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IB Physics Higher Level · Topic E.1

E.1 Structure of the atom

111 free, high-quality past-paper-style questions on Structure of the atom for IB Physics HL, each with a full mark scheme. Covers: Counting spectral lines; Emission and absorption spectra; Nuclear notation and isotopes; and more. For when you have finished the past papers, or want to practise one topic at a time.E.1 at Standard 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.456 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-22Counting spectral lines Paper 1AEasy1 mark

Hydrogen atoms are excited to the n = 4 energy level.

What is the maximum number of different spectral lines that can be emitted as the atoms return to the ground state?

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Step 1Each line corresponds to a transition from a higher level to a lower one.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2From n = 4: 4→3, 4→2, 4→1; from n = 3: 3→2, 3→1; from n = 2: 2→1.—
Step 3Total = 3 + 2 + 1 = 6 (in general n(n − 1)/2).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis counts only the direct transitions from n = 4.
  • BThis counts the levels, not the transitions between them.
  • CCorrect: every pair of levels gives one line.
  • DThis uses n²/2 = 16/2, a misremembered form of the pair-counting formula n(n − 1)/2.
2E-1A-32Emission and absorption spectra Paper 1AEasy1 mark

The spectrum of a distant star shows two dark lines at 589.0 nm and 589.6 nm. A sodium lamp in the laboratory emits bright lines at exactly these wavelengths.

What is the best explanation of the dark lines?

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Step 1Each atom can absorb only photons whose energy hf equals the difference between two of its energy levels.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The dark lines coincide with the sodium emission lines, so they correspond to transitions between the same sodium energy levels.—
Step 3Therefore sodium is present in the cooler gas the light passes through: absorption spectra reveal chemical composition.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the evidence: missing wavelengths that match sodium's lines show that sodium IS present to absorb them.
  • BA 589 nm photon has an energy of only about 2.1 eV, far less than the energy needed to ionise sodium; ionisation would also remove a continuous band, not sharp lines.
  • CThe excited atoms do re-emit, but in all directions, so the intensity along the line of sight is reduced; the lines are dark because of this, not because energy is stored permanently.
  • DCorrect: absorption at the sodium wavelengths identifies sodium in the star's atmosphere.
3E-1A-60Nuclear notation and isotopes Paper 1AEasy1 mark

A nucleus P contains 19 protons and 21 neutrons. In the options, X stands for the chemical symbol of whichever element the nucleus belongs to.

Which nucleus is an isotope of P?

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Step 1Isotopes have the same proton number Z but different neutron numbers, so an isotope of P has Z = 19.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2P itself has A = 19 + 21 = 40, so the isotope must have Z = 19 and A ≠ 40.—
Step 3Only 3919X has Z = 19: it has 19 protons and 39 − 19 = 20 neutrons, so it is an isotope of P.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis swaps the numbers of protons and neutrons: it has 21 protons (and 19 neutrons), so it is a different element.
  • BCorrect: same proton number (19), different neutron number (20 instead of 21).
  • CThis has the same nucleon number (40) as P, but 20 protons: it is a different element.
  • DThis has the same neutron number (41 − 20 = 21) as P, but 20 protons: a different element.
4E-1A-61Emission spectra and chemical composition Paper 1AEasy1 mark

The diagram shows the emission line spectra of three elements P, Q and R, and the emission spectrum of a gas-discharge lamp. Every line in each element's spectrum in this wavelength range is shown.

Which elements are present in the lamp?
I. P
II. Q
III. R

element Pelement Qelement Rlamp400450500550600650700wavelength / nm
Emission line spectra of three elements and of the lamp (wavelength scale drawn to scale).
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Step 1Each element emits its own set of wavelengths, fixed by the differences between its energy levels; if an element is present, all its lines appear.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2All three lines of P (436, 518, 612 nm) and all three lines of R (471, 556, 655 nm) are in the lamp spectrum, and together they account for every lamp line.—
Step 3Q's line at 471 nm coincides with a line of R, but its lines at 592 nm and 681 nm are missing, so Q is absent: I and III only.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis stops after finding one element whose lines all appear and assumes the lamp contains a single element; the lines at 471, 556 and 655 nm are then unexplained.
  • BThis takes the single line at 471 nm as proof that Q is present and overlooks that R's complete set of lines (471, 556 and 655 nm) appears in the lamp spectrum.
  • CCorrect: an element is present only if every one of its lines appears; P and R satisfy this, Q does not.
  • DThis treats one matching line as enough to identify an element, so Q is wrongly included.
5IBP2H-036-00-01Counting spectral lines Paper 1AEasy1 mark

A hydrogen lamp is used to calibrate a spectrometer on a space telescope. In the lamp, hydrogen atoms are excited to the n = 5 energy level and no higher. The atoms then return to the ground state by every possible route. How many different wavelengths can be emitted?

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Each pair of levels from n = 1 to 5 gives one possible transition. Every pair has a different energy difference, so each gives a different wavelength. The number of pairs is 5 × 4 / 2 = 10, so 10 different wavelengths can be emitted. The answer 4 counts only the direct jumps down from n = 5. The answer 5 counts levels, not transitions. The answer 15 also pairs each level with itself, giving n(n + 1)/2.✓ 1Answer C

Answer: C

Exam tip: For atoms excited to level n, the number of possible spectral lines is n(n − 1)/2. To check, list the jumps from each level down to every lower level.

6IBP2H-036-00-07Transitions on an energy level diagram Paper 1AEasy1 mark

Four energy levels of hydrogen are −13.6 eV (n = 1), −3.40 eV (n = 2), −1.51 eV (n = 3) and −0.85 eV (n = 4). Which transition emits the photon with the longest wavelength?

energyn = 1−13.6 eVn = 2−3.40 eVn = 3−1.51 eVn = 4−0.85 eV
Energy levels not to scale
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Wavelength is λ = hc/ΔE, so the longest wavelength comes from the smallest energy gap. The gap for 4 → 3 is 1.51 − 0.85 = 0.66 eV, which is smaller than 1.89 eV (3 → 2), 10.2 eV (2 → 1) and 12.75 eV (4 → 1). Option D has the largest gap and therefore the shortest wavelength.✓ 1Answer C

Answer: C

Common error: A larger change in n does not mean a longer wavelength. A larger energy gap gives a shorter wavelength.

