accurately drawn
Level reachedNumber of different photon energies
Show mark scheme
Marking pointMarkNotes
Step 1New energy = −13.6 + 12.09 = −1.51 eV, which is exactly a level, so the photon is absorbed.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2From −1.51 eV the electron can fall straight to −13.6 eV, or to −3.40 eV and then to −13.6 eV: three different photon energies.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe electron need not return in a single jump: it can also stop at −3.40 eV on the way down, giving two further photon energies.
  • BThis counts only the two-step route (−1.51 → −3.40 → −13.6 eV) and forgets the direct fall from −1.51 eV to −13.6 eV.
  • CCorrect: −13.6 + 12.09 = −1.51 eV; the routes −1.51 → −13.6, −1.51 → −3.40 and −3.40 → −13.6 eV give three different photon energies.
  • DThis assumes the photon lifts the electron to the highest level shown; reaching −0.85 eV would need 12.75 eV, not 12.09 eV.
54E-1A-47Energy levels & photons Paper 1AHard1 mark

An atom in its ground state absorbs a photon of wavelength λ1. The atom then returns to the ground state by emitting two photons in turn, of wavelengths λ2 and λ3, through one intermediate energy level.

What is λ1?

intermediate levelground stateλ₁λ₂λ₃
Show mark scheme
Marking pointMarkNotes
Step 1Energy is conserved: the energy absorbed equals the total energy emitted, so hc/λ1 = hc/λ2 + hc/λ3.—All 2 steps must be completed — there is no mark for a part-answer.
Step 21/λ1 = 1/λ2 + 1/λ3 = (λ2 + λ3)/λ2λ3, so λ1 = λ2λ3/(λ2 + λ3).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the photon energies add, and photon energy is proportional to 1/λ, so the reciprocals of the wavelengths add.
  • BThis adds the wavelengths as though the photon energy were proportional to the wavelength; it is proportional to 1/λ.
  • CThis is 1/λ1, not λ1: the final step of taking the reciprocal has been left out.
  • DThis subtracts one photon energy from the other; the absorbed energy equals the sum of the two emitted energies.
55E-1A-71Emission & absorption spectra Paper 1AHard1 mark

Astronomers record the spectrum of a star while one of its planets passes in front of it, and again when the planet is not in front. Both spectra show the same set of dark lines, but while the planet is in front, extra dark lines appear at wavelengths of sodium. They disappear when the planet moves away.

What can be concluded?

Show mark scheme
Marking pointMarkNotes
Step 1Dark lines at the wavelengths of sodium are produced when sodium atoms absorb photons whose energies equal their own energy-level differences.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2These lines appear only while starlight passes through the planet's atmosphere on its way to us, so the absorbing sodium atoms must be in that atmosphere.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • ALines from the star's outer layers would be present all the time; these lines appear only while the planet is in front.
  • BEmission adds light at those wavelengths and could not produce dark lines; light waves from different sources do not cancel in this way.
  • CCorrect: the extra absorption occurs only when the starlight passes through the planet's atmosphere, so sodium atoms are present there.
  • DDuring the transit we see starlight that has passed the planet, not light reflected from it; the planet's day side faces away from us.
56E-1A-73Emission & absorption spectra Paper 1AHard1 mark

The diagram shows the visible emission line spectra of three elements P, Q and R, and the emission spectrum of a sample of hot gas.

Which statement about the sample is correct?

Emission line spectra between 400 nm and 700 nm. Element P: lines at 432, 521 and 612 nm. Element Q: lines at 521 and 662 nm. Element R: lines at 468 and 578 nm. Sample: lines at 432, 468, 521, 578 and 612 nm.PQRSample400450500550600650700wavelength / nm
Wavelength scale drawn to scale
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Marking pointMarkNotes
Step 1Every line of P (432, 521, 612 nm) and every line of R (468, 578 nm) appears in the sample, so P and R are present.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Q's line at 662 nm is missing, so Q is not present; its 521 nm line is explained by P, and every line of the sample belongs to P or R.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: all the lines of P and of R are present; Q's 662 nm line is absent, and the 521 nm line is accounted for by P.
  • BQ has a line at 521 nm in the sample, but an element is present only if all its lines appear; Q's 662 nm line is missing.
  • CEvery line in the sample is a line of P or of R, so the data give no evidence for any other element.
  • DA shared wavelength does not prevent identification: each element is identified from its whole set of lines.
57E-1A-74Energy levels & photons Paper 1AHard1 mark

The diagram shows four energy levels of an atom. In a hot gas of these atoms, some atoms are in each of the levels. Photons of visible light have energies from 1.8 eV to 3.1 eV.

How many lines in the emission spectrum of the gas are in the visible region?

Four energy levels drawn to scale: −6.0 eV (ground state), −3.8 eV, −1.3 eV and −0.5 eV.0−6.0 eV (ground state)−3.8 eV−1.3 eV−0.5 eV
Energy levels drawn to scale
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Marking pointMarkNotes
Step 1Six different transitions are possible, with energies 0.8, 2.2, 2.5, 3.3, 4.7 and 5.5 eV.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Only 2.2 eV and 2.5 eV lie between 1.8 eV and 3.1 eV, so two lines are visible.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis considers only transitions that end on the ground state (2.2, 4.7 and 5.5 eV); the 2.5 eV transition from −1.3 eV to −3.8 eV is also visible.
  • BCorrect: of the six transitions only 2.2 eV and 2.5 eV are in the visible range; 0.8 eV is infrared and 3.3, 4.7 and 5.5 eV are ultraviolet.
  • CThis counts the three transitions between neighbouring levels (2.2, 2.5 and 0.8 eV); the 0.8 eV photon is infrared.
  • DSix is the total number of possible transitions; four of them give photons outside the visible range.
58IBP2S-035-00-08Ionization Paper 1AHard1 mark

In hydrogen, the shortest wavelength emitted in transitions ending on n = 1 is 91.2 nm. The energy levels are −13.6 eV for n = 1 and −3.40 eV for n = 2. What is the shortest wavelength emitted in transitions ending on n = 2?

Show mark scheme
Marking pointMarkNotes
The shortest wavelength in each series comes from an electron falling from E = 0 (the ionization limit) to the final level. Ending on n = 1 the photon energy is 13.6 eV; ending on n = 2 it is 3.40 eV. Since λ ∝ 1/E, the wavelength is 91.2 nm × (13.6/3.40) = 91.2 × 4 = 365 nm.
The distractors come from common slips:
  • 182 nm assumes λ ∝ n.
  • 22.8 nm divides by 4 instead of multiplying.
  • 122 nm is the 2 → 1 line.
✓ 1Answer D

Answer: D

Exam tip: A series limit corresponds to the ionization energy from the level on which the series ends.

59E-1B-18Rutherford scattering Paper 1BEasy8 marks

Students use a computer simulation of the Geiger–Marsden–Rutherford experiment. Alpha particles, all with the same kinetic energy, are directed at a thin gold foil in a vacuum, and a detector counts the alpha particles scattered through an angle θ in 10.0 minutes. Like a real experiment, the simulation includes random variation; the uncertainty in a count N is ±√N.

θ / °Counts in 10.0 minutes
1589664
305756
451233
60408
90106
12042
15033
α sourcethin gold foildetectorθ
Diagram not to scale
(a)

State what the very large counts at small angles show about the atoms of the foil.

(1)
(b)

Calculate the percentage uncertainty in the count at θ = 150°.

(1)
(c)

Suggest why, in a real experiment, the counting time at large angles would be made much longer than at small angles.

(1)
(d)

Explain what the counts at angles greater than 90° show about the nucleus.

(2)
(e)

The simulation is repeated at θ = 60° with a foil of twice the thickness, giving 835 counts. A student predicts that, for a thin foil, the count is proportional to the thickness. Deduce whether the data support this prediction.

(2)
(f)

Suggest why a very thin foil is used.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Most alpha particles pass through with little or no deflection, so the atom is mostly empty space✓ 1OWTTE
Part (b)
√33/33 × 100 = 17 %✓ 1Accept 17.4 %.
Part (c)
The counts at large angles are small, so their percentage uncertainty is large; a longer time gives more counts and a smaller percentage uncertainty✓ 1OWTTE
Part (d)
A few alpha particles are turned back, which needs a very large repulsive force✓ 1
so the positive charge and most of the mass of the atom are concentrated in a very small nucleus✓ 1Award MP2 for "small, dense, positively charged nucleus". OWTTE
Part (e)
Prediction: 2 × 408 = 816 ± 40; measured 835 ± 29✓ 1
The difference (19) is smaller than the combined uncertainty (≈ 69), so the data support the prediction✓ 1MP2 requires a comparison with the uncertainties. Accept a ratio of 2.05 ≈ 2 within uncertainty.
Part (f)
So that each alpha particle is deflected by at most one nucleus / is not stopped in the foil✓ 1OWTTE

Answers: (b) 17 % (the remaining parts are explanations — see the table above)

60E-1B-29Emission & absorption spectra Paper 1BEasy9 marks

A student builds a spectrometer from a smartphone camera, a narrow slit and a transmission diffraction film. The spectrum of a lamp falls across the camera sensor, and the position x of each bright line is read in pixels from the image. Each position has an uncertainty of ±2 pixels.

To calibrate the spectrometer the student uses a compact fluorescent lamp, whose line wavelengths are known. The table and the graph show the results. The student then views a hydrogen discharge tube, whose red line appears at x = 1238 pixels.

Energy levels of hydrogen: −13.6 eV, −3.40 eV, −1.51 eV, −0.85 eV. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Wavelength λ / nm404.7435.8546.1578.0611.6
Pixel position x (±2)4535488959921100
Calibration graph: wavelength lambda in nm against pixel position x, five points only: (453, 404.7), (548, 435.8), (895, 546.1), (992, 578.0), (1100, 611.6). Axes extend to x = 1300 and lambda = 700 nm.4005006007008009001000110012001300x / pixels350400450500550600650700λ / nm
Graph drawn to scale (points only)
(a)

Outline why the compact fluorescent lamp produces bright lines at only certain wavelengths.

(1)
(b)
(i)

Draw the line of best fit for the data.

(1)
(ii)

Determine the gradient of the line. State its unit.

(2)
(c)
(i)

Determine, using the graph, the wavelength of the red hydrogen line.

(1)
(ii)

Estimate the uncertainty in this wavelength that results from the uncertainty in x.

(1)
(d)

Calculate the energy, in eV, of a photon of the red hydrogen line, and hence identify the transition that emits it.

(2)
(e)

Suggest one improvement to the calibration that would make the wavelength of the red hydrogen line more reliable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Atoms «of the gas» have discrete energy levels; a photon is emitted when an electron moves to a lower level, with energy hf equal to the level difference, so only certain wavelengths occur✓ 1OWTTE
Part (b)(i)
Single straight line passing within ±2 pixels of every point✓ 1Do not accept a line joining the points.
Part (b)(ii)
Gradient from a large triangle, e.g. (611.6 − 404.7)/(1100 − 453)✓ 1
= 0.320 nm pixel−1 «nm per pixel»✓ 1Accept 0.315–0.325 nm pixel−1. Unit needed.
Part (c)(i)
Line extrapolated to x = 1238: λ = 656 nm✓ 1Accept 654–658 nm. Allow ECF from (b)(i).
Part (c)(ii)
±2 × gradient = ±0.6 nm✓ 1Allow ECF from (b)(ii). Accept ±0.6 to ±0.7 nm.
Part (d)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(656 × 10−9 × 1.60 × 10−19) = 1.90 eV✓ 1Allow ECF from (c)(i). Accept 1.89–1.90 eV.
−1.51 − (−3.40) = 1.89 eV, which agrees with the photon energy, so the transition is from −1.51 eV to −3.40 eV✓ 1MP2 needs the energy difference to be quoted.
Part (e)
Include a calibration line of known wavelength longer than 656 nm «e.g. a neon lamp», so that the hydrogen line is found by interpolation rather than by extrapolation beyond the last point✓ 1Accept: repeat the readings of each line and average. Do not accept 'more readings' without a reason. OWTTE

Answers: (b)(ii) 0.320 nm pixel−1  ·  (c)(i) 656 nm  ·  (c)(ii) ±0.6 nm  ·  (d) 1.90 eV (the remaining parts are explanations — see the table above)

61E-1B-05Emission & absorption spectra Paper 1BMedium10 marks

A student uses a calibrated spectrometer to measure the wavelengths of the seven brightest visible lines emitted by a gas-discharge tube whose label is missing. Each measured wavelength has an uncertainty of ±2 nm.

Measured wavelengths / nm: 446, 472, 493, 503, 589, 666, 707.

The table lists the brightest visible lines of four gases from a spectral database.

GasWavelengths of brightest visible lines / nm
Hydrogen410.2, 434.0, 486.1, 656.3
Helium447.1, 471.3, 492.2, 501.6, 587.6, 667.8, 706.5
Mercury404.7, 435.8, 546.1, 577.0, 579.1
Sodium498.3, 568.8, 589.0, 589.6, 615.4
(a)

Explain why the light from the tube consists of a set of separate wavelengths.

(2)
(b)

Identify the gas in the tube. Justify your answer with reference to the uncertainty in the measurements.

(2)
(c)

Determine, in eV, the photon energy for the 707 nm line and its absolute uncertainty.

(3)
(d)

The database gives four energy levels of this gas at 20.96 eV, 21.22 eV, 22.72 eV and 23.07 eV above the ground state. Deduce which transition emits the 707 nm line, and whether your answer to (c) supports this.

(2)
(e)

Suggest how the student could check that the spectrometer does not have a systematic error.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Electrons in the atoms can occupy only discrete energy levels✓ 1
A photon is emitted when an electron moves to a lower level; its energy E = hf equals the difference between the levels, so only certain frequencies/wavelengths occur✓ 1OWTTE
Part (b)
Helium✓ 1
Every measured line matches a helium line within ±2 nm; the 589 nm line alone would also match sodium, but sodium (and hydrogen, mercury) do not account for the other lines✓ 1Award MP2 for a justification that uses the ±2 nm tolerance for all the lines.
Part (c)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/707 × 10−9 = 2.81 × 10−19 J✓ 1
= 2.813 × 10−19/1.60 × 10−19 = 1.758 eV✓ 1Accept 1.76 eV.
Fractional uncertainty in E = 2/707 = 0.28 %, so E = (1.758 ± 0.005) eV✓ 1MP3 is for the uncertainty with precision matching the value. Allow ECF.
Part (d)
22.72 − 20.96 = 1.76 eV (the only pair of levels giving close to 1.76 eV)✓ 1
1.76 eV lies within 1.753–1.763 eV, so the measurement supports this transition✓ 1Allow ECF from (c).
Part (e)
Measure the lines of a lamp whose wavelengths are known (e.g. a mercury or sodium lamp) and compare the readings with the accepted values✓ 1Accept: calibrate against a known source. OWTTE

Answers: (b) helium  ·  (c) (1.758 ± 0.005) eV  ·  (d) 22.72 eV → 20.96 eV (1.76 eV) (the remaining parts are explanations — see the table above)

62E-1B-13Energy levels & photons Paper 1BMedium11 marks

A student estimates the Planck constant using six light-emitting diodes (LEDs). For each LED the potential difference is increased slowly until the LED just begins to glow; this threshold p.d. V is read from a digital voltmeter (uncertainty ±0.05 V). The peak wavelength λ of each LED (uncertainty ±5 nm) is taken from the manufacturer's data sheet. The student assumes that the energy of each photon emitted equals eV.

(e = 1.60 × 10−19 C, c = 3.00 × 108 m s−1)

LEDλ / nm1/λ / 106 m−1V / V
red6351.571.85
orange6101.641.90
amber5921.692.00
green5252.24
blue4682.142.57
violet4052.98
Threshold p.d. V against 1/λ for six LEDs, points with error bars1.41.61.82.02.22.42.61/λ / 10⁶ m⁻¹1.61.82.02.22.42.62.83.03.2V / V
Graph drawn to scale
(a)

Complete the table for the green and the violet LEDs.

(1)
(b)

Show that the model predicts a straight-line graph of V against 1/λ with gradient hc/e.

(1)
(c)

Draw the line of best fit and determine its gradient.

(3)
(d)

Draw lines of maximum and minimum gradient and determine the absolute uncertainty in the gradient.

(2)
(e)

Determine the Planck constant and its absolute uncertainty.