7IBP2H-036-01-01Rutherford scattering and the nuclear radius Paper 1AEasy1 mark

In a scattering experiment, a beam of alpha particles is directed at a thin gold foil. Almost all of the alpha particles pass through the foil with little or no deflection. What does this observation alone suggest about the atom?

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If most alpha particles meet nothing that deflects them, the massive charged part of the atom must occupy only a tiny fraction of its volume, so the atom is mostly empty space. A positive nucleus is inferred from the rare large-angle deflections, not from the undeflected majority. Energy levels come from spectra. Neutrons were not revealed by this experiment.✓ 1Answer B

Answer: B

Common error: Match each observation to its own conclusion. Undeflected particles show empty space. Rare back-scattering shows a small, dense, positive nucleus.

8IBP2H-036-01-02Rutherford scattering Paper 1AEasy1 mark

Why must the metal foil in an alpha-particle scattering experiment be extremely thin?

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The scattering formula assumes that each deflection is caused by a single close encounter with one nucleus. In a thick foil a particle could be deflected many times, and the angular distribution would no longer reveal the size and charge of a single nucleus. Charging, gravity and conduction are not the reason.✓ 1Answer A

Answer: A

Exam tip: Learn the three key conditions: a thin foil (single scattering), a vacuum (no absorption by air) and a narrow beam of particles that all have the same energy.

9IBP2S-035-02-09Rutherford scattering Paper 1AEasy1 mark

An alpha particle moves directly towards a heavy nucleus. The nucleus remains stationary.
Which statement is correct at the instant the alpha particle is at its distance of closest approach?

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In a head-on approach to a fixed nucleus, the alpha particle slows until it stops momentarily. At that instant all of its kinetic energy has become electric potential energy. The separation is at its minimum, so the Coulomb force, and therefore the acceleration, is at its greatest. The force is not zero. Zero velocity does not imply zero acceleration.✓ 1Answer B

Answer: B

Common error: An object can have zero velocity and a large acceleration at the same instant, just like a ball at the top of its flight.

10IBP3H-E1-20r13-01Excitation of atoms in a fluorescent lamp Paper 1AEasy1 mark

In a fluorescent lamp the mercury vapour emits mostly ultraviolet photons. The coating on the inside of the glass tube emits visible light. Which statement best explains how the coating produces visible light?

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A is correct: an absorbed ultraviolet photon excites the coating atom; the atom returns to its ground state in steps, giving photons of smaller energy (longer wavelength), some of the energy going to the coating as thermal energy. B is wrong because visible light has a longer wavelength than ultraviolet, so lower photon energy. C is wrong because the photon energy would then be ultraviolet, not visible. D is wrong because the coating is excited by photons, and electrons would give a range of energies, not a single visible photon with all the kinetic energy.✓ 1Answer A

Answer: A

Common error: Saying that a shorter wavelength means lower photon energy. Photon energy is inversely proportional to wavelength.

11IBP3H-E1-20r13-02Emission and absorption spectra Paper 1AEasy1 mark

White light passes through a cool sample of hydrogen gas and is then analysed with a spectrometer. The spectrum shows dark lines on a bright continuous background. What is the origin of the dark lines?

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B is correct: only photons with energy equal to an energy-level gap can excite an atom, so those wavelengths are removed from the transmitted beam (the atoms re-emit in random directions). A is wrong because only specific photon energies are absorbed. C is wrong because emission in random directions does not cancel the beam by interference. D is wrong because photons are not ionised; atoms are, and only by photons of enough energy.✓ 1Answer B

Answer: B

Exam tip: Dark absorption lines appear at the same wavelengths as the bright lines in the emission spectrum of the same element.

12IBP3H-E1-20r14-01Emission and absorption spectra Paper 1AEasy1 mark

Which statement best explains why the emission spectrum of a hot, low-density gas consists of discrete bright lines?

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B is correct: atomic energy levels are discrete, and each photon carries the energy difference between two levels, so only specific wavelengths appear. A is wrong because collisions only excite the atoms; the photon energy is set by the level gaps. C is wrong because line spectra in the visible region involve electron transitions, not nuclear transitions. D is wrong because it describes a classical model, which would give a continuous spectrum.✓ 1Answer B

Answer: B

Common error: Line spectra come from electron transitions between atomic energy levels; nuclear energy levels give much higher-energy photons.

13IBP3H-E1-20r14-02Emission and absorption spectra Paper 1AEasy1 mark

White light passes through a cool gas and is then analysed with a spectrometer, which shows dark lines at certain wavelengths. What is the best explanation of the dark lines?

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A is correct: only photons whose energy matches a level gap are absorbed, and the re-emission is in all directions so the transmitted beam is weakened at those wavelengths. B is wrong because the excited atoms re-emit photons rather than storing the energy permanently. C is wrong because the dark lines are not caused by interference. D is wrong because photons of energy between level gaps are not absorbed.✓ 1Answer A

Answer: A

Exam tip: Absorption needs an exact match between the photon energy and an energy-level gap; a photon with slightly more energy is not absorbed.

14IBP3H-E1-20r14-04Rutherford scattering Paper 1AEasy1 mark

In Rutherford's alpha-particle scattering experiment a very small fraction of the alpha particles was deflected through angles greater than 90°. Which conclusion does this observation support?

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D is correct: a large deflection requires a strong repulsive force from a small, massive, positively charged centre. A follows from the observation that most particles pass straight through, not from large-angle scattering. B is wrong because uniform charge would give only small deflections. C is a conclusion from atomic spectra, not from scattering.✓ 1Answer D

Answer: D

Exam tip: Link each observation to its own conclusion: undeflected particles mean mostly empty space; large-angle scattering means a small, dense, charged nucleus.

15IBP3H-E1-20r33-01Rutherford scattering Paper 1AEasy1 mark

In a scattering experiment, a narrow beam of α particles is directed at a very thin gold foil. A very small fraction of the α particles is deflected through angles greater than 90°. What can be deduced from this observation?

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A large deflection needs a strong repulsive force on a massive, positive α particle, which is only possible if the positive charge and most of the mass sit in a tiny nucleus, so that key is right. Atomic electrons have a tiny fraction of the mass of an α particle and cannot turn it through a large angle, so electron collisions do not explain the result. A negatively charged centre would attract the α particles and would not scatter them backwards. Absorption and re-emission is not what is observed; the particles are scattered by a force, not absorbed.✓ 1Answer C

Answer: C

Common error: Do not attribute large-angle scattering to electrons: their mass is about 1/7300 of the mass of an α particle.

16IBP3H-E1-20r33-02Emission and absorption spectra Paper 1AEasy1 mark

White light from a filament lamp passes through cool sodium vapour and is then analysed with a spectrometer. Which statement describes the spectrum that is seen?

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Sodium atoms absorb only photons whose energy equals the gap between two of their energy levels. These are the same photon energies that sodium emits in downward transitions, so the transmitted light shows dark lines at those wavelengths against a continuous background; that is the correct description. Bright lines on a dark background would be an emission spectrum from hot sodium, not what is seen here. A spectrum with no lines ignores absorption by the vapour. Dark lines at wavelengths sodium does not emit contradict the fact that absorption and emission involve the same level gaps.✓ 1Answer A

Answer: A

Exam tip: Absorption lines and emission lines of one element appear at the same wavelengths because both come from the same energy-level differences.

17IBP3H-E1-20r33-03Excitation of atoms in a fluorescent lamp Paper 1AEasy1 mark

A red LED in an optical communications link emits light of wavelength 620 nm. Use h = 6.63 × 10−34 J s and c = 3.00 × 108 m s−1. What is the energy of one photon of this light?

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The photon energy is E = hc/λ = (6.63 × 10−34 × 3.00 × 108) / (620 × 10−9) = 3.2 × 10−19 J. Using 620 as if it were in metres gives a value 109 times too small. Calculating h/λ leaves out the factor c and has the wrong units. Using hλ/c puts the wavelength in the numerator and gives an absurdly small energy.✓ 1Answer D

Answer: D

Common error: Convert wavelengths in nanometres to metres (×10⁻⁹) before substituting into E = hc/λ.

18E-1A-01Energy levels and spectra Paper 1AMedium1 mark

A cool gas contains hydrogen atoms that are all in the ground state. The energy levels of hydrogen are given by E = −13.6/n² eV.

Photons of three different energies pass through the gas:
I. 10.2 eV
II. 11.0 eV
III. 14.0 eV

Which photons can be absorbed by the hydrogen atoms?

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Step 1From n = 1 the possible excitation energies are 13.6(1 − 1/n²): 10.2 eV, 12.1 eV, 12.8 eV, … up to 13.6 eV. 10.2 eV matches the 1 → 2 transition, so I is absorbed.—All 4 steps must be completed — there is no mark for a part-answer.
Step 211.0 eV lies between 10.2 eV and 12.1 eV and equals no energy-level difference, so II is not absorbed.—
Step 314.0 eV exceeds the ionisation energy 13.6 eV: the photon is absorbed and the electron leaves the atom with 0.4 eV of kinetic energy, because the final state of a free electron is not restricted to discrete values.—
Step 4So I and III only.✓ 1Answer C

Answer: C  ·  4 stages of work, one mark

Every option, and why

  • AThis assumes that only photons matching a transition between two bound levels can be absorbed, and forgets that any photon above 13.6 eV can ionise the atom.
  • BThis assumes any photon with at least 10.2 eV can be absorbed but that one above 13.6 eV is "too energetic"; 11.0 eV matches no level difference.
  • CCorrect: a bound–bound transition needs an exact energy match (I); above the ionisation energy any energy can be absorbed (III).
  • DThis treats absorption as possible for any photon energy above the first excitation energy; 11.0 eV is not a level difference.
19E-1A-02The Bohr model Paper 1AMedium1 mark

In the Bohr model of hydrogen, the electron's angular momentum is nh/2π and its orbital radius is proportional to n².

What is the ratio (speed of the electron in n = 2) / (speed in n = 1)?

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Step 1Angular momentum mvr = nh/2π, so v = nh/(2πmr).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2With r ∝ n², v ∝ n/n² = 1/n.—
Step 3v2/v1 = 1/2.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis would need v ∝ 1/n².
  • BCorrect: v ∝ 1/n, so the electron in the second orbit moves at half the speed.
  • CThis inverts the ratio — the outer electron moves more slowly.
  • DThis is the ratio of the radii, not the speeds.
20E-1A-03Nuclear radius and density Paper 1AMedium1 mark

Nucleus X has nucleon number 27 and nucleus Y has nucleon number 216.

Which row is correct?

RY/RXρY/ρX
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Step 1R = R0A1/3, so RY/RX = (216/27)1/3 = 81/3 = 2.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Density = mass/volume ∝ A/R³ ∝ A/A = constant.—
Step 3So the density ratio is 1: nuclear matter has the same density whatever the nucleus.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the radius scales as the cube root of A, and the density is independent of A.
  • BY has 8 times the mass but also 8 times the volume, so the density is unchanged.
  • CThe radius ratio is the cube root of 8, not 8.
  • DThis takes R ∝ A (no cube root) and also assumes that density rises with mass, ignoring that the volume grows by the same factor 8.
21E-1A-27Transitions on an energy level diagram Paper 1AMedium1 mark

The diagram shows some energy levels of hydrogen and four transitions A, B, C and D in which photons are emitted.

Which transition produces the photon of longest wavelength?

0 (n = ∞)−13.6 eV n = 1-3.40 eV n = 2-1.51 eV n = 3-0.85 eV n = 4ABCDenergy
Some energy levels of the hydrogen atom (not to scale) with four emission transitions marked.
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Step 1Photon energy = difference between the levels; wavelength λ = hc/ΔE, so the longest wavelength comes from the smallest energy difference.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2A: 3.40 − 0.85 = 2.55 eV; B: 13.6 − 1.51 = 12.1 eV; C: 1.51 − 0.85 = 0.66 eV; D: 13.6 − 3.40 = 10.2 eV.—
Step 3C has the smallest energy (0.66 eV), giving λ ≈ 1.9 μm (infrared).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • A2.55 eV — a visible (blue) photon, not the longest wavelength.
  • B12.1 eV — the largest energy gap of the four, giving the shortest wavelength (ultraviolet).
  • CCorrect: the smallest energy gap gives the lowest frequency and longest wavelength.
  • D10.2 eV — the ultraviolet Lyman-α line.
22E-1A-30Rutherford scattering Paper 1AMedium1 mark

Alpha particles are scattered by a thin foil. For alpha-particle kinetic energies below about 25 MeV the number scattered through large angles agrees with the Rutherford model, which assumes only an electric repulsion between the alpha particle and a point nucleus. At higher kinetic energies fewer alpha particles are scattered through large angles than the model predicts.

Which conclusion is best supported by these observations?