(2)
(f)

The accepted value of the Planck constant is 6.63 × 10−34 J s. Comment on your result.

(1)
(g)

When extended, the line of best fit does not pass through the origin. Suggest one reason for this systematic error.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
green: 1.90; violet: 2.47 (× 106 m−1)✓ 1Both needed. Accept 3 s.f.
Part (b)
eV = hf = hc/λ, so V = (hc/e) × (1/λ), which is y = mx with m = hc/e✓ 1
Part (c)
Straight line through all the error bars✓ 1
Gradient from a large triangle✓ 1
Gradient = 1.28 × 10−6 V m✓ 1Accept 1.22–1.36 × 10−6 V m. Unit and power of ten needed.
Part (d)
Steepest and shallowest lines drawn through all the error bars✓ 1
Uncertainty = (max − min)/2, e.g. (1.44 − 1.12)/2 = ±0.16 × 10−6 V m✓ 1Accept ±0.08 to ±0.19 × 10−6 V m from the candidate's lines.
Part (e)
h = e × gradient/c = 1.60 × 10−19 × 1.28 × 10−6/3.00 × 108 = 6.8 × 10−34 J s✓ 1Allow ECF from (c).
Same percentage uncertainty as the gradient: h = (6.8 ± 0.9) × 10−34 J s✓ 1MP2 is for matching precision. Allow ECF from (d).
Part (f)
6.63 × 10−34 J s lies within the range of the result, so the result is consistent with the accepted value✓ 1Allow ECF.
Part (g)
Every V differs from hc/eλ by a similar amount: e.g. the photons are emitted over a range of wavelengths, not only at the peak λ, and electrons can gain extra energy from thermal motion, so the LED starts to emit at a p.d. slightly below hc/eλ✓ 1Accept any systematic effect that shifts every V by a similar amount, e.g. the onset of the glow is judged by eye; p.d. across the LED's internal resistance. OWTTE

Answers: (a) 1.90 and 2.47 × 106 m−1  ·  (c) 1.28 × 10−6 V m  ·  (d) ±0.16 × 10−6 V m  ·  (e) (6.8 ± 0.9) × 10−34 J s (the remaining parts are explanations — see the table above)

63E-1B-21Energy levels & photons Paper 1BMedium12 marks

A student investigates the energy levels of argon with an electron-collision tube. Electrons leave a hot cathode and are accelerated through a potential difference V towards a wire grid in argon gas at low pressure. Beyond the grid, a small potential difference opposes the motion of the electrons towards a collector, and the current I at the collector is recorded as V is increased slowly.

The current falls sharply at a series of values of V. The value at which the n-th fall occurs is Vn; each value has an uncertainty of ±1.0 V because the minima are broad. The student's model is Vn = V0 + nΔE/e, where ΔE is the energy needed to excite an argon atom from its ground state and V0 is a constant. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

n123456
Vn / V15.526.438.749.862.273.1
Graph of V_n against n for n = 1 to 6 with error bars of ±1.0 V; points only. V axis 0 to 80 V, n axis 0 to 7.01234567n020406080Vₙ / V
Graph drawn to scale
(a)

Outline why the current at the collector falls sharply when V reaches V1.

(2)
(b)

Draw the line of best fit for the data.

(1)
(c)

Determine the gradient of your line and hence ΔE, in eV.

(2)
(d)

Draw lines of maximum and minimum gradient and hence determine the absolute uncertainty in ΔE.

(2)
(e)

The line of best fit does not pass through the origin. Identify the type of error that this shows, and explain why it does not affect your value of ΔE.

(2)
(f)

An excited argon atom returns to its ground state by emitting one photon. Calculate the wavelength of this photon.

(2)
(g)

A spectral database lists a strong ultraviolet line of argon at 106.7 nm. Comment on whether your results are consistent with this line.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
An electron whose kinetic energy is at least ΔE can transfer this energy to an argon atom in a collision, raising the atom to an excited state✓ 1
The electron is then left with too little kinetic energy to overcome the opposing p.d. and reach the collector, so fewer electrons arrive and I falls✓ 1OWTTE
Part (b)
Single straight line passing through all the error bars✓ 1Do not accept a line forced through the origin.
Part (c)
Gradient from a large triangle (at least half the line): ≈ 11.6 V✓ 1Accept 11.3–11.9 V.
ΔE = e × gradient, so ΔE ≈ 11.6 eV✓ 1Allow ECF from the gradient. Unit eV needed.
Part (d)
Steepest and shallowest straight lines drawn, each passing through all the error bars✓ 1
Uncertainty = (maximum − minimum gradient)/2, e.g. (11.9 − 11.3)/2 = ±0.3 eV✓ 1Accept ±0.2 to ±0.7 eV from the candidate's lines.
Part (e)
A systematic (offset / zero) error✓ 1Accept: the cathode–grid contact p.d. or the opposing p.d. shifts every reading.
Every Vn is shifted by the same amount V0, which changes the intercept but not the gradient, and ΔE is found from the gradient✓ 1OWTTE
Part (f)
E = 11.6 × 1.60 × 10−19 = 1.86 × 10−18 J✓ 1Allow ECF from (c).
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/1.86 × 10−18 = 1.07 × 10−7 m✓ 1Accept 104–110 nm. Award [2] for CNA.
Part (g)
Fractional uncertainty in λ equals that in ΔE (≈ 3 %), so λ = (107 ± 3) nm; 106.7 nm lies within this range, so the results are consistent with the line✓ 1Allow ECF from (d) and (f). A conclusion with no reference to an uncertainty scores 0.

Answers: (c) ΔE ≈ 11.6 eV  ·  (d) ≈ ±0.3 eV  ·  (f) 1.07 × 10−7 m (107 nm) (the remaining parts are explanations — see the table above)

64E-1B-30Energy levels & photons Paper 1BMedium11 marks

A single-photon detector produces one count for each photon that it detects. A student directs light of wavelength 785 nm from a laser diode through calibrated filters onto the detector, so that the power P reaching the detector is known, and records the count rate R. The uncertainty in each value of R is ±0.10 × 106 s−1. The fraction of the incident photons that the detector counts is its efficiency η.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1; 1 pW = 1.0 × 10−12 W)

P / pW0.200.400.600.801.001.20
R / 106 s−1 (±0.10)0.701.131.722.162.763.20
Graph of count rate R in units of 10^6 per second against power P in pW, six points with error bars of plus or minus 0.10: (0.20, 0.70), (0.40, 1.13), (0.60, 1.72), (0.80, 2.16), (1.00, 2.76), (1.20, 3.20). Axes start at the origin.00.20.40.60.811.21.4P / pW00.511.522.533.5R / 10⁶ s⁻¹
Graph drawn to scale (points only)
(a)

Show that the energy of a photon of wavelength 785 nm is about 2.5 × 10−19 J.

(1)
(b)
(i)

Draw the line of best fit for the data.

(1)
(ii)

Determine the gradient of your line.

(2)
(iii)

Draw lines of maximum and minimum gradient and hence determine the absolute uncertainty in the gradient.

(2)
(c)

The gradient equals η/E, where E is the energy of one photon. Determine η and its absolute uncertainty.

(2)
(d)

The line of best fit does not pass through the origin. Suggest a reason for this, and explain why it does not affect the value of η found in (c).

(2)
(e)

The manufacturer states that η = 0.70 ± 0.02 at 785 nm. Comment on whether the student's result agrees with this value.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/785 × 10−9 «= 2.53 × 10−19 J»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(i)
Single straight line passing through all the error bars✓ 1Do not accept a line forced through the origin.
Part (b)(ii)
Gradient calculated from a large triangle «at least half the line»✓ 1
= 2.55 × 106 s−1 pW−1 «= 2.55 × 1018 s−1 W−1»✓ 1Accept 2.47–2.63 × 106 s−1 pW−1. Unit needed.
Part (b)(iii)
Steepest and least steep straight lines each passing through all the error bars✓ 1
Uncertainty = (max − min)/2, e.g. (2.70 − 2.38)/2 = ±0.2 × 106 s−1 pW−1✓ 1Accept ±0.10 to ±0.25 × 106 s−1 pW−1 from the candidate's lines.
Part (c)
η = gradient × E = 2.55 × 1018 × 2.53 × 10−19 = 0.65✓ 1Allow ECF from (a) and (b)(ii). No unit.
Same fractional uncertainty as the gradient: η = 0.65 ± 0.04✓ 1Accept ±0.03 to ±0.06. Allow ECF from (b)(iii). MP2 is for matching precision of value and uncertainty.
Part (d)
Counts are recorded even with no laser light reaching the detector, e.g. stray light from the room or counts produced inside the detector itself «background/dark counts»✓ 1OWTTE
This adds the same count rate to every reading, which shifts the intercept but not the gradient; η is found from the gradient only✓ 1
Part (e)
The student's range «0.61 to 0.69» overlaps the manufacturer's range «0.68 to 0.72», so the results agree within the uncertainties✓ 1Allow ECF from (c). A comment with no reference to the uncertainties scores 0.

Answers: (b)(ii) 2.55 × 106 s−1 pW−1  ·  (b)(iii) ±0.2 × 106 s−1 pW−1  ·  (c) η = 0.65 ± 0.04 (the remaining parts are explanations — see the table above)

65E-1B-31Emission & absorption spectra Paper 1BMedium11 marks

A gas-leak sensor uses a laser tuned to 1653.7 nm, a wavelength at which methane molecules absorb strongly. To test the sensor, a student fills a cell with air containing 1.0 % methane. Mirrors reflect the beam back and forth so that the path length L in the gas can be changed. The fraction T of the laser power transmitted is measured with an uncertainty of ±0.010. The measurement is repeated with the laser tuned to 1652.0 nm, where methane does not absorb.

The student's model is T = e−kL, where k is a constant that is proportional to the concentration of methane.

L / mT at 1653.7 nm (±0.010)ln TT at 1652.0 nm (±0.010)
2.00.826−0.1910.979
4.00.692−0.3680.981
6.00.5880.978
8.00.490−0.7130.980
10.00.4180.981
12.00.347−1.060.979
Graph of ln T against path length L in m, four points with error bars: (2.0, −0.191), (4.0, −0.368), (8.0, −0.713), (12.0, −1.058). The points for L = 6.0 m and 10.0 m are not plotted. The vertical axis runs from −1.2 to 0.02468101214L / m−1.2−1.0−0.8−0.6−0.4−0.20ln T
Graph drawn to scale (four points with error bars only)
(a)

Explain why methane absorbs strongly at 1653.7 nm but hardly at all at 1652.0 nm.

(2)
(b)

Complete the table for L = 6.0 m and L = 10.0 m, and plot these two points on the graph.

(2)
(c)
(i)

Draw the line of best fit for the data.

(1)
(ii)

Determine k.

(2)
(d)

The line of best fit does not pass through the origin. Using the data at 1652.0 nm, suggest a reason for this, and explain why it does not affect the value of k.

(2)
(e)

The sensor is used along an open path of 40.0 m above a gas pipeline. After correcting for the effect identified in (d), the fraction transmitted at 1653.7 nm is 0.55. Determine the concentration of methane in the air above the pipeline.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A photon is absorbed only if its energy hc/λ equals the difference between two energy levels of the molecule✓ 1
The photon energy at 1652.0 nm is different and does not match any energy difference of methane, so it is not absorbed✓ 1OWTTE
Part (b)
−0.531 «L = 6.0 m» and −0.872 «L = 10.0 m»✓ 1Both needed, with negative signs. Accept 3 or 4 s.f.
Both points plotted correctly, to within half a small square✓ 1Allow ECF from MP1. Error bars not required.
Part (c)(i)
Single straight line through all the error bars✓ 1Do not accept a line forced through the origin.
Part (c)(ii)
Gradient from a large triangle = −0.0862 m−1✓ 1
k = −gradient = 0.0862 m−1✓ 1Accept 0.083–0.090 m−1. Unit needed. Award [2] for CNA.
Part (d)
At 1652.0 nm T ≈ 0.98 for every L: about 2 % of the power is lost by reflection at the windows of the cell «or by another loss that does not depend on L», whatever the path length✓ 1OWTTE
This multiplies every T by the same factor, so ln T is shifted by the same amount «ln 0.98 ≈ −0.02» for all L: the intercept changes but the gradient, and so k, does not✓ 1
Part (e)
k = −ln 0.55/40.0 = 0.0149 m−1✓ 1
Concentration = 1.0 % × 0.0149/0.0862 = 0.17 %✓ 1Allow ECF from (c)(ii). Award [2] for CNA.

Answers: (b) −0.531, −0.872  ·  (c)(ii) k = 0.0862 m−1  ·  (e) 0.17 % (the remaining parts are explanations — see the table above)

66E-1B-32Energy levels & photons Paper 1BMedium11 marks

A sodium lamp emits, among others, the four lines in the table. They were measured with three different spectrometers, for the ultraviolet, the visible and the infrared. A spectral database states that a sodium atom in one particular excited level can return to the ground state either by emitting the 330.2 nm photon, or by emitting the other three photons one after another. The table gives the photon energies E and their uncertainties ΔE.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Lineλ / nmE / eVΔE / eV
ultraviolet330.2 ± 0.13.7650.001
yellow589.0 ± 0.22.11060.0007
infrared 11138 ± 1
infrared 22206 ± 20.56350.0005
Blank axes for an energy-level diagram: energy above the ground state from 0 to 4.0 eV.00.511.522.533.54E / eV
Axes for (d): energy E above the ground state, drawn to scale
(a)

Complete the table for the infrared 1 line.

(2)
(b)

Determine the sum of the photon energies of the yellow, infrared 1 and infrared 2 lines, and its absolute uncertainty.

(2)
(c)

Deduce whether the data support the statement in the database.

(2)
(d)

The yellow photon is emitted in the last step to the ground state, and the infrared 2 photon in the first step. Draw, on the axes, the energy levels involved and arrows for all four transitions.

(2)
(e)

Identify the measurement that contributes most to the uncertainty in (b), and suggest how this contribution could be reduced.

(1)
(f)

Explain how these results support the model that the atom has discrete energy levels.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λe = 6.63 × 10−34 × 3.00 × 108/(1138 × 10−9 × 1.60 × 10−19) = 1.092 eV✓ 1Accept 1.0924 eV.
ΔE = E × Δλ/λ = 1.092 × 1/1138 = 0.001 eV✓ 1Accept 0.0010 eV. Fractional uncertainty in E = fractional uncertainty in λ.
Part (b)
Sum = 2.1106 + 1.092 + 0.5635 = 3.766 eV✓ 1Allow ECF from (a).
Uncertainty = 0.0007 + 0.001 + 0.0005 ≈ ±0.002 eV✓ 1Absolute uncertainties add for a sum. Allow ECF from (a).
Part (c)
Ranges: sum 3.764 to 3.768 eV; ultraviolet photon 3.764 to 3.766 eV✓ 1OR difference 0.0011 eV is less than the combined uncertainty 0.0032 eV. Allow ECF from (b).
The ranges overlap, so the energies are equal within uncertainty and the data support the statement✓ 1MP2 only scores if MP1 scores.
Part (d)
Levels drawn at 0, about 2.11 eV, about 3.20 eV and about 3.77 eV✓ 1Accept ±0.05 eV on the scale.
Downward arrows: top → ground «ultraviolet»; top → 3.20 eV «infrared 2»; 3.20 eV → 2.11 eV «infrared 1»; 2.11 eV → ground «yellow»✓ 1All four arrows, pointing downwards, needed.
Part (e)
The infrared 1 «1138 nm» line «±0.001 eV»: measure it with a spectrometer of higher resolution / repeat and average the readings✓ 1Both identification and suggestion needed. OWTTE
Part (f)
Each line has a single photon energy, so the atom loses only definite amounts of energy✓ 1
The energy of the single photon equals the sum of the three photons emitted in steps, as expected if the atom moves between the same fixed levels by different routes✓ 1OWTTE

Answers: (a) 1.092 ± 0.001 eV  ·  (b) 3.766 ± 0.002 eV (the remaining parts are explanations — see the table above)

67E-1B-33Emission & absorption spectra Paper 1BHard12 marks

The cold, thin gas between the stars contains a small amount of sodium. Sodium atoms in their ground state absorb photons of wavelength 589.0 nm, so the spectrum of a distant star shows a narrow, dark interstellar sodium line at this wavelength.