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Step 1A higher kinetic energy lets an alpha particle approach closer to the nucleus before it is turned back.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The electric model fails only when the particle reaches a distance at which another interaction acts: the short-range strong nuclear force at the nuclear surface.—
Step 3So the distance of closest approach at the threshold energy (about 25 MeV) is an estimate of the nuclear radius.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThe Rutherford model already includes the effect of the kinetic energy on the deflection; faster particles are expected to be deflected less, and the model predicts this correctly up to 25 MeV.
  • BCorrect: deviations begin when the alpha particles reach the range of the strong nuclear force at the edge of the nucleus.
  • CThe electrons affect only distant, low-angle encounters; large-angle scattering happens deep inside the electron cloud at all energies.
  • DRecoil of a massive nucleus is small and does not switch on suddenly at one energy; it cannot explain a deviation that starts at a definite kinetic energy.
23E-1A-31Rutherford scattering Paper 1AMedium1 mark

In a Geiger–Marsden–Rutherford type experiment, a narrow beam of alpha particles is directed at a thin metal foil in a vacuum, and the number of alpha particles scattered through angles greater than 90° in a fixed time is counted.

Which change, made on its own, increases this number?

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Step 1A deflection through more than 90° needs a strong electric repulsion from a nucleus acting on an alpha particle that passes very close to it.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2A nucleus of greater proton number has a greater positive charge, so it repels an alpha particle more strongly at the same distance.—
Step 3With the same number of nuclei in the beam's path, more alpha particles are turned back through more than 90°.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AFor the same path past a nucleus, a faster alpha particle is deflected through a smaller angle, so fewer are scattered through more than 90° — the misconception is that a more energetic particle rebounds more.
  • BA thinner foil contains fewer nuclei in the path of the beam, so fewer close encounters occur — the misconception is that a thinner foil "lets more bounce off".
  • CCorrect: a larger nuclear charge gives a stronger Coulomb repulsion for the same approach, so more large-angle scattering.
  • DA proton has charge +e, half that of an alpha particle (+2e), so the repulsion is weaker and fewer are scattered through large angles; the misconception is that a lighter particle "bounces back more easily".
24E-1A-62The Bohr model Paper 1AMedium1 mark

In the Bohr model of hydrogen, the angular momentum of the electron is mvr = nh/2π and the energy levels are E = −13.6/n² eV.

The electron is in an orbit in which its angular momentum is L. What is the minimum energy needed to ionise the atom from this state?

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Marking pointMarkNotes
Step 1From L = nh/2π, the quantum number is n = 2πL/h.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ionisation takes the electron from E = −13.6/n² eV to E = 0, so it needs 13.6/n² eV = 13.6(h/2πL)² eV.—
Step 3Check: for L = 3h/2π this gives 13.6/9 = 1.51 eV, the ionisation energy from n = 3.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: n = 2πL/h and the ionisation energy from level n is 13.6/n² eV.
  • BThis multiplies by n² instead of dividing (the quantum number has been inverted); for n = 3 it gives 122 eV, more than the ionisation energy from the ground state.
  • CThis omits the square: it uses 13.6/n eV (4.53 eV for n = 3).
  • DThis is 13.6 − 13.6/n² eV, the energy needed to raise the electron from the ground state to level n (12.1 eV for n = 3), not to remove it from level n.
25IBP2H-036-00-02The Bohr model Paper 1AMedium1 mark

In the Bohr model of hydrogen, the electron moves from the n = 1 orbit to the n = 3 orbit. By what factors do the orbital radius and the orbital speed of the electron change?

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In the Bohr model r ∝ n2, so the radius becomes 9 times larger. From mvr = nh/2π, v = nh/(2πmr) ∝ n/n2 = 1/n. The speed therefore falls to 1/3 of its value. Option A wrongly assumes r ∝ n. Option C wrongly assumes v ∝ 1/r. Option D forgets that r increases faster than n.✓ 1Answer B

Answer: B

Common error: Speed does not scale as 1/r. Combine mvr = nh/2π with r ∝ n² to get v ∝ 1/n.

26IBP2H-036-00-03Nuclear notation and isotopes Paper 1AMedium1 mark

Beams of the ion 4820Ca2+ are used in accelerators to make superheavy elements. What are the numbers of protons, neutrons and electrons in one of these ions?

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Z = 20 gives 20 protons. The number of neutrons is A − Z = 48 − 20 = 28. The 2+ charge means the ion has lost two electrons, so it has 20 − 2 = 18 electrons. Option A treats the nucleon number as the neutron number. Option B adds electrons instead of removing them. Option D swaps the proton and electron counts.✓ 1Answer C

Answer: C

Exam tip: The ionic charge changes only the electron count. The nucleus, and so Z and A, is unchanged.

27IBP2H-036-00-04Nuclear radius and closest approach Paper 1AMedium1 mark

In a scattering experiment, alpha particles approach a heavy nucleus head-on. The initial kinetic energy of the alpha particles is doubled. Recoil of the nucleus can be ignored. What happens to the distance of closest approach?

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At closest approach, all of the kinetic energy has become electric potential energy: Ek = kqQ/d. This means d ∝ 1/Ek, so doubling Ek halves d. Dividing by √2 confuses kinetic energy with speed. Dividing by 4 wrongly uses the inverse-square force law in place of the 1/r potential energy. Doubling has the proportionality the wrong way round.✓ 1Answer A

Answer: A

Common error: Use energy conservation with potential energy kqQ/r, not the force kqQ/r². The distance depends on 1/Ek.

28IBP2H-036-00-06Emission and absorption spectra Paper 1AMedium1 mark

When an exoplanet passes in front of its star, some starlight passes through the planet's cool atmosphere before reaching a telescope. Sodium vapour is present in the atmosphere. Which statement about the observed spectrum is correct?

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Sodium atoms absorb only photons whose energies equal the differences between their energy levels. These are the same differences that produce the sodium emission lines. The excited atoms re-emit in all directions, so fewer photons of those wavelengths travel towards the telescope. Dark lines therefore appear at the emission wavelengths. Option A describes an emission spectrum seen against a dark background. Option C wrongly separates absorption from emission energies. Option D ignores the fact that absorption depends on energy matching, not on the gas being hot.✓ 1Answer B

Answer: B

Common error: Absorption lines are not missing because the energy is destroyed. The absorbed photons are re-emitted in random directions, so few of them continue along the line of sight.