A student takes from a database the stellar parallax angle p of eight hot stars that lie in almost the same direction in the sky, and the strength W of the interstellar sodium line in the spectrum of each star. The student assumes that W is proportional to the number of sodium atoms per unit area along the line of sight from the Earth to the star. Each value of W has an uncertainty of ±0.5 pm; the uncertainties in p are negligible. The graph shows W against the distance d of the star for six of the stars.

Starp / 10−3 arc-secondd / pcW / pm (±0.5)
18.701150.9
26.251601.7
34.553.6
43.572804.6
52.863506.8
62.274418.4
71.8511.3
81.4967114.0
Graph of interstellar sodium line strength W in pm against distance d in parsec, six points with error bars0100200300400500600700800d / pc0246810121416W / pm
Graph drawn to scale (six points with error bars only)
(a)

Complete the table for stars 3 and 7, and plot these two points on the graph.

(2)
(b)

Draw the line of best fit for the data.

(1)
(c)

Determine the gradient of your line. State its unit.

(2)
(d)
(i)

Determine the intercept of your line on the d-axis. By drawing lines of maximum and minimum gradient, estimate the absolute uncertainty in this intercept.

(2)
(ii)

Outline what the intercept suggests about the interstellar gas close to the Sun.

(1)
(e)

The student used only stars with surface temperatures above 15 000 K. Suggest why cooler stars, such as the Sun, are not suitable for this investigation.

(2)
(f)

A more distant star in the same direction has an interstellar sodium line with W = 19.5 pm. Its parallax angle is too small to be measured. Estimate the distance to this star, and comment on the reliability of your estimate.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/p: 220 pc «star 3» and 541 pc «star 7»✓ 1Both needed. Accept 540 pc.
Both points plotted correctly, to within half a small square✓ 1Allow ECF from MP1. Error bars not required.
Part (b)
Single straight line passing through all the error bars✓ 1Do not accept a line forced through the origin.
Part (c)
Gradient from a large triangle «at least half the line» using two points read from the line, e.g. (14.0 − 0)/(665 − 80)✓ 1
= 0.024 pm pc−1✓ 1Accept 0.022–0.026 pm pc−1. Unit needed.
Part (d)(i)
Intercept on the d-axis ≈ 80 pc✓ 1Accept 65–95 pc. Allow ECF from (b).
Uncertainty = half the difference between the intercepts of the steepest and least steep lines through all the error bars ≈ ±15 pc✓ 1Accept ±8 to ±25 pc from the candidate's lines.
Part (d)(ii)
There is almost no sodium «gas» within about 80 pc of the Sun: W only begins to increase beyond this distance «the Sun lies in a region of very low-density gas»✓ 1Allow ECF from (d)(i). OWTTE
Part (e)
The atmosphere of a cool star contains many neutral sodium atoms, which absorb at 589.0 nm and produce a strong line of the star's own at the same wavelength✓ 1
so the interstellar line cannot be separated from it «W would not measure only the interstellar sodium»; in a hot star almost all the sodium in the atmosphere is ionised, so the star produces no 589.0 nm line of its own✓ 1Award MP2 for either idea. Award [1] max for: cool stars are much less luminous, so distant ones are too faint to give good spectra. OWTTE
Part (f)
d ≈ 80 + 19.5/0.024 ≈ 890 pc✓ 1Allow ECF from (c) and (d)(i). Accept 850–950 pc.
Not reliable: the estimate extrapolates the line far beyond the largest distance measured «about 670 pc», and the gas need not be spread uniformly beyond it «e.g. a dense cloud would increase W»✓ 1Accept: the proportionality assumed for W may not hold for strong lines. OWTTE

Answers: (a) 220 pc, 541 pc  ·  (c) 0.024 pm pc−1  ·  (d)(i) (80 ± 15) pc  ·  (f) 890 pc (the remaining parts are explanations — see the table above)

68E-2-03Emission & absorption spectra Paper 2Easy9 marks

A student looks through a spectroscope at two light sources: a filament lamp and a discharge tube containing helium gas. Figure 1 shows the two spectra between 400 nm and 700 nm.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Figure 1: the visible spectrum of a filament lamp is continuous from 400 nm to 700 nm; the helium discharge tube gives bright lines at 447, 471, 492, 502, 588 and 668 nm on a dark background.filament lamphelium discharge tube400450500550600650700wavelength / nm
Figure 1 — wavelength scale drawn to scale
(a)

The two spectra.

(i)

Describe how the spectrum of the helium tube differs from that of the filament lamp.

(2)
(ii)

Explain how the helium spectrum provides evidence that atoms have discrete energy levels.

(3)
(b)

The yellow line in the helium spectrum has a wavelength of 587.6 nm.

(i)

Calculate the energy, in eV, of a photon of this wavelength.

(2)
(ii)

This photon is emitted when an electron in a helium atom moves from a level of energy −1.51 eV to a lower level. Calculate the energy of the lower level.

(1)
(c)

A second discharge tube contains an unknown gas. Outline how its spectrum could be used to identify the gas.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The filament lamp gives a continuous spectrum, containing all visible wavelengths «colours»✓ 1
The helium tube gives a line spectrum: bright lines at a few particular wavelengths, with dark regions between them✓ 1
Part (a)(ii)
Each line consists of photons of one wavelength/frequency and therefore of one energy, E = hf✓ 1
A photon is emitted when an electron in an atom moves from a higher to a lower energy level; its energy equals the difference between the two levels✓ 1
Only certain photon energies are emitted, so only certain energy differences exist: the energy levels are discrete✓ 1OWTTE
Part (b)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/587.6 × 10−9 = 3.38 × 10−19 J✓ 1
E = 3.38 × 10−19/1.60 × 10−19 = 2.12 eV✓ 1Award [2] for CNA. Accept 2.11–2.12 eV.
Part (b)(ii)
−1.51 − 2.12 = −3.63 eV✓ 1Allow ECF from (b)(i). Accept −3.62 eV. The answer must be negative.
Part (c)
Measure the wavelengths of its lines and compare them with the known line spectra of the elements; each element has its own unique set of lines✓ 1OWTTE

Answers: (b)(i) 2.12 eV  ·  (b)(ii) −3.63 eV (the remaining parts are explanations — see the table above)

69E-2-56Energy levels & photons Paper 2Easy12 marks

A helium–neon laser contains a mixture of helium and neon gases in a glass tube. A potential difference across the tube accelerates electrons, which collide with helium atoms and excite them. Excited helium atoms then collide with neon atoms and excite them. The laser emits red light of wavelength 632.8 nm. The figure shows some energy levels of the two atoms.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)

Energy levels of helium and neon. Helium: ground state 0 and an excited level at 20.61 eV. Neon: ground state 0 and excited levels at 16.62 eV, 18.70 eV and 20.66 eV. The levels above 16 eV are drawn to scale; the vertical scale is broken between 0 and 16 eV.heliumneon20.61 eV020.66 eV18.70 eV16.62 eV0 (ground state)scale broken
Energy levels above 16 eV drawn to scale
(a)

Calculate, in eV, the energy of a photon of wavelength 632.8 nm.

(2)
(b)

Draw, on the figure, an arrow to show the transition that emits the laser light.

(1)
(c)
(i)

State the minimum kinetic energy, in eV, that an electron needs to excite a helium atom from its ground state to the 20.61 eV level.

(1)
(ii)

Calculate the minimum speed of such an electron.

(2)
(d)

Outline why an excited helium atom can transfer its energy to a neon atom in a collision, raising the neon atom to the 20.66 eV level.

(2)
(e)

Neon atoms in the 18.70 eV level fall to the 16.62 eV level. Determine whether the photons emitted in this transition are visible.

(2)
(f)

The laser emits a light power of 2.0 mW. The potential difference across the tube is 1.8 kV and the current is 6.5 mA. Calculate the efficiency of the laser.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/632.8 × 10−9 = 3.14 × 10−19 J✓ 1
= 1.96 eV✓ 1Award [2] for CNA.
Part (b)
Downward arrow on the neon diagram from 20.66 eV to 18.70 eV✓ 1Arrow must start and end on the levels. Do not accept an upward arrow.
Part (c)(i)
20.61 eV✓ 1Accept 3.30 × 10−18 J.
Part (c)(ii)
½mev² = 20.61 × 1.60 × 10−19 J✓ 1
v = √(2 × 20.61 × 1.60 × 10−19/9.11 × 10−31) = 2.7 × 106 m s−1✓ 1Award [2] for CNA.
Part (d)
The energy of the excited helium atom «20.61 eV» is almost equal to the energy needed to raise neon to its 20.66 eV level✓ 1
The small difference «0.05 eV» is supplied by the kinetic energy of the colliding atoms✓ 1OWTTE
Part (e)
ΔE = 18.70 − 16.62 = 2.08 eV, so λ = hc/ΔE = 598 nm✓ 1ALTERNATIVE: compare 2.08 eV with visible photon energies, about 1.8 eV to 3.1 eV.
This is between 400 nm and 700 nm, so the photons are visible «orange-yellow»✓ 1Conclusion must be consistent with the calculation.
Part (f)
Input power = IV = 6.5 × 10−3 × 1.8 × 103 = 11.7 W✓ 1
Efficiency = 2.0 × 10−3/11.7 = 1.7 × 10−4 «0.017 %»✓ 1Award [2] for CNA.

Answers: (a) 1.96 eV  ·  (c)(i) 20.61 eV  ·  (c)(ii) 2.7 × 106 m s−1  ·  (e) 598 nm — visible  ·  (f) 1.7 × 10−4 (the remaining parts are explanations — see the table above)

70E-2-59Emission & absorption spectra Paper 2Easy10 marks

The colours of fireworks are produced by metal compounds that are heated strongly when a shell bursts. Lithium compounds give red light: lithium atoms emit visible lines at 610.4 nm and 670.8 nm. Sodium compounds give yellow light: sodium atoms emit visible light at 589.0 nm.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, g = 9.81 m s−2)

A blank spectrum strip from 400 nm to 700 nm, dark, with a wavelength scale.400450500550600650700wavelength / nm
Wavelength scale drawn to scale
(a)

Calculate, in eV, the energy of a photon of wavelength 670.8 nm.

(2)
(b)

The 670.8 nm photon is emitted when a lithium atom falls from its first excited level to the ground state. The 610.4 nm photon is emitted when the atom falls from a higher level to this first excited level. Determine the energy of this higher level above the ground state.

(2)
(c)

Explain why hot lithium atoms emit light of only certain wavelengths.

(2)
(d)

A firework contains both a lithium compound and a sodium compound. Draw, on the strip, the visible spectrum of the light emitted by its hot atoms.

(2)
(e)
(i)

A shell is launched vertically upwards at 52 m s−1 and bursts at its highest point. Calculate the height of the burst, ignoring air resistance.

(1)
(ii)

State the effect of air resistance on the height of the burst.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/670.8 × 10−9 = 2.97 × 10−19 J✓ 1
= 1.85 eV✓ 1Award [2] for CNA.
Part (b)
Energy of a 610.4 nm photon = hc/λ = 2.04 eV✓ 1
Energy of the level = 1.85 + 2.04 = 3.89 eV «above the ground state»✓ 1Allow ECF from (a). Accept 3.88–3.89 eV. Award [2] for CNA.
Part (c)
The electrons in an atom can have only certain «discrete» energies / energy levels✓ 1
A photon is emitted when an electron falls to a lower level; its energy hf equals the difference between the levels, so only certain frequencies and wavelengths are emitted✓ 1OWTTE
Part (d)
Three narrow bright lines, at 589 nm, 610 nm and 671 nm on the scale✓ 1Accept ±5 nm.
No other lines and the rest of the strip left dark «not a continuous spectrum»✓ 1MP2 only scores if MP1 scores.
Part (e)(i)
h = u²/2g = 52²/(2 × 9.81) = 138 m✓ 1
Part (e)(ii)
The height is smaller «air resistance decelerates the shell in addition to gravity»✓ 1

Answers: (a) 1.85 eV  ·  (b) 3.89 eV  ·  (e)(i) 138 m (the remaining parts are explanations — see the table above)

71E-2-62Energy levels & photons Paper 2Easy11 marks

Each pixel of a phone display contains red, green and blue organic light-emitting diodes (OLEDs). In each OLED, electrons passing through a thin layer of material move from a higher energy level to a lower one, and in each such transition one photon is emitted. The photons emitted by the green OLED have energy 2.33 eV and those emitted by the blue OLED have energy 2.66 eV. The red OLED emits light of wavelength 630 nm.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Blank axes: relative intensity from 0 to 1 against wavelength from 400 nm to 700 nm.400450500550600650700wavelength / nm00.20.40.60.81relative intensity
Axes for (d), drawn to scale
(a)

Calculate the wavelength of the light emitted by the green OLED.

(2)
(b)

Explain why the three OLEDs must be made of different materials.

(2)
(c)

The current in the blue OLED is 1.50 μA and the potential difference across it is 3.2 V. It emits a light power of 1.0 μW.

(i)

Calculate the number of electrons passing through the blue OLED each second.

(1)
(ii)

Determine the fraction of these electrons that produce a photon that leaves the OLED.

(2)
(iii)

Calculate the efficiency of the blue OLED.

(2)
(d)

All three OLEDs of a pixel are switched on together. Sketch, on the axes, the spectrum of the light emitted. Each OLED emits over a narrow band of wavelengths.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 2.33 × 1.60 × 10−19 = 3.73 × 10−19 J✓ 1
λ = hc/E = 5.34 × 10−7 m «534 nm»✓ 1Award [2] for CNA.
Part (b)
The photon energy «and so the colour» equals the difference between the energy levels of the material✓ 1
Each colour needs a different energy difference, and each material has its own energy levels✓ 1OWTTE
Part (c)(i)
N = I/e = 1.50 × 10−6/1.60 × 10−19 = 9.4 × 1012 s−1✓ 1
Part (c)(ii)
Photons per second = 1.0 × 10−6/(2.66 × 1.60 × 10−19) = 2.3 × 1012 s−1✓ 1
Fraction = 2.3 × 1012/9.4 × 1012 = 0.25✓ 1Accept 0.24–0.25. Allow ECF from (c)(i). Award [2] for CNA.
Part (c)(iii)
Electrical power = IV = 1.50 × 10−6 × 3.2 = 4.8 × 10−6 W✓ 1
Efficiency = 1.0/4.8 = 0.21 «21 %»✓ 1Award [2] for CNA.
Part (d)
Three separate narrow peaks, at about 467 nm, 534 nm and 630 nm✓ 1Accept ±10 nm. Allow ECF from (a).
Intensity close to zero between the peaks «not a continuous spectrum»✓ 1

Answers: (a) 534 nm  ·  (c)(i) 9.4 × 1012 s−1  ·  (c)(ii) 0.25  ·  (c)(iii) 0.21 (the remaining parts are explanations — see the table above)

72IBP2S-035-01-11Energy levels & photons Paper 2Easy7 marks

Low-pressure sodium street lamps emit almost all of their light at a wavelength of 589 nm.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

State what is meant by a photon.

(1)
(b)

Outline why the sodium atoms emit light of one particular wavelength rather than a continuous range.

(2)
(c)

Calculate, in eV, the energy of a photon of wavelength 589 nm.

(2)
(d)

One lamp emits 35 W of light at this wavelength. Calculate the number of photons emitted per second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A photon is a discrete packet (quantum) of electromagnetic energy .✓ 1
Part (b)
Electrons in sodium atoms can only occupy discrete energy levels, and a photon is emitted when an electron moves down between two levels .✓ 1
The photon energy equals the fixed energy difference between those levels, so E = hc/λ fixes the wavelength .✓ 1
Part (c)
E = hc/λ = (6.63×10−34 × 3.00×108)/(589×10−9) = 3.38×10−19 J .✓ 1
E = 3.38×10−19/1.60×10−19 = 2.11 eV .✓ 1
Part (d)
N = P/E = 35/(3.38×10−19) . ECF from (c).✓ 1
N = 1.04×1020 s−1 .✓ 1

Exam tip: Keep the photon energy in joules for the photon-rate calculation. The eV value is useful for comparing with energy levels, but P/E needs SI units.

73IBP2S-035-00-11Energy levels & photons Paper 2Easy5 marks

A low-pressure sodium lamp lights a car park. Almost all of its visible output is at a wavelength of 589 nm, and the lamp emits 35 W of light at this wavelength.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1)

(a)

Outline why the light from the lamp contains only certain wavelengths.