29IBP2H-036-01-03Nuclear radius and density Paper 1AMedium1 mark

Nucleus P has a nucleon number 27 times that of nucleus Q. Which row gives the ratio \(\frac{\text{radius of P}}{\text{radius of Q}}\) and the ratio \(\frac{\text{density of P}}{\text{density of Q}}\)?

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R = R0A1/3, so the radius ratio is 271/3 = 3. Mass is proportional to A and volume is proportional to R3, which is also proportional to A. The density is therefore the same for all nuclei, so the ratio is 1.✓ 1Answer A

Answer: A

Common error: Volume scales with R³, not R². Using an area gives a density that wrongly appears to depend on A.

30IBP2H-036-01-04Nuclear radius and closest approach Paper 1AMedium1 mark

An alpha particle with kinetic energy E is fired head-on at a stationary gold nucleus. Its distance of closest approach is d. A proton, also with kinetic energy E, is then fired head-on at the same nucleus. Recoil of the nucleus is negligible. What is the distance of closest approach of the proton?

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At closest approach, E = kQq/d, so d = kQq/E. The proton has half the charge of the alpha particle and the same kinetic energy, so its closest approach is d/2. Mass does not appear in the expression, so the factor of 4 in mass is irrelevant.✓ 1Answer B

Answer: B

Common error: When the kinetic energy is fixed, only charge matters. Do not include the mass ratio, and do not assume the distance is unchanged because the energy is the same.

31IBP2H-036-01-05Energy levels and spectra Paper 1AMedium1 mark

An atom used in a laboratory gas-discharge lamp has only four energy levels: −9.0 eV, −4.0 eV, −2.0 eV and −0.5 eV. Which photon energy can not be emitted by this atom?

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An emitted photon's energy equals the difference between two levels. The six possible differences are 5.0, 7.0, 8.5, 2.0, 3.5 and 1.5 eV. No pair of levels is 4.0 eV apart. The value 4.0 eV is the magnitude of a single level, not a difference between two levels.✓ 1Answer C

Answer: C

Common error: Photon energies are always differences between levels. A level value on its own is not a photon energy, unless the other level is 0 eV.

32IBP2H-036-01-06Emission and absorption spectra Paper 1AMedium1 mark

A space telescope observes a hot star through a cool cloud of interstellar gas. Which best describes the spectrum of the light that reaches the telescope?

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Atoms in the cool cloud absorb photons whose energies match their own level differences. They then re-emit those photons in random directions. Light at those wavelengths is therefore depleted along the line of sight, giving dark lines on a continuous background. The positions of the dark lines identify the elements in the cloud.✓ 1Answer B

Answer: B

Exam tip: Absorption and emission lines of the same element occur at the same wavelengths. The difference is whether the gas lies in front of a hotter continuous source, or is itself hot and seen against a dark background.

33IBP2S-035-01-07Energy levels and spectra Paper 1AMedium1 mark

Hydrogen energy levels are given by En = −13.6/n2 eV. An astronomer compares photon X, emitted in the transition n = 3 → n = 2, with photon Y, emitted in the transition n = 2 → n = 1. What is \(\frac{\text{wavelength of X}}{\text{wavelength of Y}}\)?

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EX = 13.6(1/4 − 1/9) = 13.6 × 5/36 eV and EY = 13.6(1 − 1/4) = 13.6 × 27/36 eV. Since λ ∝ 1/E, λX/λY = EY/EX = 27/5 = 5.4. Option A is the inverted ratio. Option B is the ratio of the initial quantum numbers, 3/2. Option C is (3/2)2, a ratio of single-level energies rather than of energy differences.✓ 1Answer D

Answer: D

Common error: Wavelength depends on the energy difference between the two levels, not on the ratio of the quantum numbers.

34IBP2S-035-00-07The Bohr model Paper 1AMedium1 mark

Radio telescopes detect hydrogen atoms in interstellar space with electrons in extremely high levels. Using the Bohr model, En = −13.6/n2 eV, what is the ionization energy of a hydrogen atom with n = 100?

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The ionization energy is |En| = 13.6/1002 = 13.6/10 000 = 1.36×10−3 eV. The energy depends on 1/n2, not 1/n (which gives 0.136 eV). The value 13.6 eV applies only to n = 1.✓ 1Answer B

Answer: B

Exam tip: In the Bohr model, energies scale as 1/n². High-n atoms are bound extremely weakly.

35IBP2S-035-02-05Nuclear radius and closest approach Paper 1AMedium1 mark

An alpha particle with kinetic energy E travels directly towards a stationary nucleus. Its distance of closest approach is d.
A proton with the same kinetic energy E travels directly towards an identical stationary nucleus. What is the proton's distance of closest approach?

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At closest approach all the kinetic energy has become electric potential energy: \(E = \frac{kqQ}{d}\), so \(d = \frac{kqQ}{E}\). With E and Q fixed, d is proportional to the charge q of the projectile. The proton has half the charge of an alpha particle, so its closest approach is d/2. Mass does not appear in the equation, so the factor 1/4 from the mass ratio is wrong.✓ 1Answer B

Answer: B

Common error: The closest-approach equation depends on kinetic energy, not speed, so the projectile's mass plays no part when E is fixed.

36IBP2S-035-02-06Nuclear radius and density Paper 1AMedium1 mark

Nucleus P has nucleon number 8. Nucleus Q has nucleon number 125.
What is the value of \(\frac{\text{radius of Q}}{\text{radius of P}}\)?

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Nuclear radius follows R = R0A1/3. The ratio is (125/8)1/3 = 5/2 = 2.5. The value 15.6 is the ratio of nucleon numbers, which assumes R ∝ A. The value 6.25 corresponds to A2/3. The value 3.95 is the square root of the nucleon-number ratio.✓ 1Answer A

Answer: A

Exam tip: Choose nucleon numbers that are perfect cubes (8, 27, 64, 125, 216) to check radius ratios without a calculator.

37IBP2S-035-02-08Rutherford scattering and the nuclear radius Paper 1AMedium1 mark

When alpha particles of very high kinetic energy are scattered by nuclei, the number scattered through large angles is less than the Rutherford model predicts.
What is the best explanation?

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The Rutherford model assumes that only the Coulomb force acts between the alpha particle and the nucleus. At high energies the distance of closest approach becomes comparable to the nuclear radius. The short-range strong nuclear force then also acts, so the scattering no longer follows the Coulomb prediction. The energy at which this deviation begins can be used to estimate nuclear size. Coulomb's law remains valid well inside the atom. Fast alpha particles do not capture electrons, and electron shielding is negligible this close to the nucleus.✓ 1Answer A

Answer: A

Common error: The deviation does not mean Coulomb's law has failed. A second, short-range force has started to act.