(1)
(b)

Calculate the energy of one photon of wavelength 589 nm.

(2)
(c)

Calculate the number of photons emitted per second at this wavelength.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Sodium atoms have discrete energy levels, so electrons moving down between levels emit photons of only certain energies ΔE, and therefore only certain wavelengths λ = hc/ΔE✓ 1
Part (b)
E = hc/λ = (6.63×10−34 × 3.00×108)/(589×10−9)✓ 1
E = 3.38×10−19 J✓ 1
Part (c)
N = P/E = 35/(3.38×10−19)✓ 1
N = 1.0×1020 s−1 (accept 1.04×1020 s−1; allow ECF from (b))✓ 1

Exam tip: Convert nm to m (×10⁻⁹) before substituting into E = hc/λ.

74IBP2S-035-02-11Rutherford scattering Paper 2Easy5 marks

A science centre builds an interactive exhibit that simulates the Geiger–Marsden experiment. On screen, alpha particles are fired at a thin foil of gold-197, \(^{197}_{79}\text{Au}\).

(a)

State what happens to most of the alpha particles. Outline what this shows about the structure of the atom.

(2)
(b)

A very small fraction of the alpha particles are deflected through angles greater than 90°. Explain what this shows about the nucleus.

(2)
(c)

Determine the number of neutrons in one nucleus of gold-197.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Most alpha particles pass straight through the foil undeflected, or are deflected only through very small angles . This shows that most of the atom is empty space, with the nucleus occupying only a tiny fraction of its volume .✓ 2
Part (b)
A very large repulsive force is needed, so the positive charge must be concentrated in a very small region . Most of the atom's mass must also be concentrated there, because the alpha particle is turned back by an object much more massive than itself .✓ 2
Part (c)
197 − 79 = 118 neutrons .✓ 1

Exam tip: When asked what large-angle scattering shows, mention both charge and mass being concentrated in a tiny region. Each idea can earn a separate mark.

75IBP2S-035-03-11Nuclear notation Paper 2Easy5 marks

Cobalt-60, 6027Co, has been used as a gamma-ray source in radiotherapy machines for treating tumours.
Elementary charge e = 1.60 × 10−19 C.

(a)

State the number of protons and the number of neutrons in a nucleus of 6027Co.

(2)
(b)

The stable isotope of cobalt is 5927Co. Outline how a 5927Co nucleus differs from a 6027Co nucleus.

(1)
(c)

Calculate the charge of a 6027Co nucleus.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
27 protons✓ 1
60 − 27 = 33 neutrons✓ 1
Part (b)
It has one fewer neutron (32 instead of 33) but the same number of protons✓ 1
Part (c)
Q = Ze = 27 × 1.60 × 10−19✓ 1
Q = +4.32 × 10−18 C (positive)✓ 1

Common error: The charge of a nucleus depends only on Z. Neutrons add mass but no charge, so do not multiply e by the nucleon number. The nucleus is positively charged.

76IBP2S-035-03-12Nuclear notation Paper 2Easy4 marks

A sample of magnesium from a meteorite is analysed. The table gives the stable isotopes found.

Nuclide | Relative abundance / % | Atomic mass / u
2412Mg | 79.0 | 23.985
2512Mg | 10.0 | 24.986
2612Mg | 11.0 | 25.983

(a)

State one feature that is the same for the nuclei of all three nuclides.

(1)
(b)

State the number of neutrons in a 2612Mg nucleus.

(1)
(c)

Calculate the mean atomic mass of magnesium in this sample.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Same number of protons (12) / same proton number / same nuclear charge✓ 1
Part (b)
26 − 12 = 14✓ 1
Part (c)
Uses weighted mean: 0.790 × 23.985 + 0.100 × 24.986 + 0.110 × 25.983✓ 1
= 24.3 u✓ 1

Exam tip: Convert each percentage abundance to a fraction before multiplying. As a check, the mean must lie between the lightest and heaviest isotope masses and be closest to the most abundant one.

77E-2-01Rutherford scattering Paper 2Medium12 marks

In an experiment of the Geiger–Marsden–Rutherford type, a narrow beam of alpha particles, all with the same kinetic energy, is directed at a very thin gold foil, 19779Au, inside an evacuated chamber. A detector that can be moved around the foil counts the alpha particles scattered through different angles.

Figure 1 shows three alpha particles, P, Q and R, approaching one gold nucleus. R is moving directly towards the nucleus.

Figure 1: three alpha particles P, Q and R approach a gold nucleus along parallel lines; R is aimed directly at the nucleus, Q passes close to it and P further away.gold nucleusPQRalpha particles approaching the nucleus
Figure 1 — diagram not to scale
(a)

The gold atom.

(i)

State the number of protons and the number of neutrons in a nucleus of 19779Au.

(1)
(ii)

Outline why the electrons in the gold atoms have a negligible effect on the paths of the alpha particles.

(1)
(b)

Draw, on Figure 1, the paths of alpha particles P, Q and R as they pass the nucleus and move away from it.

(3)
(c)

The detector shows that most alpha particles pass through the foil with very small deflections, and that a very small fraction are scattered through more than 90°. Explain what each of these observations shows about the structure of the atom.

(3)
(d)

The gold foil is replaced by an aluminium foil, 2713Al, of the same thickness. The number of nuclei per unit volume is almost the same in aluminium as in gold. Predict, with a reason, how the fraction of alpha particles scattered through more than 90° changes.

(2)
(e)

Suggest why the foil must be very thin and why the chamber is evacuated.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
79 protons and 197 − 79 = 118 neutrons✓ 1Both values needed.
Part (a)(ii)
The mass of an electron is tiny compared with that of an alpha particle «about 1/7300», so collisions with electrons hardly change the alpha particle's velocity✓ 1Accept: the electrons' negative charge is spread thinly through the atom, so the force it exerts is small. OWTTE
Part (b)
R slows down, stops short of the nucleus and returns back along its line of approach✓ 1Do not accept a path that touches or passes through the nucleus.
P and Q both curve away from the nucleus, the bending occurring close to it, and then continue in straight lines✓ 1Do not accept paths that bend towards the nucleus.
Q is deflected through a clearly larger angle than P✓ 1MP3 only scores if MP2 scores.
Part (c)
Most pass with very small deflections: the atom is mostly empty space / the nucleus occupies a tiny fraction of the volume of the atom✓ 1
A few are scattered through large angles: the positive charge of the atom is concentrated in a very small region, where the repulsive «electric» force is very large✓ 1
Turning back a fast, massive alpha particle shows that this small region «the nucleus» also contains most of the mass of the atom✓ 1Award [1 max] for 'the nucleus is small, dense and positive' with no link to the observations.
Part (d)
The fraction decreases✓ 1
The aluminium nucleus has a much smaller charge «13e compared with 79e», so the repulsive force at a given distance is smaller and fewer alpha particles are turned through large angles✓ 1MP2 only scores if MP1 scores. Do not accept answers based only on the mass of the foil.
Part (e)
Thin foil: so that each alpha particle is deflected by at most one nucleus «and the observed angle is due to a single encounter»✓ 1Accept: so that the alpha particles are not absorbed or slowed significantly in the foil.
Vacuum: alpha particles have a range of only a few centimetres in air; collisions with air molecules would absorb or deflect them✓ 1OWTTE

Answers: (a)(i) 79 protons, 118 neutrons (the remaining parts are explanations — see the table above)

78E-2-16Emission & absorption spectra Paper 2Medium12 marks

Figure 1 shows four energy levels, E1 to E4, of the atoms of an element Q. E1 is the ground state.

When Q is very hot it emits a line spectrum. In the visible region, 400 nm to 700 nm, there are three lines, at 478 nm, 497 nm and 654 nm (Figure 2, upper strip).

In a second experiment, white light is passed through a tube of Q vapour at room temperature and the transmitted light is examined with a spectroscope. The lower strip of Figure 2 shows the spectrum of the white light before it enters the tube.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Figure 1: energy levels of element Q: E1 = −6.00 eV (ground state), E2 = −3.50 eV, E3 = −1.60 eV, E4 = −0.90 eV.0E₁−6.00 eVE₂−3.50 eVE₃−1.60 eVE₄−0.90 eV
Figure 1 — energy levels drawn to scale
Figure 2: upper strip, emission spectrum of hot Q with bright lines at 478, 497 and 654 nm; lower strip, a continuous visible spectrum on which the candidate draws the absorption lines.emission spectrum of hot Qlight transmitted through cool Q (draw the dark lines here)400450500550600650700wavelength / nm
Figure 2 — wavelength scale drawn to scale
(a)

Calculate the wavelength of the photon emitted when an atom of Q moves from E2 to E1.

(2)
(b)

Identify the transitions that produce the emission lines at 478 nm and 654 nm.

(2)
(c)

The absorption spectrum of cool Q.

(i)

Draw, on the lower strip of Figure 2, the dark lines that are seen in the visible spectrum of the light transmitted through the cool vapour.

(2)
(ii)

Explain your answer to (c)(i).

(3)
(iii)

Atoms that absorb a photon re-emit a photon of the same wavelength. Explain why the absorption lines nevertheless appear dark.

(2)
(d)

Suggest how the visible absorption spectrum would change if the vapour were heated to a very high temperature, with the white light still passing through it.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = −3.50 − (−6.00) = 2.50 eV = 4.00 × 10−19 J✓ 1
λ = hc/ΔE = 6.63 × 10−34 × 3.00 × 108/4.00 × 10−19 = 4.97 × 10−7 m «497 nm»✓ 1Award [2] for CNA.
Part (b)
478 nm: photon energy hc/λ = 2.60 eV, the transition E4 → E2✓ 1
654 nm: photon energy 1.90 eV, the transition E3 → E2✓ 1Award [1 max] for both transitions identified with no supporting energies.
Part (c)(i)
One dark line, at 497 nm «directly below the 497 nm emission line»✓ 1
No dark lines at 478 nm or at 654 nm✓ 1MP2 only scores if MP1 scores.
Part (c)(ii)
At room temperature almost all the atoms of Q are in the ground state E1✓ 1
So only photons of energy En − E1 can be absorbed; the 478 nm and 654 nm lines need atoms already in E2, and there are almost none✓ 1
The other absorptions from E1 «to E3 and E4, about 283 nm and 244 nm» are in the ultraviolet, so they are not seen✓ 1Allow ECF from (c)(i).
Part (c)(iii)
The re-emitted photons travel in all «random» directions✓ 1
So far fewer photons of that wavelength continue towards the observer, and the line looks dark compared with the neighbouring wavelengths✓ 1OWTTE
Part (d)
Collisions would raise some atoms to E2, so dark lines could also appear at 478 nm and 654 nm✓ 1Accept: extra absorption lines due to transitions from excited states.

Answers: (a) 497 nm (the remaining parts are explanations — see the table above)

79E-2-17Emission & absorption spectra Paper 2Medium12 marks

The spectrum of a star S shows dark absorption lines on a continuous background. Table 1 gives the wavelengths of the absorption lines of S. Table 2 gives laboratory wavelengths of some lines of four elements.

(c = 3.00 × 108 m s−1, Wien's displacement constant = 2.9 × 10−3 m K)

Table 1: wavelengths of the absorption lines of star S / nm
393.49 · 396.97 · 434.18 · 486.28 · 589.17 · 589.77 · 656.48
Table 2: elementLaboratory wavelengths / nm
Hydrogen434.05 · 486.13 · 656.28
Helium447.15 · 587.56 · 667.82
Sodium588.99 · 589.59
Calcium393.37 · 396.85
(a)

Outline why the dark lines of an element in the spectrum of a star occur at wavelengths that also appear as bright lines in the laboratory emission spectrum of that element.

(2)
(b)

Deduce, with reasons, which of the four elements in Table 2 are present in the outer layers of S.

(3)
(c)

All the lines of S are shifted from their laboratory wavelengths.

(i)

Show that the fractional shift Δλ/λ of the hydrogen line at 656.28 nm is about 3.0 × 10−4.

(1)
(ii)

Determine the speed of S along the line of sight, and state whether S is moving towards or away from the Earth.

(2)
(iii)

State why every line in the spectrum of S is shifted by the same fraction of its wavelength.

(1)
(d)

The continuous spectrum of S has its maximum intensity at a wavelength of 480 nm. Estimate the surface temperature of S.

(1)
(e)

The spectrometer measures wavelengths with an uncertainty of ±0.05 nm. A second star moves directly away from the Earth at 15 km s−1. Determine whether the spectrometer can detect this motion using the calcium line at 393.37 nm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
An atom absorbs or emits only photons whose energy equals the difference between two of its energy levels, E = hc/λ✓ 1
Absorption between two levels involves the same energy difference, and so the same wavelength, as emission between them; each element has its own set of levels, so its lines identify it✓ 1OWTTE
Part (b)
Hydrogen and calcium are present: each of their laboratory lines has a matching line of S, slightly «0.1–0.2 nm» longer✓ 1
Sodium is present: both of its lines are matched, by 589.17 nm and 589.77 nm✓ 1
Helium is not present: S has no lines near 447.15, 587.56 or 667.82 nm «589.17 nm is the sodium line at 588.99 nm, not the helium line at 587.56 nm»✓ 1Award [1 max] for the three correct elements with no reasons.
Part (c)(i)
Δλ/λ = (656.48 − 656.28)/656.28 = 3.05 × 10−4✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
v = cΔλ/λ = 3.00 × 108 × 3.05 × 10−4 = 9.1 × 104 m s−1✓ 1Accept 8.9 × 104 to 9.3 × 104 m s−1 from other lines; allow use of 3.0 × 10−4 «9.0 × 104 m s−1».
Away from the Earth, because the observed wavelengths are longer «red-shifted»✓ 1
Part (c)(iii)
Δλ/λ ≈ v/c depends only on the speed of the source, not on the wavelength, and all the light from S comes from a source with the same speed✓ 1OWTTE
Part (d)
T = 2.9 × 10−3/480 × 10−9 = 6.0 × 103 K✓ 1Accept 6040 K.
Part (e)
Δλ = λv/c = 393.37 × 1.5 × 104/3.00 × 108 = 0.020 nm✓ 1
0.020 nm is smaller than the uncertainty of 0.05 nm, so the motion cannot be detected «reliably»✓ 1Award MP2 for a conclusion consistent with the candidate's Δλ.

Answers: (c)(i) 3.05 × 10−4  ·  (c)(ii) 9.1 × 104 m s−1, away from the Earth  ·  (d) 6.0 × 103 K  ·  (e) Δλ = 0.020 nm — not detectable (the remaining parts are explanations — see the table above)

80E-2-18Nuclear notation Paper 2Medium14 marks

Copper has two stable isotopes, copper-63 and copper-65. In a mass spectrometer, copper atoms are ionized so that every ion carries a charge of +e. The ions pass through a velocity selector and then through a narrow slit S into a region of uniform magnetic field of flux density 0.400 T directed into the page, where they move in semicircles and strike a detector plate (Figure 1).

In the velocity selector the ions move between two parallel plates 2.00 cm apart with a potential difference of 640 V between them. A uniform magnetic field of flux density 0.160 T, perpendicular to both the electric field and the velocity of the ions, also acts in the selector.

Masses of the ions: 63Cu+ 62.93 u, 65Cu+ 64.93 u. (1 u = 1.661 × 10−27 kg, e = 1.60 × 10−19 C)

Figure 1: copper ions leave a velocity selector, travel up the page through slit S in a detector plate and enter a uniform magnetic field directed into the page; the detector plate extends on both sides of S.×××××××××××××××××××××××××××××××××××××××××××××××××××××××Svelocity selectordetector plateuniform magnetic field, 0.400 T, directed into the pageion beam
Figure 1 — diagram not to scale
(a)

The ions.

(i)

State the number of protons, neutrons and electrons in one 6529Cu+ ion.

(2)
(ii)

Outline why the two isotopes of copper cannot be separated by chemical reactions.

(1)
(b)

The velocity selector.

(i)

Show that the ions leave the velocity selector with a speed of about 2.0 × 105 m s−1.

(2)
(ii)

Explain why ions of both isotopes leave the velocity selector with the same speed.

(2)
(c)

The region of uniform magnetic field.