38IBP3H-E1-20r13-03Transitions on an energy level diagram Paper 1AMedium1 mark

A lamp used for greenhouse lighting contains a vapour whose atoms have these four energy levels: −8.0 eV, −3.0 eV, −1.5 eV and −0.5 eV. Atoms in the n = 4 level (−0.5 eV) make transitions to lower levels. Which transition produces the photon of longest wavelength?

energyn = 1−8.0 eVn = 2−3.0 eVn = 3−1.5 eVn = 4−0.5 eV
Energy levels not to scale
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Wavelength is longest for the smallest energy difference. The gaps are 4→3: 1.0 eV, 3→2: 1.5 eV, 2→1: 5.0 eV and 4→1: 7.5 eV. So 4→3 is correct. The 4→1 option assumes the biggest jump gives the longest wavelength, the 2→1 and 3→2 options use a larger gap than 4→3.✓ 1Answer C

Answer: C

Common error: Linking a larger energy jump to a longer wavelength. Since E = hc/λ, the larger the energy change, the shorter the wavelength.

39IBP3H-E1-20r13-04Counting spectral lines Paper 1AMedium1 mark

A sample of gas has atoms that can occupy only the levels n = 1, 2, 3 and 4, with all energy gaps different. Atoms are excited to n = 4. What is the maximum number of different photon energies that can be emitted as the atoms return to the ground state?

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The possible transitions are 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1, which is 6 different gaps. 3 counts only the transitions starting at n = 4, 4 counts levels instead of transitions, and 10 comes from counting five levels (n = 0 to 4).✓ 1Answer D

Answer: D

Exam tip: With N levels, the number of different gaps is N(N − 1)/2. Check by listing the pairs once.

40IBP3H-E1-20r13-05Transitions on an energy level diagram Paper 1AMedium1 mark

The energy levels of hydrogen are given by En = −13.6/n² eV. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed. A hydrogen atom makes a transition from n = 2 to n = 1. What is the wavelength of the emitted photon?

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ΔE = 13.6 − 3.40 = 10.2 eV = 1.63 × 10−18 J, and λ = hc/ΔE = 1.22 × 10−7 m, so A is correct. B uses 13.6 eV (the ionisation energy) instead of the gap. C uses 3.40 eV (the energy of n = 2) instead of the gap. D leaves ΔE in eV without converting to joules.✓ 1Answer A

Answer: A

Common error: Using the energy of one level as the photon energy. The photon energy is the difference between the two levels.

41IBP3H-E1-20r13-08Excitation energy from electron collisions Paper 1AMedium1 mark

Sodium vapour lamps on an airport apron emit strongly at a wavelength of 589 nm. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed. What is the energy difference between the two atomic levels involved in this emission?

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ΔE = hc/λ = (6.63 × 10−34 × 3.00 × 108)/(589 × 10−9) = 3.38 × 10−19 J = 2.1 eV, so D is correct. 0.34 eV uses h/2π in place of h, 13 eV uses 2πh, and 3.4 eV forgets to convert joules to electronvolts (it is the J value × 1019).✓ 1Answer D

Answer: D

Exam tip: To convert joules to electronvolts divide by 1.60 × 10⁻¹⁹; quote the answer in the unit the options use.

42IBP3H-E1-20r14-05Rutherford scattering and the nuclear radius Paper 1AMedium1 mark

An alpha particle of kinetic energy 6.0 MeV is fired directly at the nucleus of an aluminium atom (atomic number 13). Ignoring recoil of the nucleus, what is the distance of closest approach? (k = 8.99×109 N m2 C−2, e = 1.60×10−19 C)

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B is correct: at closest approach all the kinetic energy is electric potential energy, so d = k(2e)(13e)/Ek = 8.99×109 × 2 × 13 × 1.60×10−19 / (6.0×106) = 6.2×10−15 m. A omits the charge of the alpha particle (uses 13e×e). C uses only half the energy. D squares the atomic number (169) instead of multiplying by 2 × 13.✓ 1Answer B

Answer: B

Common error: The alpha particle carries charge +2e, so the product of charges is 2 × Z × e².

43IBP3H-E1-20r14-06Nuclear radius and density Paper 1AMedium1 mark

The nuclear radius is given by R = R0A1/3. What is the approximate ratio of the radius of a uranium-238 nucleus to the radius of a helium-4 nucleus?

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D is correct: RU/RHe = (238/4)1/3 = 59.51/3 = 3.9. A is the ratio of the mass numbers, ignoring the cube root. B is (59.5)2/3, the ratio of cross-sectional areas. C is the square root of 59.5, using the power 1/2 instead of 1/3.✓ 1Answer D

Answer: D

Common error: Radius scales as the cube root of the nucleon number, not linearly, and not as its square root.

44IBP3H-E1-20r14-08Nuclear radius and closest approach Paper 1AMedium1 mark

A proton and an alpha particle, each with the same kinetic energy, are fired directly at the same stationary heavy nucleus. Ignoring recoil, what is (closest approach of the alpha particle) ÷ (closest approach of the proton)?

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C is correct: d = k(q1)(Ze)/Ek, so for the same energy and target d is proportional to the projectile charge; the alpha has charge 2e and the proton e, so the ratio is 2. A assumes the heavier alpha particle gets closer. B assumes the same energy gives the same distance regardless of charge. D squares the charge ratio or uses the mass ratio.✓ 1Answer C

Answer: C

Exam tip: For head-on approach, closest distance depends on the charges and the kinetic energy, not on the mass of the projectile.

45IBP3H-E1-20r33-04The hydrogen spectrum and a diffraction grating Paper 1AMedium1 mark

Light of wavelength 656 nm is incident normally on a diffraction grating with 500 lines per millimetre. What is the angle between the two first-order maxima, one on each side of the central maximum?

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The slit spacing is d = 1/(500 × 103) = 2.00 × 10−6 m, so sin θ = λ/d = 0.328 and θ = 19.1° on each side of the central maximum. The angle between the two first-order maxima is therefore 2θ = 38.3°. Giving 19.1° only measures from the central maximum to one maximum. 41.0° results from using order n = 2 for the first maximum (sin θ = 0.656), and 82.0° doubles that angle as well.✓ 1Answer B

Answer: B

Exam tip: Check what the question measures the angle from: θ is measured from the central maximum, so the separation of the two first-order maxima is 2θ.