(i)

Draw, on Figure 1, the paths of the 63Cu+ ions and of the 65Cu+ ions after they pass through S. Label each path.

(2)
(ii)

Determine the distance between the points at which the two types of ion strike the detector plate.

(3)
(d)

The current at the detector due to 63Cu+ ions is 6.4 × 10−10 A. Calculate the number of 63Cu+ ions arriving at the detector each second.

(1)
(e)

Suggest why the whole spectrometer must be evacuated.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
29 protons and 65 − 29 = 36 neutrons✓ 1
28 electrons «one electron has been removed from the neutral atom»✓ 1
Part (a)(ii)
Isotopes have the same number of protons and so the same number and arrangement of electrons in the atom; their chemical behaviour is identical✓ 1OWTTE
Part (b)(i)
E = V/d = 640/0.0200 = 3.20 × 104 V m−1✓ 1
For undeflected ions qE = qvB, so v = E/B = 3.20 × 104/0.160 = 2.00 × 105 m s−1✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
Only ions for which the electric force qE is balanced by the magnetic force qvB travel straight through to the exit slit; others are deflected✓ 1
The condition v = E/B does not depend on the mass «or charge» of the ion, so both isotopes emerge with the same speed✓ 1
Part (c)(i)
Both paths are semicircles that start at S and curve to the LEFT, ending on the detector plate✓ 1F = qvB on a positive ion moving up the page in a field into the page is to the left «Fleming's left-hand rule». Do not accept paths curving to the right.
The 65Cu+ path has the larger radius and both paths are labelled✓ 1Any visibly larger radius is acceptable.
Part (c)(ii)
qvB = mv2/r gives r = mv/qB; for 63Cu+ r = 62.93 × 1.661 × 10−27 × 2.00 × 105/(1.60 × 10−19 × 0.400) = 0.327 m✓ 1
For 65Cu+ r = 0.337 m✓ 1
Distance = 2 × (0.3370 − 0.3266) = 2.1 × 10−2 m✓ 1Accept 0.020–0.021 m. Award [3] for CNA. Allow ECF from (b)(i). Award [2 max] for 1.0 × 10−2 m «difference in radii, not diameters».
Part (d)
N = I/e = 6.4 × 10−10/1.60 × 10−19 = 4.0 × 109 s−1✓ 1
Part (e)
Collisions with gas molecules would deflect the ions or change their speed «or neutralise them», so they would not follow the predicted paths✓ 1OWTTE

Answers: (a)(i) 29 p, 36 n, 28 e  ·  (b)(i) 2.00 × 105 m s−1  ·  (c)(ii) 2.1 × 10−2 m  ·  (d) 4.0 × 109 s−1 (the remaining parts are explanations — see the table above)

81E-2-33Energy levels & photons Paper 2Medium13 marks

In a fluorescent lamp, mercury atoms emit ultraviolet photons of wavelength 254 nm. The inside of the glass tube is coated with a phosphor, which absorbs these photons and emits visible light.

The diagram shows four energy levels of an ion in the phosphor. The energies are measured from the ground state.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, Wien's displacement constant = 2.90 × 10−3 m K)

Four energy levels of a phosphor ion, measured from the ground state: 0, 2.10 eV, 2.65 eV and 4.89 eV, drawn to scale.energy / eV0ground state2.10 eV2.65 eV4.89 eV
Diagram drawn to scale
(a)

Show that the energy of a photon of wavelength 254 nm is about 4.9 eV.

(1)
(b)

The filament of a filament lamp is at a temperature of 2900 K.

(i)

Calculate the wavelength at which the spectrum of the filament has its maximum intensity.

(1)
(ii)

Hence suggest why a fluorescent lamp gives out much more visible light than a filament lamp with the same input power.

(2)
(c)

An ion in the ground state absorbs a 254 nm photon.

(i)

Draw arrows on the diagram to show this absorption and one way in which the ion can return to the ground state by emitting exactly two photons.

(2)
(ii)

Calculate the wavelength of the photon emitted in the transition from the 4.89 eV level to the 2.10 eV level.

(2)
(iii)

Determine whether a transition from the 2.65 eV level to the 2.10 eV level produces visible light.

(2)
(d)

Suggest why a lamp without the phosphor coating would give out very little visible light.

(1)
(e)

Some ions return to the ground state through the 2.65 eV and 2.10 eV levels in turn, emitting three photons. Determine the percentage of the energy of the absorbed ultraviolet photon that is emitted as visible light by these ions.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 6.63 × 10−34 × 3.00 × 108/254 × 10−9 = 7.83 × 10−19 J «= 4.89 eV»✓ 1Must see full substitution OR answer to 3 s.f. (4.89 eV).
Part (b)(i)
λmax = 2.90 × 10−3/2900 = 1.0 × 10−6 m «1000 nm»✓ 1
Part (b)(ii)
The filament gives a continuous spectrum whose peak is in the infrared, so most of its energy is emitted as infrared and only a small fraction between 400 nm and 700 nm✓ 1Allow ECF from (b)(i).
In the fluorescent lamp the energy is emitted at a few particular wavelengths: the ultraviolet line of mercury is converted by the phosphor into visible lines, so a large fraction of the energy is visible✓ 1OWTTE
Part (c)(i)
upward arrow from the ground state to the 4.89 eV level✓ 1
two downward arrows: from 4.89 eV to 2.65 eV OR to 2.10 eV, and then to the ground state✓ 1Arrows must start and end on levels. Do not award MP2 for a route with three emissions.
Part (c)(ii)
ΔE = 2.79 eV = 2.79 × 1.60 × 10−19 = 4.46 × 10−19 J✓ 1
λ = hc/ΔE = 4.46 × 10−7 m «446 nm»✓ 1Award [2] for CNA.
Part (c)(iii)
ΔE = 0.55 eV, so λ = hc/ΔE = 2.3 × 10−6 m✓ 1ALTERNATIVE: compare 0.55 eV with the photon energies of visible light, about 1.8–3.1 eV.
this is longer than 700 nm «infrared», so the light is not visible✓ 1Conclusion must be consistent with the calculation.
Part (d)
most of the energy is emitted by mercury as ultraviolet radiation, which is invisible «and is absorbed by the glass»✓ 1OWTTE
Part (e)
visible photons: 4.89 − 2.65 = 2.24 eV «555 nm» and 2.10 eV «592 nm»; the 0.55 eV photon is infrared✓ 1Allow ECF from (c)(iii).
(2.24 + 2.10)/4.89 × 100 = 89 %✓ 1Award [2] for CNA.

Answers: (a) 4.89 eV  ·  (b)(i) 1.0 × 10−6 m  ·  (c)(ii) 446 nm  ·  (c)(iii) 2.3 × 10−6 m — infrared, not visible  ·  (e) 89 % (the remaining parts are explanations — see the table above)

82E-2-38Rutherford scattering Paper 2Medium13 marks

Figure 1 shows two models of a gold atom, 19779Au. In model T the positive charge +79e is spread uniformly through a sphere of radius 1.4 × 10−10 m, the size of the whole atom, with the electrons embedded in it. In model N the same charge is concentrated in a nucleus of radius about 7 × 10−15 m.

In a Geiger–Marsden–Rutherford experiment alpha particles of kinetic energy 5.0 MeV are directed at a thin gold foil. Ignore the effect of the electrons on the alpha particles. (mass of an alpha particle = 6.64 × 10−27 kg, k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)

Figure 1: two models of a gold atom. Model T: positive charge +79e spread uniformly through a sphere the size of the atom with electrons embedded in it. Model N: the same charge concentrated in a tiny central nucleus, with the electrons outside it.Model TModel N++++++++++++++++++nucleuspositive charge +79e spreadthrough the whole atompositive charge +79e in atiny central nucleuselectron
Figure 1 — diagram not to scale
(a)

State the composition and the charge of an alpha particle.

(1)
(b)

Model T.

(i)

Show that the speed of the alpha particles is about 1.6 × 107 m s−1.

(1)
(ii)

The largest electric force on an alpha particle in model T acts when it is at the surface of the sphere of positive charge, where the charge acts as if it were at the centre. Calculate this force.

(2)
(iii)

An alpha particle crosses the atom in a time of about 2R/v, where R is the radius of the atom and v the speed of the alpha particle. Using your answer to (b)(ii), estimate the largest possible angle, in degrees, through which one atom of model T could deflect the alpha particle.

(3)
(c)

In model N an alpha particle can come within 5.0 × 10−14 m of the centre of a gold nucleus. Calculate the electric force on the alpha particle at this distance.

(1)
(d)

About 1 alpha particle in 8000 is deflected through more than 90° by the foil. Explain, with reference to your answers to (b) and (c), why model N can account for this observation but model T cannot.

(3)
(e)

Estimate the fraction of the cross-sectional area of an atom that is occupied by a nucleus in model N, and hence suggest why most alpha particles pass through the foil almost undeflected.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
two protons and two neutrons AND charge +2e «+3.2 × 10−19 C»✓ 1Both needed. Accept 'a helium-4 nucleus' for the composition.
Part (b)(i)
v = √(2Ek/m) = √(2 × 5.0 × 1.60 × 10−13/6.64 × 10−27) «= 1.55 × 107 m s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
F = kq1q2/r2 = 8.99 × 109 × (2 × 1.60 × 10−19) × (79 × 1.60 × 10−19)/(1.4 × 10−10)2✓ 1Charges 2e and 79e must be used.
F = 1.9 × 10−6 N✓ 1Award [2] for CNA. Award [1 max] for 9.3 × 10−7 N «charge of the alpha particle taken as e».
Part (b)(iii)
Δt = 2 × 1.4 × 10−10/1.6 × 107 = 1.8 × 10−17 s✓ 1
largest change of momentum «impulse» Δp = FΔt = 1.9 × 10−6 × 1.8 × 10−17 = 3.4 × 10−23 N s✓ 1Allow ECF from (b)(ii).
angle ≈ Δp/p = 3.4 × 10−23/(6.64 × 10−27 × 1.6 × 107) = 3.2 × 10−4 rad ≈ 0.018°✓ 1Accept 0.017°–0.021° «or the answer in rad». Allow ECF. Any answer of the order of 10−2 degrees scores if the method is correct.
Part (c)
F = 8.99 × 109 × 3.2 × 10−19 × 1.264 × 10−17/(5.0 × 10−14)2 = 15 N✓ 1Accept 14–15 N «≈ 8 × 106 × the answer to (b)(ii)». Award [0] for an answer in which the distance is not squared.
Part (d)
Model T: the largest deflection by one atom is a tiny fraction of a degree «(b)(iii)», so even many atoms in the foil cannot turn an alpha particle through more than 90°✓ 1Allow ECF from (b)(iii).
Model N: near the nucleus the force is about 107 times larger «15 N compared with 1.9 × 10−6 N», because the same charge is concentrated in a much smaller region✓ 1Allow ECF from (c).
so a single close encounter with a nucleus «which also contains most of the mass of the atom» can reverse the direction of an alpha particle, giving the large-angle scattering observed✓ 1OWTTE
Part (e)
(7 × 10−15/1.4 × 10−10)2 ≈ 2.5 × 10−9✓ 1Accept 10−9 to 10−8.
the nucleus occupies a tiny fraction of the atom, so most alpha particles pass far from every nucleus, where the force on them is small «the atom is mostly empty space»✓ 1OWTTE

Answers: (b)(i) 1.55 × 107 m s−1  ·  (b)(ii) 1.9 × 10−6 N  ·  (b)(iii) ≈ 3.2 × 10−4 rad ≈ 0.018°  ·  (c) 15 N  ·  (e) ≈ 2.5 × 10−9 (the remaining parts are explanations — see the table above)

83E-2-39Energy levels & photons Paper 2Medium14 marks

The light of an aurora is emitted by atoms high in the atmosphere after they have been excited by collisions with fast electrons. Two of the brightest lines are emitted by oxygen atoms: a green line of wavelength 557.7 nm and a red line of wavelength 630.0 nm. Figure 1 shows three energy levels of an oxygen atom, measured from the ground state E1.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)

Figure 1: three energy levels of an oxygen atom drawn to scale, measured from the ground state: E1 = 0, E2 = 1.97 eV, E3 = 4.19 eV.energy above the ground stateE₁0 eV (ground state)E₂1.97 eVE₃4.19 eV
Figure 1 — energy levels drawn to scale
(a)

Calculate, in eV, the energy of a photon of the green light.

(2)
(b)

Draw, on Figure 1, arrows to show the transitions that emit the red line and the green line. Label each arrow with its wavelength.

(2)
(c)

An atom in level E3 can also return directly to the ground state. Determine whether the photon emitted in this transition is visible.

(2)
(d)

Electrons from space spiral along the lines of the Earth's magnetic field as they descend into the upper atmosphere.

(i)

An electron of kinetic energy 1.0 keV moves at right angles to a uniform magnetic field of flux density 5.0 × 10−5 T. Calculate the radius of its circular path.

(3)
(ii)

Explain why the magnetic field does not change the kinetic energy of the electron.

(2)
(iii)

Estimate the largest number of oxygen atoms that one such electron could excite to level E3.

(1)
(e)

An oxygen atom typically stays in level E2 for about 100 s before it emits a photon, but stays in level E3 for less than 1 s. The red line is emitted strongly only above a height of about 200 km, where the air is very thin, whereas the green line is also emitted lower down. Suggest an explanation for this difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/557.7 × 10−9 = 3.57 × 10−19 J✓ 1
= 3.57 × 10−19/1.60 × 10−19 = 2.23 eV✓ 1Award [2] for CNA.
Part (b)
red «630.0 nm, 1.97 eV»: downward arrow from E2 to E1✓ 1Arrows must start and end on levels.
green «557.7 nm, 2.23 eV ≈ 4.19 − 1.97»: downward arrow from E3 to E2✓ 1Award [1 max] if both arrows point upwards or the labels are swapped.
Part (c)
ΔE = 4.19 eV, so λ = hc/ΔE = 1.989 × 10−25/(4.19 × 1.60 × 10−19) = 2.97 × 10−7 m «297 nm»✓ 1ALTERNATIVE: compare 4.19 eV with the photon energies of visible light, about 1.8 eV to 3.1 eV.
this is shorter than 400 nm «ultraviolet», so the photon is not visible✓ 1The conclusion must be consistent with the calculation.
Part (d)(i)
v = √(2Ek/m) = √(2 × 1.0 × 103 × 1.60 × 10−19/9.11 × 10−31) = 1.9 × 107 m s−1✓ 1
evB = mv2/r, so r = mv/eB✓ 1
r = 9.11 × 10−31 × 1.87 × 107/(1.60 × 10−19 × 5.0 × 10−5) = 2.1 m✓ 1Award [3] for CNA. Allow ECF for the speed. Award [2 max] if the energy is not converted from keV to J.
Part (d)(ii)
the magnetic force is always perpendicular to the velocity of the electron✓ 1
so the force does no work on the electron «W = Fs cos 90° = 0» and its speed and kinetic energy stay constant✓ 1OWTTE
Part (d)(iii)
1.0 × 103/4.19 ≈ 240✓ 1Accept 230–240. Assumes that each collision transfers exactly 4.19 eV.
Part (e)
at lower heights the air is denser, so an excited atom collides with other particles much more often✓ 1
an atom in E2 usually loses its energy in a collision before the 100 s needed to emit the red photon, whereas an atom in E3 emits the green photon within 1 s, before it is likely to collide✓ 1OWTTE. Award [1 max] for 'more collisions lower down' with no reference to the different times.

Answers: (a) 2.23 eV  ·  (c) 297 nm — ultraviolet, not visible  ·  (d)(i) 2.1 m  ·  (d)(iii) ≈ 240 (the remaining parts are explanations — see the table above)

84E-2-40Emission & absorption spectra Paper 2Medium12 marks

A water company tests drinking water for lead using the apparatus in Figure 1. A lamp whose cathode is made of lead emits the line spectrum of lead. The light passes through a flame at a temperature of 2500 K into which the water sample is sprayed, so that any lead in the water forms a vapour of free lead atoms. A filter passes only the lead line of wavelength 283.3 nm to a detector, which measures the percentage of this light that is absorbed in the flame. The 283.3 nm line is absorbed by lead atoms in the ground state; even the lowest excited level of a lead atom is about 1 eV above the ground state.

The table shows the results for standard solutions of known lead concentration, measured in the same session. The legal limit for lead in drinking water is 10 μg L−1. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, kB = 1.38 × 10−23 J K−1)

Lead concentration / μg L−10.05.010.015.020.0
Light absorbed / %0.11.93.95.67.7
Figure 1: atomic absorption apparatus. Light from a lamp with a lead cathode passes through a flame into which the water sample is sprayed, then through a filter that passes only 283.3 nm, to a detector.lamp withlead cathodeflame: samplesprayed infilter for283.3 nmdetector
Figure 1 — diagram not to scale
(a)

Calculate, in eV, the energy of a photon of wavelength 283.3 nm.

(2)
(b)

Explain why lead atoms in the flame absorb the 283.3 nm light from the lamp, whereas atoms of other elements in the flame do not.

(2)
(c)

The lead atoms in the flame.

(i)

Show that the mean kinetic energy of an atom in the flame is about 0.3 eV.

(1)
(ii)

Hence explain why almost all the lead atoms in the flame are in the ground state, and why this is necessary for the method to work.

(2)
(d)

Suggest why a lamp containing lead is used rather than a white-light source.

(1)
(e)

A sample of drinking water absorbs 4.1 % of the 283.3 nm light. Determine whether the water meets the legal limit.

(3)
(f)

Suggest why the standard solutions must be measured in the same session as the sample.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/283.3 × 10−9 = 7.02 × 10−19 J✓ 1
= 4.39 eV✓ 1Award [2] for CNA.
Part (b)
a photon is absorbed only if its energy equals the difference between two energy levels of the atom; the lead atoms in the flame have the same levels as those in the lamp that emitted the light✓ 1
each element has its own set of energy levels, so atoms of other elements have no pair of levels separated by 4.39 eV and do not absorb this wavelength✓ 1OWTTE
Part (c)(i)
Ek = 3/2 kBT = 1.5 × 1.38 × 10−23 × 2500 = 5.17 × 10−20 J «= 0.32 eV»✓ 1Must see full substitution OR answer to 3 s.f. «0.323 eV».
Part (c)(ii)
0.3 eV is less than the ≈ 1 eV needed to raise a lead atom even to its lowest excited level «and far less than 4.39 eV», so only a small fraction of collisions can excite an atom✓ 1Allow ECF from (c)(i).
only ground-state atoms absorb the 283.3 nm line, so with «almost» every lead atom in the ground state the fraction absorbed depends only on the number of lead atoms, i.e. on the concentration✓ 1OWTTE
Part (d)
the lead line has exactly the wavelength «and very narrow width» that lead atoms absorb; with white light almost all of the light reaching the detector would be at wavelengths that lead does not absorb, so the change would be too small to measure✓ 1OWTTE
Part (e)
from the calibration data, absorption ≈ 0.38 % per μg L−1✓ 1Accept a gradient from a graph or any pair of data points, 0.37–0.39 % per μg L−1.
concentration = 4.1/0.38 ≈ 10.8 μg L−1✓ 1Accept 10.5–11.1 μg L−1. Accept interpolation between 10.0 and 15.0 μg L−1.
this is greater than 10 μg L−1, so the water does not meet the legal limit✓ 1The conclusion must be consistent with the value.
Part (f)
the conditions in the flame «temperature, rate at which the sample is sprayed, lamp brightness» can change between sessions, so the fraction of the light absorbed for a given concentration changes «a systematic error»✓ 1OWTTE

Answers: (a) 4.39 eV  ·  (c)(i) 0.32 eV  ·  (e) ≈ 10.8 μg L−1 — above the limit (the remaining parts are explanations — see the table above)

85E-2-54Energy levels & photons Paper 2Medium11 marks

To correct for the blurring caused by the atmosphere, a large telescope makes an artificial star. A laser of output power 20 W and wavelength 589.0 nm sends a beam vertically upwards into a thin layer of sodium atoms that lies 90 km above the telescope. The sodium atoms absorb photons from the beam and then re-emit photons of the same wavelength, equally in all directions, so that a small glowing spot is seen from the ground. The figure shows some energy levels of a sodium atom.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Energy levels of a sodium atom measured from the ground state, drawn to scale: 0 (ground state), 2.11 eV, 3.19 eV and 3.76 eV.0 (ground state)2.11 eV3.19 eV3.76 eV
Energy levels drawn to scale
(a)
(i)

Show that the energy of a photon of wavelength 589.0 nm is about 3.4 × 10−19 J.

(1)
(ii)

Calculate the number of photons emitted by the laser each second.

(1)
(b)

Explain why the laser must be tuned very precisely to 589.0 nm, and why the nitrogen and oxygen molecules of the atmosphere hardly absorb the beam.

(2)
(c)

A fraction 0.030 of the photons in the beam are absorbed in the sodium layer, and each absorbed photon is re-emitted. The telescope mirror has a diameter of 8.2 m. Determine the number of re-emitted photons collected by the mirror each second. Ignore absorption in the lower atmosphere.

(3)
(d)
(i)

Draw, on the figure, an arrow to show the absorption of a laser photon by a sodium atom.

(1)
(ii)

Outline why the re-emitted photons have the same wavelength as the laser photons.

(1)
(e)

Sodium atoms in the ground state also absorb photons of wavelength 589.6 nm, which raise them to another excited level just below the 2.11 eV level (not shown on the figure). Determine the difference in energy, in eV, between these two excited levels.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/589.0 × 10−9 «= 3.38 × 10−19 J»✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
N = 20/3.38 × 10−19 = 5.9 × 1019 s−1✓ 1Accept 5.9–6.0 × 1019 s−1.
Part (b)
A photon is absorbed only if its energy exactly equals the difference between two energy levels of the atom; 589.0 nm matches the sodium ground state → 2.11 eV transition✓ 1
Nitrogen and oxygen have different energy levels, with no energy difference equal to the photon energy «2.11 eV», so they do not absorb this wavelength✓ 1OWTTE
Part (c)
Photons absorbed «and re-emitted» per second = 0.030 × 5.9 × 1019 = 1.8 × 1018 s−1✓ 1Allow ECF from (a)(ii).
Fraction reaching the mirror = πr²/4πd² = π × 4.1²/(4π × (9.0 × 104)²) = 5.2 × 10−10✓ 1
Number collected = 9.2 × 108 s−1✓ 1Accept 9.2–9.4 × 108 s−1. Award [3] for CNA. Award [2 max] if the diameter is used as the radius.
Part (d)(i)
Single upward arrow from the ground state to the 2.11 eV level✓ 1Arrow must start and end on the levels.
Part (d)(ii)
The excited atom can only return from the 2.11 eV level to the ground state «no level between them on the figure», so it emits a photon of the same energy and so the same wavelength✓ 1OWTTE
Part (e)
Photon energies hc/λ = 2.1106 eV «589.0 nm» and 2.1084 eV «589.6 nm»✓ 1OR ΔE = hc(1/589.0 − 1/589.6)/(e × 10−9). Energies must be kept to at least 4 s.f.
ΔE = 2.1 × 10−3 eV✓ 1Accept 2.1–2.2 × 10−3 eV. Award [1 max] for 0 eV «energies rounded to 3 s.f. before subtracting».

Answers: (a)(ii) 5.9 × 1019 s−1  ·  (c) 9.2 × 108 s−1  ·  (e) 2.1 × 10−3 eV (the remaining parts are explanations — see the table above)

86E-2-58Emission & absorption spectra Paper 2Medium10 marks

A museum tests a painting said to date from 1890 with a portable X-ray fluorescence analyser. X-ray photons from the analyser remove electrons from the innermost energy levels of atoms in the paint. Electrons from higher levels then fall into the vacancies, and the atoms emit X-ray photons. The graph shows the spectrum recorded from an area of white paint; photon energies are measured with an uncertainty of ±0.15 keV. The table gives the energies of the strongest X-ray lines of some elements.

White pigments: lead white has been used for centuries; titanium white was first produced as a pigment in about 1920. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

ElementLine energies / keV
calcium3.69
titanium4.51
iron6.40
copper8.05
zinc8.64
mercury9.99
lead10.55 and 12.61
X-ray spectrum from the white paint: number of photons detected against photon energy from 2 to 14 keV. There are four peaks, at 3.7 keV (small), 4.5 keV (largest), 10.6 keV and 12.6 keV, on a low background.2468101214photon energy / keV020040060080010001200number of photons detected
Graph drawn to scale
(a)

Calculate the wavelength of a photon of energy 10.55 keV.

(2)
(b)

Explain why the X-ray photons emitted by atoms of each element have particular energies.

(2)
(c)

Identify, with reasons, the elements present in the white paint.

(2)
(d)
(i)

Determine whether this area of white paint can have been painted in 1890.

(2)
(ii)

Suggest why your conclusion in (d)(i) does not prove that the whole painting is a forgery.

(1)
(e)

Suggest why the photons from the analyser must each have an energy greater than 12.61 keV for the 12.61 keV line of lead to be produced.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 10.55 × 103 × 1.60 × 10−19 = 1.69 × 10−15 J✓ 1
λ = hc/E = 1.18 × 10−10 m✓ 1Award [2] for CNA.
Part (b)
Each element has its own set of discrete energy levels «inner levels»✓ 1
The photon energy equals the difference between the two levels involved, so only particular photon energies are emitted by each element✓ 1OWTTE
Part (c)
Titanium «4.5 keV» and lead «both 10.55 and 12.61 keV lines present»✓ 1
Calcium «3.7 keV»; mercury is absent «no peak at 9.99 keV, which differs from 10.55 keV by more than the uncertainty»✓ 1Award [1 max] for the elements listed with no reference to the peak energies.
Part (d)(i)
Titanium is present, so the white paint contains titanium white✓ 1Allow ECF from (c).
Titanium white was not available until about 1920, so this paint cannot date from 1890✓ 1
Part (d)(ii)
The area may have been retouched/restored later «with modern paint»✓ 1OWTTE
Part (e)
The photon must remove an inner electron completely from the atom, which needs more energy than the difference between the inner level and the higher level from which an electron falls✓ 1OWTTE

Answers: (a) 1.18 × 10−10 m (the remaining parts are explanations — see the table above)

87E-2-60Nuclear notation Paper 2Medium12 marks

The ion thruster of a space probe uses xenon. Natural xenon is a mixture of isotopes with nucleon numbers from 124 to 136, for example 13154Xe. Xenon atoms are ionised by collisions with electrons, and each atom loses one electron. The ions are then accelerated from rest through a potential difference of 1100 V and leave the thruster as a beam. The current carried by the beam is 1.76 A. The thruster gives out a blue glow.

(e = 1.60 × 10−19 C, 1 u = 1.661 × 10−27 kg, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1; for (b) to (e) take the mass of a xenon ion to be 131 u, the mean value for natural xenon)

(a)
(i)

State one quantity that is the same for the nuclei of all the isotopes of xenon and one quantity that is different.

(1)
(ii)

State, with a reason, which isotope of xenon gives ions that leave the thruster with the greatest speed.

(1)
(b)

Show that the speed of an ion of mass 131 u in the beam is about 4.0 × 104 m s−1.

(2)
(c)

Calculate the number of ions that leave the thruster each second.

(1)
(d)

Determine the force exerted on the probe by the thruster.

(2)
(e)

The energy needed to ionise one xenon atom is 12.1 eV. Determine whether ionising the xenon uses a significant fraction of the power carried by the beam.

(2)
(f)
(i)

The strongest line in the blue glow, emitted by xenon ions, has a wavelength of 484.4 nm. Calculate the energy, in eV, of a photon of this line.

(1)
(ii)

The grids that accelerate the ions are made of molybdenum and are slowly worn away by the beam. Explain how engineers can use the spectrum of the glow to detect this wear.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Same: number of protons «54» / charge of the nucleus; different: number of neutrons «70 to 82» / nucleon number / mass✓ 1Both needed.
Part (a)(ii)
124Xe: every ion has the same charge, so it gains the same kinetic energy qV; the ion of smallest mass has the greatest speed✓ 1Isotope and reason both needed.
Part (b)
Ek = qV = 1.60 × 10−19 × 1100 = 1.76 × 10−16 J✓ 1
v = √(2 × 1.76 × 10−16/(131 × 1.661 × 10−27)) «= 4.02 × 104 m s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
N = I/e = 1.76/1.60 × 10−19 = 1.1 × 1019 s−1✓ 1
Part (d)
Force = rate of change of momentum of the ions = (N/t)mv✓ 1Newton's third law: the force on the probe equals that on the ions.
= 1.1 × 1019 × 2.18 × 10−25 × 4.0 × 104 = 0.096 N✓ 1Allow ECF from (b) and (c). Accept 0.095–0.097 N. Award [2] for CNA.
Part (e)
Power to ionise = 1.1 × 1019 × 12.1 × 1.60 × 10−19 = 21 W; beam power = IV = 1940 W✓ 1Allow ECF from (c).
Fraction = 1.1 %, so ionisation uses only a small fraction of the beam power✓ 1Conclusion must be consistent with the calculation.
Part (f)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(484.4 × 10−9 × 1.60 × 10−19) = 2.57 eV✓ 1
Part (f)(ii)
Molybdenum atoms have their own energy levels and so their own unique set of spectral lines✓ 1
If molybdenum lines appear in the spectrum of the glow, molybdenum atoms from the grids are in the beam «brighter lines indicate faster wear»✓ 1OWTTE

Answers: (c) 1.1 × 1019 s−1  ·  (d) 0.096 N  ·  (e) 1.1 %  ·  (f)(i) 2.57 eV (the remaining parts are explanations — see the table above)

88E-2-61Emission & absorption spectra Paper 2Medium9 marks

A lightning stroke heats a narrow channel of air to about 28 000 K. A spectrometer pointed at the stroke records a continuous spectrum with bright lines on it. The lines include one at 500.5 nm emitted by nitrogen ions N+, one at 777.4 nm emitted by oxygen atoms, and one at 656.3 nm emitted by hydrogen atoms.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, Wien's displacement constant = 2.9 × 10−3 m K, speed of sound in air = 340 m s−1)

(a)
(i)

Calculate the wavelength at which the continuous spectrum of the channel has its maximum intensity.

(1)
(ii)

Hence state why most of the energy radiated by the channel is not visible.

(1)
(b)

Explain how the bright lines are produced in the hot channel.

(2)
(c)
(i)

Show that the energy of a photon of wavelength 777.4 nm is about 1.6 eV.

(1)
(ii)

Visible photons have energies from about 1.8 eV to 3.1 eV. State whether the 777.4 nm line can be seen by eye.

(1)
(d)

Air contains almost no hydrogen gas. Suggest why the spectrum contains a hydrogen line.

(1)
(e)
(i)

The stroke transfers a charge of 15 C through a potential difference of 1.0 × 108 V. Calculate the energy transferred.

(1)
(ii)

An observer hears the thunder 4.5 s after seeing the flash. Estimate the distance of the stroke from the observer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λmax = 2.9 × 10−3/28 000 = 1.0 × 10−7 m✓ 1
Part (a)(ii)
The peak is in the ultraviolet «shorter than 400 nm», so most of the energy is radiated at invisible wavelengths✓ 1Allow ECF from (a)(i).
Part (b)
Collisions in the hot gas raise electrons in the atoms/ions to higher energy levels✓ 1
When electrons fall to lower levels, photons are emitted with energies equal to the differences between discrete levels, giving lines at particular wavelengths✓ 1OWTTE
Part (c)(i)
E = 6.63 × 10−34 × 3.00 × 108/(777.4 × 10−9 × 1.60 × 10−19) «= 1.60 eV»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
No: 1.6 eV is below 1.8 eV «infrared»✓ 1Allow ECF from (c)(i).
Part (d)
Water vapour in the air is broken up in the hot channel, releasing hydrogen atoms✓ 1OWTTE
Part (e)(i)
W = qV = 15 × 1.0 × 108 = 1.5 × 109 J✓ 1
Part (e)(ii)
d = 340 × 4.5 ≈ 1500 m «the light arrives almost instantly»✓ 1Accept 1.5 km.

Answers: (a)(i) 1.0 × 10−7 m  ·  (e)(i) 1.5 × 109 J  ·  (e)(ii) 1500 m (the remaining parts are explanations — see the table above)

89IBP2S-035-01-12Energy levels & photons Paper 2Medium10 marks

An engineer is testing a gas for a new type of discharge lamp. The table gives the four lowest energy levels of the gas atoms.