46IBP3H-E1-20r33-05The Bohr model Paper 1AMedium1 mark

In the Bohr model of the hydrogen atom, what is the orbital angular momentum of the electron when it is in the n = 2 energy level? (h is the Planck constant.)

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In the Bohr model the angular momentum is quantised as mvr = nh/2π, so for n = 2 it equals 2h/2π = h/π. The value h/2π is the angular momentum of the n = 1 level. The value 2h leaves out the factor 1/2π. The value h/4π divides by n instead of multiplying by it.✓ 1Answer A

Answer: A

Common error: Angular momentum in the Bohr model increases in proportion to n; it is not divided by n.

47IBP3H-E1-20r33-06Counting spectral lines Paper 1AMedium1 mark

The atoms of a gas in a discharge lamp have only four energy levels, at −5.0 eV, −3.1 eV, −1.6 eV and −0.8 eV. Atoms excited to the −0.8 eV level can return to lower levels by any sequence of downward transitions. What is the maximum number of different photon energies in the emission spectrum of the gas?

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Every pair of levels gives one possible downward transition: from −0.8 eV there are 3 (to each lower level), from −1.6 eV there are 2, and from −3.1 eV there is 1, a total of 6, and the six energy gaps (4.2, 3.4, 2.3, 1.9, 1.5 and 0.8 eV) are all different. Counting only transitions from the top level gives 3, which ignores the cascades. Counting the number of levels gives 4, which confuses levels with transitions. Counting 12 includes upward transitions, which are absorption, not emission.✓ 1Answer C

Answer: C

Exam tip: For N levels the number of possible downward transitions is N(N − 1)/2.

48E-1A-17Energy levels and spectra Paper 1AHard1 mark

An atom has three energy levels E1 < E2 < E3. The transition from E3 to E1 emits a photon of wavelength λ31 and the transition from E2 to E1 emits a photon of wavelength λ21.

What is the wavelength of the photon emitted in the transition from E3 to E2?

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Step 1Photon energies add, not wavelengths: E3 − E2 = (E3 − E1) − (E2 − E1).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2With E = hc/λ: 1/λ32 = 1/λ31 − 1/λ21 = (λ21 − λ31)/(λ31λ21).—
Step 3λ32 = λ31λ21/(λ21 − λ31) — positive because λ21 > λ31. (For hydrogen: 102.6 nm and 121.6 nm give 656 nm.)✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the reciprocals of the wavelengths subtract because the energies subtract.
  • BThis subtracts the wavelengths as if they were energies; for hydrogen it would give 19 nm instead of 656 nm.
  • CThis adds the reciprocals: it is the wavelength for a photon carrying the sum of the two energies.
  • DThe subtraction is the wrong way round: λ31 < λ21, so this is negative.
49E-1A-25The Bohr model Paper 1AHard1 mark

In the Bohr model of hydrogen the radius of the orbit with quantum number n is n²r1, where r1 is the radius of the smallest orbit, and the angular momentum of the electron is mvr = nh/2π.

What is the de Broglie wavelength of the electron in the orbit with quantum number n?

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Step 1λ = h/mv and, from the quantisation condition, mv = nh/(2πr).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2So λ = 2πr/n: exactly n wavelengths fit around the orbit.—
Step 3With r = n²r1: λ = 2πnr1.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the speed to be proportional to n (inverting v ∝ 1/n), so the wavelength wrongly falls with n.
  • BThis assumes one wavelength fits the smallest orbit and that the wavelength is the same in every orbit.
  • CCorrect: λ = 2πr/n = 2πnr1.
  • DThis is the circumference of orbit n, i.e. one wavelength per orbit; the quantisation condition fits n wavelengths.
50E-1A-33Rutherford scattering Paper 1AHard1 mark

An alpha particle of kinetic energy E is directed head-on at a nucleus and reaches a distance of closest approach d. A proton of kinetic energy E/2 is then directed head-on at the same nucleus. The nucleus remains at rest in both cases.

What is the distance of closest approach of the proton?

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Step 1At closest approach all the kinetic energy has become electric potential energy: Ek = kqQ/d, so d ∝ q/Ek.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The proton has half the charge of the alpha particle (factor 1/2) and half the kinetic energy (factor 2).—
Step 3The two factors cancel: the distance of closest approach is again d.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis accounts only for the smaller charge of the proton and ignores the halved kinetic energy.
  • BCorrect: d ∝ q/Ek, and both are halved.
  • CThis accounts only for the halved kinetic energy and ignores the smaller charge of the proton.
  • DThis takes d ∝ 1/(qEk): the charge dependence is inverted, so both changes appear to double d.
51E-1A-34Nuclear radius and closest approach Paper 1AHard1 mark

Alpha particles are fired head-on at nuclei. In a simple model, deviations from Rutherford scattering begin at the kinetic energy Edev for which the distance of closest approach equals the nuclear radius R = R0A1/3.

Gold has Z = 79 and A = 197; aluminium has Z = 13 and A = 27. What is Edev(gold)/Edev(aluminium)?

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Step 1At closest approach the kinetic energy has become electric potential energy: Edev = k(2e)(Ze)/R, so Edev ∝ Z/R.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2With R = R0A1/3: Edev ∝ Z/A1/3.—
Step 3Ratio = (79/13) × (27/197)1/3 = 6.08 × 0.516 = 3.1: the larger charge of gold raises the energy, its larger radius lowers it.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the radius as proportional to A instead of A1/3: (79/13) × (27/197) = 0.83.
  • BCorrect: Edev ∝ Z/A1/3, giving 3.1.
  • CThis compares only the nuclear charges (79/13) and ignores the larger radius of the gold nucleus.
  • DThe radius ratio has been inverted: (79/13) × (197/27)1/3 = 11.8, as if a larger nucleus needed the alpha particle to come closer.
52IBP2H-036-00-05Energy levels and spectra Paper 1AHard1 mark

In the Bohr model, the energy levels of a one-electron ion with proton number Z are En = −13.6 Z2/n2 eV. Singly ionised helium, He+ (Z = 2), emits a photon in the transition n = 4 → n = 2. Which transition in hydrogen emits a photon of the same wavelength?