Level | Energy / eV
n = 1 | −8.20
n = 2 | −4.10
n = 3 | −2.30
n = 4 | −1.20

(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

−1.20 eVn = 4−2.30 eVn = 3−4.10 eVn = 2−8.20 eVn = 1
(a)

Determine, in J, the ionisation energy of an atom in its ground state.

(2)
(b)

A photon of energy 6.50 eV is incident on a ground-state atom. Explain whether the photon can be absorbed.

(2)
(c)

Visible light has wavelengths from 400 nm to 700 nm. Determine which transitions between these four levels produce visible light.

(3)
(d)

Calculate the longest wavelength that can be emitted in transitions between these four levels.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The ionisation energy is the energy needed to raise the electron from −8.20 eV to 0, so it is 8.20 eV .✓ 1
8.20 × 1.60×10−19 = 1.31×10−18 J .✓ 1
Part (b)
To be absorbed, the photon would have to raise the electron to a level at −8.20 + 6.50 = −1.70 eV .✓ 1
There is no level at −1.70 eV, and 6.50 eV is less than the ionisation energy, so the photon is not absorbed .✓ 1
Part (c)
Visible photon energies range from hc/(700 nm) = 1.78 eV to hc/(400 nm) = 3.11 eV .✓ 1
The energy differences are: 4→3: 1.10 eV; 3→2: 1.80 eV; 4→2: 2.90 eV; 2→1: 4.10 eV; 3→1: 5.90 eV; 4→1: 7.00 eV .✓ 1
Only 3→2 (1.80 eV, about 690 nm) and 4→2 (2.90 eV, about 429 nm) are visible .✓ 1
Part (d)
The longest wavelength comes from the smallest energy gap, 4→3, which is 1.10 eV .✓ 1
ΔE = 1.10 × 1.60×10−19 = 1.76×10−19 J .✓ 1
λ = hc/ΔE = 1.989×10−25/1.76×10−19 = 1.13×10−6 m, which is in the infrared .✓ 1

Exam tip: Convert the visible wavelength range into an energy range once. Then you can compare each level gap with it directly instead of calculating six wavelengths.

90IBP2S-035-01-14Energy levels & photons Paper 2Medium6 marks

A pulse oximeter clipped to a patient's finger measures blood oxygen using two LEDs. One emits red light of wavelength 660 nm and the other emits infrared radiation of wavelength 940 nm. Each LED emits a power of 2.0 mW.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

Calculate, in eV, the energy of a photon emitted by the red LED.

(2)
(b)

Without further calculation, explain which LED emits more photons per second.

(2)
(c)

Calculate the number of photons emitted per second by the infrared LED.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = (6.63×10−34 × 3.00×108)/(660×10−9) = 3.01×10−19 J .✓ 1
E = 3.01×10−19/1.60×10−19 = 1.88 eV .✓ 1
Part (b)
The infrared photons have a longer wavelength, so each carries less energy, because E = hc/λ .✓ 1
For the same power (energy per second), more photons must be emitted per second, so the infrared LED emits more photons per second .✓ 1
Part (c)
E = (6.63×10−34 × 3.00×108)/(940×10−9) = 2.12×10−19 J .✓ 1
N = P/E = 2.0×10−3/2.116×10−19 = 9.5×1015 s−1. Accept 9.4×1015 s−1 to 9.5×1015 s−1, because rounding E to 2.12×10−19 J first gives 9.4×1015 s−1 .✓ 1

Exam tip: In an 'explain' answer, link three ideas: wavelength, energy per photon, and power as energy per second.

91IBP2S-035-01-16Energy levels & photons Paper 2Medium7 marks

A ruby laser used in dermatology clinics removes dark tattoo ink. Chromium ions in the ruby absorb pump photons of wavelength 550 nm, which raise them from the ground state to a pump level. The ions then drop quickly, without emitting light, to a lower metastable level. Laser light of wavelength 694 nm is emitted when ions fall from the metastable level to the ground state.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

Sketch an energy-level diagram for the chromium ion showing the three levels and the three transitions described. (Describe your sketch in words.)

(2)
(b)

Calculate, in eV, the energy of a laser photon.

(2)
(c)

Determine the energy, in eV, transferred non-radiatively by each ion as it drops from the pump level to the metastable level.

(2)
(d)

A single laser pulse delivers 1.5 J of energy at 694 nm. Calculate the number of photons in the pulse.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The diagram shows three horizontal lines: the ground state at the bottom, the metastable level above it, and the pump level at the top. An upward arrow from the ground state to the pump level is labelled 'absorption, 550 nm', and a downward arrow from the pump level to the metastable level is labelled 'non-radiative' .✓ 1
A downward arrow from the metastable level to the ground state is labelled 'emission, 694 nm'. This arrow is shorter than the pump arrow, because the metastable level lies below the pump level .✓ 1
Part (b)
E = hc/λ = (6.63×10−34 × 3.00×108)/(694×10−9) = 2.87×10−19 J .✓ 1
E = 2.87×10−19/1.60×10−19 = 1.79 eV .✓ 1
Part (c)
Pump photon energy = hc/(550×10−9) = 3.62×10−19 J, which is 2.26 eV .✓ 1
Energy transferred = 2.26 − 1.79 = 0.47 eV (7.5×10−20 J) . ECF from (b).✓ 1
Part (d)
N = 1.5/2.87×10−19 = 5.2×1018 . ECF from (b).✓ 1

Model sketch

pump levelmetastable levelground stateabsorbs 550 nmnon-radiativeemits 694 nm

Common error: The pump photon and the laser photon have different energies. The emitted photon matches the metastable-to-ground gap, not the gap that the pump photon bridged.

92IBP2S-035-00-12Energy levels & photons Paper 2Medium7 marks

A germicidal lamp, used to sterilise hospital air, contains mercury vapour. The table gives five energy levels of a mercury atom (simplified).

Level : Energy / eV
E0 (ground state) : −10.4
E1 : −5.5
E2 : −3.7
E3 : −2.7
E4 : −1.6

(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C. Visible light: 400 nm to 700 nm.)

−1.6 eVE₄−2.7 eVE₃−3.7 eVE₂−5.5 eVE₁−10.4 eVE₀ (ground state)
(a)

Determine, in J, the ionization energy of a mercury atom in its ground state.

(2)
(b)

Show that the transition from E3 to E1 produces visible light.

(2)
(c)

The germicidal action is due to the transition from E1 to E0. Calculate the wavelength of the emitted photon.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Ionization energy = 0 − (−10.4) = 10.4 eV✓ 1
10.4 × 1.60×10−19 = 1.66×10−18 J✓ 1
Part (b)
ΔE = −2.7 − (−5.5) = 2.8 eV = 4.48×10−19 J✓ 1
λ = hc/ΔE = 1.989×10−25/4.48×10−19 = 4.4×10−7 m, i.e. 444 nm, which lies between 400 nm and 700 nm✓ 1
Part (c)
ΔE = −5.5 − (−10.4) = 4.9 eV✓ 1
ΔE = 4.9 × 1.60×10−19 = 7.84×10−19 J✓ 1
λ = hc/ΔE = 2.54×10−7 m (254 nm, ultraviolet)✓ 1

Common error: Use the difference between the two levels, not the energy of one level. Here, using 5.5 eV alone gives the wrong wavelength.

93IBP2S-035-00-13Emission & absorption spectra Paper 2Medium6 marks

A cloud of hydrogen gas lies between Earth and a very hot star. The spectrum of the star, observed from Earth, shows a dark line at a wavelength of 656 nm. This line corresponds to hydrogen atoms being excited from the n = 2 level to the n = 3 level.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

Outline how the dark line is produced.

(2)
(b)

Calculate, in eV, the energy difference between the n = 2 and n = 3 levels.

(2)
(c)

A second cloud of very cold hydrogen shows no dark line at 656 nm, although its hydrogen atoms absorb strongly in the ultraviolet. Explain this observation.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Atoms in the cloud absorb photons whose energy equals the n = 2 to n = 3 energy difference, removing them from the beam✓ 1
The atoms then re-emit this energy in random directions (or by other transitions), so less light at 656 nm reaches Earth✓ 1
Part (b)
ΔE = hc/λ = (6.63×10−34 × 3.00×108)/(656×10−9) = 3.03×10−19 J✓ 1
ΔE = 3.03×10−19/1.60×10−19 = 1.9 eV✓ 1
Part (c)
In cold gas almost all atoms are in the ground state (n = 1), so few are in n = 2 to absorb 656 nm photons✓ 1
Absorption from n = 1 needs larger energy differences (at least 10.2 eV), so the absorption lines are in the ultraviolet✓ 1

Exam tip: An absorption line needs atoms already in the lower level of the transition. Think about which level is populated.

94IBP2S-035-02-14Rutherford scattering Paper 2Medium7 marks

A university outreach laboratory runs a scattering experiment in a sealed vacuum chamber. Alpha particles from a source strike a thin gold foil. A detector is rotated to different scattering angles θ around the foil, and the count rate is recorded at each angle.

θ / ° : 10, 30, 60, 90, 150
count rate / min−1 : 75 400, 970, 70, 17, 5

Background radiation is negligible.

α sourcethin gold foildetectorθ
Diagram not to scale
(a)

State how the count rate varies with scattering angle.

(1)
(b)

Explain why the count rate at 150° is evidence against the Thomson model of the atom.

(2)
(c)

Outline, in terms of the paths of the alpha particles relative to the nuclei, why the count rate at 10° is much greater than at 150°.

(2)
(d)

The foil is replaced by a gold foil twice as thick, which is still thin. The same source and detector are used, in the same positions. Estimate the new count rate at 60°, justifying your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The count rate decreases very rapidly (non-linearly, by orders of magnitude) as the scattering angle increases, e.g. from 75 400 min−1 at 10° to 5 min−1 at 150° .✓ 1
Part (b)
In the Thomson model the positive charge is spread through the whole atom, so the force on an alpha particle is always small and each deflection is tiny. Many small deflections in random directions cannot add up to a deflection as large as 150° . A non-zero count at 150° requires a very large force, which only positive charge concentrated in a tiny, massive nucleus can provide .✓ 2
Part (c)
A large deflection requires the alpha particle to pass very close to a nucleus, almost head-on, where the electric repulsion is very large . The nuclei are tiny compared with the atoms, so very few alpha particles pass that close; most pass far from any nucleus, feel only a weak force and are deflected through small angles .✓ 2
Part (d)
Each alpha particle is still scattered at most once, and there are twice as many nuclei along its path, so the probability of scattering through 60° doubles . The new count rate is 2 × 70 ≈ 140 min−1 .✓ 2

Exam tip: In data questions, use the actual numbers in your answer: for example, 'falls from 75 400 to 5 min⁻¹'. This makes a trend statement precise.

95IBP2S-035-03-13Nuclear notation Paper 2Medium6 marks

In industry, helium leak detectors are used to test welded vacuum chambers. Helium gas sprayed outside the chamber enters through any leak and is ionised to singly charged ions. A mass spectrometer then separates the ions. The detector is set to respond to 42He+ ions, but a small number of 32He+ ions are also present.
Mass of a 32He atom = 3.016 u; 1 u = 1.661 × 10−27 kg; elementary charge e = 1.60 × 10−19 C.

(a)

State the number of protons, neutrons and electrons in a 32He+ ion.

(2)
(b)

Explain why 32He and 42He cannot be separated chemically but can be separated in a mass spectrometer.

(2)
(c)

Calculate the charge-to-mass ratio of a 32He+ ion. The mass of the missing electron may be neglected.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
2 protons and 1 neutron✓ 1
1 electron✓ 1
Part (b)
Both have the same proton number, so their neutral atoms have the same number and arrangement of electrons, which gives identical chemical behaviour✓ 1
They have different numbers of neutrons and so different masses. Ions with the same charge and speed in the same magnetic field therefore follow paths of different radius (r = mv/qB)✓ 1
Part (c)
m = 3.016 × 1.661 × 10−27 = 5.01 × 10−27 kg✓ 1
q/m = 1.60 × 10−19 / 5.01 × 10−27 = 3.19 × 107 C kg−1 (accept 3.19–3.20 × 107 C kg−1)✓ 1

Common error: A singly charged ion carries a charge of +e, not +2e. The charge comes from the one missing electron, not from the proton number.

96E-2-02Energy levels & photons Paper 2Hard15 marks

In a gas-discharge lamp, electrons are accelerated by a potential difference and then collide with atoms of a gas. A collision can transfer energy from an electron to an atom and excite the atom.

The figure shows the four lowest energy levels, E1 to E4, of an atom of the gas. Before a collision every atom is in the ground state E1. Ignore the kinetic energy given to the atom as a whole in a collision.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Energy-level diagram of the gas atom: E1 = −9.00 eV (ground state), E2 = −6.60 eV, E3 = −4.80 eV, E4 = −2.60 eV, with the zero of energy shown dashed.0E₁−9.00 eVE₂−6.60 eVE₃−4.80 eVE₄−2.60 eVenergy levels of the gas atom
Energy levels drawn to scale
(a)

Exciting the atoms.

(i)

State the minimum potential difference through which an electron, initially at rest, must be accelerated to be able to excite an atom from E1 to E2.

(1)
(ii)

In the lamp, each electron strikes an atom with a kinetic energy of 5.00 eV. Deduce which of the levels E2, E3 and E4 can be reached.

(2)
(iii)

A photon of energy 5.00 eV passes through the gas. Explain why it is not absorbed, although an electron of kinetic energy 5.00 eV can excite an atom.

(2)
(b)

The light emitted by the lamp.

(i)

Draw, on the figure, arrows to show all the transitions that produce photons in this lamp.

(2)
(ii)

Determine which of these transitions emit visible light (wavelengths from 400 nm to 700 nm).

(3)
(c)

The electrical power input to the lamp is 12 W. 3.0 % of this power is emitted as photons from the transition E2 → E1. Calculate the number of these photons emitted per second.

(2)
(d)

The accelerating potential difference is increased so that each electron strikes an atom with a kinetic energy of 7.00 eV. Predict, with a reason, the change in the visible spectrum of the lamp.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
2.40 V✓ 1Accept 2.4 V. Do not accept 2.40 eV.
Part (a)(ii)
Energies needed from the ground state: E2 2.40 eV, E3 4.20 eV, E4 6.40 eV✓ 1
E2 and E3 can be reached; E4 cannot, because 6.40 eV > 5.00 eV✓ 1MP2 only scores if MP1 scores.
Part (a)(iii)
A photon is absorbed only if its whole energy equals the difference between two levels; 5.00 eV matches no difference from E1 «2.40, 4.20 or 6.40 eV»✓ 1
An electron can transfer only part of its kinetic energy to the atom and keep the rest «e.g. it keeps 0.80 eV after exciting E3»✓ 1OWTTE
Part (b)(i)
Downward arrows from E3 to E2 and from E3 to E1✓ 1
Downward arrow from E2 to E1, and no arrow starting at E4✓ 1Award [1 max] if the arrows point upwards. Allow ECF from (a)(ii).
Part (b)(ii)
λ = hc/ΔE used, e.g. λ = 6.63 × 10−34 × 3.00 × 108/(2.40 × 1.60 × 10−19)✓ 1
Wavelengths 518 nm «E2 → E1», 691 nm «E3 → E2» and 296 nm «E3 → E1»✓ 1Accept 517–518 nm, 690–691 nm and 296 nm.
E2 → E1 and E3 → E2 are visible; E3 → E1 is ultraviolet✓ 1Allow ECF from (b)(i).
Part (c)
Energy of one photon = 2.40 × 1.60 × 10−19 = 3.84 × 10−19 J✓ 1
N = 0.030 × 12/3.84 × 10−19 = 9.4 × 1017 s−1✓ 1Award [2] for CNA. Award [1 max] for 3.1 × 1019 s−1 «3.0 % not applied».
Part (d)
7.00 eV > 6.40 eV, so E4 can now be reached✓ 1
New transitions E4 → E3 «2.20 eV», E4 → E2 «4.00 eV» and E4 → E1 «6.40 eV» give 565 nm, 311 nm and 194 nm✓ 1Accept 564–565 nm, 310–311 nm, 194 nm.
Only the 565 nm line is visible, so one new «yellow-green» line appears in the visible spectrum; the other two are ultraviolet✓ 1Do not award MP3 for 'three new visible lines'.

Answers: (a)(i) 2.40 V  ·  (b)(ii) 518 nm and 691 nm visible; 296 nm ultraviolet  ·  (c) 9.4 × 1017 s−1  ·  (d) new visible line at 565 nm (the remaining parts are explanations — see the table above)

97E-2-55Rutherford scattering Paper 2Hard13 marks

Rutherford backscattering is used in the semiconductor industry to analyse very thin surface layers. A beam of alpha particles, 42He nuclei, each of kinetic energy 2.0 MeV, is directed at a silicon wafer, 2814Si, coated with a very thin layer of gold, 19779Au. A detector counts the alpha particles that are scattered back through about 170°.

(mass of an alpha particle = 6.64 × 10−27 kg, 1 u = 1.661 × 10−27 kg, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)

A beam of alpha particles travels from the left to a silicon wafer coated with a very thin gold layer on the side facing the beam. A detector placed just above the incoming beam receives alpha particles scattered back through about 170 degrees. Everything is in a vacuum chamber.vacuumsourcealpha particles, 2.0 MeVsilicon wafergold layerdetector
Diagram NOT to scale
(a)

The alpha particles are accelerated from rest by a potential difference. State the charge of an alpha particle in terms of e, and calculate the potential difference needed to give it a kinetic energy of 2.0 MeV.

(2)
(b)

Calculate the speed of the alpha particles in the beam.

(2)
(c)

As they travel through the gold and the silicon, the alpha particles gradually lose kinetic energy. Outline how this energy is lost.

(2)
(d)

Suggest why the gold layer must be very thin if the alpha particles scattered back by gold nuclei are all to reach the detector with almost the same kinetic energy.

(1)
(e)
(i)

An alpha particle is scattered straight back by a gold nucleus that is initially at rest. Assume that the alpha particle rebounds with almost the same speed. Estimate, in MeV, the kinetic energy given to the gold nucleus.

(3)
(ii)

Alpha particles scattered straight back by silicon nuclei return with much less kinetic energy than those scattered back by gold nuclei. Explain why.

(2)
(f)

Suggest how measuring the kinetic energies of the backscattered alpha particles allows the elements in a sample to be identified.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Charge = +2e «two protons, Z = 2»✓ 1
V = W/q = 2.0 × 106 eV/2e = 1.0 × 106 V «1.0 MV»✓ 1Allow ECF from MP1. Accept 3.2 × 10−13 J/3.2 × 10−19 C. Award [1 max] for 2.0 × 106 V «charge taken as e».
Part (b)
v = √(2Ek/m) = √(2 × 2.0 × 1.60 × 10−13/6.64 × 10−27)✓ 1Energy must be converted to J.
= 9.8 × 106 m s−1✓ 1Award [2] for CNA.
Part (c)
The alpha particles ionise atoms along their path: the electric force between an alpha particle and the electrons of nearby atoms removes electrons from these atoms✓ 1Accept 'excite or ionise'.
The energy needed for each ionisation comes from the kinetic energy of the alpha particle, so it loses kinetic energy in many small steps✓ 1OWTTE
Part (d)
Alpha particles scattered by gold nuclei at different depths travel different distances through the gold and so lose different amounts of energy; in a very thin layer these differences are negligible✓ 1OWTTE
Part (e)(i)
Momentum of alpha particle p = mv = 6.64 × 10−27 × 9.8 × 106 = 6.5 × 10−20 N s✓ 1Allow ECF from (b).
Change of momentum of the alpha particle ≈ 2p, so the gold nucleus gains momentum ≈ 1.3 × 10−19 N s «conservation of momentum»✓ 1
Ek = p²/2M = (1.3 × 10−19)²/(2 × 197 × 1.661 × 10−27) = 2.6 × 10−14 J ≈ 0.16 MeV✓ 1Accept 0.15–0.17 MeV. Award [3] for CNA.
Part (e)(ii)
The silicon nucleus is much less massive «28 u compared with 197 u» and receives a similar momentum, so its kinetic energy p²/2M is much larger✓ 1
Kinetic energy is conserved in the collision, so the alpha particle keeps correspondingly less kinetic energy✓ 1OWTTE. Award [1 max] for 'silicon is lighter' with no reasoning.
Part (f)
The energy kept by the alpha particle depends on the mass of the nucleus it strikes, so each energy measured corresponds to a particular element «larger energy ↔ heavier nucleus»✓ 1OWTTE

Answers: (a) 1.0 × 106 V  ·  (b) 9.8 × 106 m s−1  ·  (e)(i) 0.16 MeV (the remaining parts are explanations — see the table above)

98E-2-57Emission & absorption spectra Paper 2Hard11 marks

A very hot star of surface temperature 39 000 K and radius 9.0 R☉ lies inside a large cloud of hydrogen gas. Ultraviolet photons from the star ionise hydrogen atoms in the cloud. When a free electron later recombines with a proton, it is captured into an excited level and returns to the ground state in one or more steps. The cloud glows, and its spectrum contains a strong red line of wavelength 656 nm. The figure shows some energy levels of the hydrogen atom.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, Wien's displacement constant = 2.9 × 10−3 m K, σ = 5.67 × 10−8 W m−2 K−4, R☉ = 6.96 × 108 m, L☉ = 3.83 × 1026 W)

Energy levels of the hydrogen atom drawn to scale: −13.6 eV (ground state), −3.40 eV, −1.51 eV, −0.85 eV, with the zero of energy shown dashed.0−13.6 eV (ground state)−3.40 eV−1.51 eV−0.85 eV
Energy levels drawn to scale
(a)

Calculate the wavelength at which the spectrum of the star has its maximum intensity.

(1)
(b)

Calculate the longest wavelength of a photon that can ionise a hydrogen atom in its ground state.

(2)
(c)

The Sun has a surface temperature of 5800 K. Explain, using your answers to (a) and (b), why a gas cloud glows in this way around the hot star but not around a star like the Sun.

(2)
(d)

An electron is captured into the −0.85 eV level and returns to the ground state in steps, one of which emits the 656 nm photon. Draw, on the figure, arrows to show these steps.

(2)
(e)

Determine the luminosity of the star, in terms of L☉.

(2)
(f)

Explain why the spectrum of the glowing cloud consists of bright lines on a dark background, whereas the spectrum of the star is continuous with dark lines.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λmax = 2.9 × 10−3/39 000 = 7.4 × 10−8 m «74 nm»✓ 1
Part (b)
E = 13.6 × 1.60 × 10−19 = 2.18 × 10−18 J✓ 1
λ = hc/E = 9.14 × 10−8 m «91 nm»✓ 1Accept 91 nm. Award [2] for CNA.
Part (c)
The hot star's peak «74 nm» is shorter than 91 nm, so a large fraction of its photons can ionise hydrogen✓ 1
The Sun's peak «500 nm» is far longer than 91 nm, so it emits very few ionising photons: very little of the hydrogen is ionised and there is no recombination glow✓ 1Allow ECF from (a) and (b).
Part (d)
A downward arrow from −1.51 eV to −3.40 eV «1.89 eV, 656 nm»✓ 1Arrows must start and end on levels.
Downward arrows from −0.85 eV to −1.51 eV and from −3.40 eV to −13.6 eV, completing the only route that includes the 656 nm step✓ 1Award [1 max] if any arrow points upwards.
Part (e)
L = σ4πR²T⁴ = 5.67 × 10−8 × 4π × (9.0 × 6.96 × 108)² × 39 000⁴ = 6.5 × 1031 W✓ 1
= 1.7 × 105 L☉✓ 1Award [2] for CNA.
Part (f)
The cloud is a thin gas seen against a dark sky: its atoms emit only photons whose energies equal differences between their levels, so only bright lines appear✓ 1
The star's dense surface emits a continuous spectrum; cooler gas in its outer layers absorbs photons of particular energies, removing those wavelengths «dark lines»✓ 1OWTTE

Answers: (a) 7.4 × 10−8 m  ·  (b) 91 nm  ·  (e) 1.7 × 105 L☉ (the remaining parts are explanations — see the table above)

99E-2-63Emission & absorption spectra Paper 2Hard11 marks

During a total solar eclipse in 1869, astronomers observed a bright green emission line of wavelength 530.3 nm in the light from the Sun's corona, the very hot, thin outer atmosphere of the Sun. The line matched no line of any element known in the laboratory, and it was thought to come from a new element. In 1939 the line was shown to come from iron ions that have lost 13 electrons. The common isotope of iron is 5626Fe.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, kB = 1.38 × 10−23 J K−1, 1 u = 1.661 × 10−27 kg)

(a)

State the number of protons, neutrons and electrons in one of these iron ions.

(2)
(b)

Calculate, in eV, the energy of a photon of the green line.

(2)
(c)

Explain why the green line was not found in the spectrum of iron produced in laboratories at the time, and why it was attributed to a new element.

(2)
(d)
(i)

Removing the thirteenth electron from an iron ion requires 361 eV. Show that the temperature at which the average kinetic energy of the particles is 361 eV is about 3 × 106 K.

(2)
(ii)

The temperature of the corona is about 2 × 106 K. Suggest why the corona nevertheless contains many of these ions.

(1)
(e)

The ions in the corona move randomly at high speeds, so the green line is spread over a small range of wavelengths. Taking the temperature of the corona as 2 × 106 K and the mass of an iron ion as 56 u, estimate the change in wavelength Δλ of the green line emitted by an ion that moves directly away from the observer with the mean kinetic energy of the particles.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
26 protons and 30 neutrons✓ 1
13 electrons✓ 1
Part (b)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/530.3 × 10−9 = 3.75 × 10−19 J✓ 1
= 2.34 eV✓ 1Award [2] for CNA.
Part (c)
Each atom or ion has its own set of energy levels, and so its own set of lines; removing 13 electrons gives an ion with energy levels different from those of iron atoms✓ 1
Such ions could not be produced in laboratories, so the line matched no known spectrum and seemed to need a new element✓ 1OWTTE
Part (d)(i)
361 × 1.60 × 10−19 = 3/2 kBT✓ 1
T = 2 × 361 × 1.60 × 10−19/(3 × 1.38 × 10−23) «= 2.79 × 106 K»✓ 1Must see full substitution OR answer to 3 s.f.
Part (d)(ii)
The particles have a range of kinetic energies; many have far more than the average, enough to ionise «and electrons from collisions build up the ionisation step by step»✓ 1OWTTE
Part (e)
½mv² = 3/2 kBT gives v = √(3 × 1.38 × 10−23 × 2 × 106/(56 × 1.661 × 10−27)) = 3.0 × 104 m s−1✓ 1
Δλ = λv/c = 530.3 × 10−9 × 3.0 × 104/3.00 × 108 = 5.3 × 10−11 m «0.053 nm»✓ 1Allow ECF for v. Accept 5.0–5.5 × 10−11 m. Award [2] for CNA.

Answers: (b) 2.34 eV  ·  (d)(i) 2.8 × 106 K  ·  (e) 5.3 × 10−11 m (the remaining parts are explanations — see the table above)

100IBP2S-035-01-15Energy levels & photons Paper 2Hard8 marks

In a compact fluorescent lamp, excited mercury atoms emit ultraviolet photons of wavelength 254 nm. A coating on the inside of the glass absorbs each ultraviolet photon and emits one visible photon of wavelength 545 nm. The rest of the energy becomes internal energy of the coating.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

Explain, with reference to energy levels, how the coating can emit a photon of longer wavelength than the photon it absorbs.

(2)
(b)

Calculate, in eV, the energy of an ultraviolet photon.

(2)
(c)

Show that about 47% of the energy of each ultraviolet photon is re-emitted as visible light.

(2)
(d)

The lamp takes 18 W of electrical power. Of this power, 40% is converted into ultraviolet photons that are absorbed by the coating. Determine the power emitted as visible light by the coating.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The absorbed ultraviolet photon raises an electron in the coating to a high energy level .✓ 1
The electron returns in more than one step, with part of the energy transferred non-radiatively as internal energy. The emitted photon therefore corresponds to a smaller energy gap, so it has a longer wavelength .✓ 1
Part (b)
E = hc/λ = (6.63×10−34 × 3.00×108)/(254×10−9) = 7.83×10−19 J .✓ 1
E = 7.83×10−19/1.60×10−19 = 4.89 eV .✓ 1
Part (c)
Evisible = hc/(545×10−9) = 3.65×10−19 J, which is 2.28 eV .✓ 1
Fraction = 3.65×10−19/7.83×10−19 = 0.466, which is about 47% (or 254/545 = 0.466) .✓ 1
Part (d)
Ultraviolet power absorbed = 0.40 × 18 = 7.2 W .✓ 1
Visible power = 7.2 × 0.466 = 3.4 W . ECF from (c).✓ 1

Exam tip: When each absorbed photon produces exactly one emitted photon, the energy fraction is simply the ratio of the wavelengths, λUV/λvisible. Use this to check your answer to (c).

101IBP2S-035-00-16Energy levels & photons Paper 2Hard8 marks

In a prototype display panel, electrons are accelerated from rest through a potential difference of 4.5 V. They then collide with atoms of a gas Q, all initially in the ground state. Assume that each atom undergoes at most one collision with an electron, and that any excited atom returns to the ground state before it could be struck again. The energy levels of Q are −8.0 eV (ground state), −5.9 eV, −4.1 eV and −1.3 eV.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

−1.3 eV−4.1 eV−5.9 eV−8.0 eVground state
(a)

State the kinetic energy, in eV, of an electron just before a collision.

(1)
(b)

Determine all the photon energies that the gas can emit.

(3)
(c)

Calculate the longest wavelength of light emitted.

(2)
(d)

The electron beam is replaced by a beam of photons, each of energy 4.5 eV. Explain why the gas now emits no light.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
4.5 eV✓ 1
Part (b)
Excitation is possible to −5.9 eV (needs 2.1 eV) and to −4.1 eV (needs 3.9 eV), but not to −1.3 eV (needs 6.7 eV, which is more than 4.5 eV)✓ 1
Emission from −4.1 eV gives 3.9 eV (to the ground state) and 1.8 eV (to −5.9 eV)✓ 1
Emission from −5.9 eV gives 2.1 eV, so the photon energies are 1.8 eV, 2.1 eV and 3.9 eV✓ 1
Part (c)
The longest wavelength corresponds to the smallest photon energy, 1.8 eV = 2.88×10−19 J✓ 1
λ = hc/E = 1.989×10−25/2.88×10−19 = 6.9×10−7 m✓ 1
ECF from (b)—
Part (d)
A photon is absorbed only if its whole energy equals a level difference from the ground state (2.1, 3.9 or 6.7 eV), or is at least the ionisation energy of 8.0 eV; 4.5 eV is neither✓ 1
An electron can give up only part of its kinetic energy in a collision, but a photon cannot give up part of its energy, so the photons pass through unabsorbed and no atoms are excited✓ 1

Common error: Only levels the electron can reach take part, but once an atom is excited, cascades between excited levels add further lines.

102IBP2S-035-03-16Nuclear notation Paper 2Hard6 marks

Deep-space probes are powered by radioisotope generators containing plutonium dioxide, PuO2. The plutonium is 23894Pu and the oxygen is 168O. A generator contains 4.8 kg of PuO2.
Assume that the molar mass of each nuclide in grams per mole is numerically equal to its nucleon number. NA = 6.02 × 1023 mol−1.

(a)

State the number of neutrons in a 23894Pu nucleus.

(1)
(b)

Determine the total number of protons in one formula unit of PuO2.

(1)
(c)

Calculate the number of 23894Pu nuclei in the generator.

(2)
(d)

Determine the total number of neutrons in the 4.8 kg of PuO2.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
238 − 94 = 144✓ 1
Part (b)
94 + 2 × 8 = 110✓ 1
Part (c)
Molar mass of PuO2 = 238 + 2 × 16 = 270 g mol−1, so amount = 4800 / 270 = 17.8 mol✓ 1
Number of Pu nuclei = 17.8 × 6.02 × 1023 = 1.07 × 1025✓ 1
Part (d)
Neutrons per formula unit = 144 + 2 × 8 = 160 (allow error carried forward from the Pu neutron number)✓ 1
Total = 160 × 1.07 × 1025 = 1.7 × 1027 (allow error carried forward from the number of Pu nuclei)✓ 1

Common error: Remember the two oxygen atoms in each formula unit. They count in the molar mass and they count in the particle total. Leaving them out changes the answer noticeably.