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For He+: ΔE = 13.6 × 22 × (1/22 − 1/42) = 54.4 × 3/16 = 10.2 eV.
For hydrogen n = 2 → 1: ΔE = 13.6 × (1 − 1/4) = 10.2 eV.
Equal photon energies mean equal wavelengths (λ = hc/ΔE ≈ 122 nm), so B is correct.
Option A assumes the same quantum numbers give the same photon, ignoring the factor Z2. In hydrogen this transition gives only 2.55 eV.
Option C gives 12.75 eV.
Option D gives 1.89 eV.
✓ 1Answer B

Answer: B

Exam tip: Because Z² = 4, He⁺ level 2n has exactly the same energy as hydrogen level n. A He⁺ transition between levels 2a and 2b therefore matches the hydrogen transition between levels a and b.

53IBP2H-036-00-08The Bohr model Paper 1AHard1 mark

In the Bohr model, the electron's orbit can be pictured as a whole number of de Broglie wavelengths fitting around the circumference. How does the de Broglie wavelength of the electron in the n = 3 orbit of hydrogen compare with its value in the n = 1 orbit?

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The fitting condition is nλ = 2πr, so λ = 2πr/n. Since r ∝ n2, λ ∝ n2/n = n, and the wavelength for n = 3 is 3 times larger. The same result follows from λ = h/mv with v ∝ 1/n. Option A assumes the wavelength shrinks because more waves fit around the orbit. Option B assumes the wavelength is fixed. Option D forgets to divide the circumference by n.✓ 1Answer C

Answer: C

Exam tip: Two independent routes give the same answer: λ = 2πr/n and λ = h/(mv). Using both is a quick way to check your reasoning.

54IBP2H-036-01-07The Bohr model Paper 1AHard1 mark

In the Bohr model of hydrogen, the electron moves from the n = 2 orbit to the n = 4 orbit. What is \(\frac{\text{de Broglie wavelength in } n=4}{\text{de Broglie wavelength in } n=2}\)?

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The quantisation condition mvr = nh/2π means an orbit holds n whole de Broglie wavelengths, so nλ = 2πr. Bohr radii scale as r ∝ n2, so λ = 2πr/n ∝ n. Going from n = 2 to n = 4 doubles the wavelength. Equivalently, v ∝ 1/n, so the momentum halves and λ = h/p doubles.✓ 1Answer B

Answer: B

Common error: Option 4 comes from using r ∝ n² without dividing by n, the number of wavelengths that fit around the orbit.

55IBP2H-036-01-08Transitions on an energy level diagram Paper 1AHard1 mark

The hydrogen energy levels are En = −13.6/n2 eV. What is \(\frac{\text{longest wavelength in the Lyman series}}{\text{longest wavelength in the Balmer series}}\)?

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The longest wavelength corresponds to the smallest energy gap. For the Lyman series this is 2→1: 13.6(1 − 1/4) = 13.6 × 3/4. For the Balmer series this is 3→2: 13.6(1/4 − 1/9) = 13.6 × 5/36. Since λ ∝ 1/ΔE, the ratio is (5/36)/(3/4) = 5/27.✓ 1Answer A

Answer: A

Common error: Wavelength is inversely proportional to energy, so invert the energy ratio. Taking the energy ratio directly gives 27/5. Using only the lower-level numbers gives 1/4.

56IBP2S-035-01-06The Bohr model Paper 1AHard1 mark

In the Bohr model of hydrogen, the orbital radius is proportional to n2 and the electron's orbital speed is proportional to 1/n. What is \(\frac{\text{orbital period in } n=2}{\text{orbital period in } n=1}\)?

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The period is T = 2πr/v. With r ∝ n2 and v ∝ 1/n, this gives T ∝ n2/(1/n) = n3. The ratio is therefore 23 = 8. Option B uses the radius scaling alone. Option A uses the speed scaling alone. Option D multiplies by n4, as if both the radius and the inverse speed went as n2.✓ 1Answer C

Answer: C

Exam tip: Build the period from T = 2πr/v and combine the powers of n. Do not guess from one scaling alone.

57IBP2S-035-02-07Nuclear radius and density Paper 1AHard1 mark

Nucleus X has nucleon number A. Nucleus Y has nucleon number 8A.
Which row gives the ratio of radius and the ratio of nuclear density, Y : X?

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R = R0A1/3, so the radius ratio is 81/3 = 2. The mass is proportional to A, and the volume is proportional to R3, which is proportional to A. Density is therefore independent of A, giving a density ratio of 1. A ratio of 8 ignores the increase in volume. The rows with radius ratio 8 wrongly assume R ∝ A.✓ 1Answer A

Answer: A

Exam tip: All nuclei have approximately the same density, about 10¹⁷ kg m⁻³. This follows directly from R ∝ A¹/³.

58IBP2S-035-03-10Nuclear radius and closest approach Paper 1AHard1 mark

Alpha particles with the same initial kinetic energy are directed head-on at a gold nucleus, 19779Au, and at an aluminium nucleus, 2713Al. Treat each nucleus as a stationary point charge that does not recoil. What is \(\frac{\text{distance of closest approach to gold}}{\text{distance of closest approach to aluminium}}\)?

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At closest approach, all the alpha particle's kinetic energy has become electric potential energy: Ek = k(2e)(Ze)/d. So d = 2kZe2/Ek. With the same Ek, d is proportional to Z. The ratio is 79/13 ≈ 6.1.
The value 7.3 uses nucleon numbers (197/27), but the electric repulsion depends only on charge. The value 2.5 is the square root of the correct ratio. The value 0.16 is the inverted ratio.
✓ 1Answer C

Answer: C

Common error: The repulsion depends on the nuclear charge Ze, so use the proton number, not the nucleon number.

59IBP3H-E1-20r13-06The Bohr model and muonic hydrogen Paper 1AHard1 mark

Muonic hydrogen consists of a proton and a negative muon. The muon has the same charge as an electron and a mass about 207 times as great. Using the Bohr model with the proton assumed to be stationary, how do the ground-state orbital radius and the magnitude of the ground-state energy of muonic hydrogen compare with those of ordinary hydrogen?

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In the Bohr model rn ∝ n²/m and En ∝ −m/n², so with m 207 times larger the radius is 207 times smaller and the energy magnitude 207 times larger. C is correct. A gets the radius right but the energy wrong, B reverses both (the heavier particle binds more tightly, not less) and D wrongly takes a square root.✓ 1Answer C

Answer: C

Exam tip: A heavier orbiting particle is held closer and bound more strongly: radius ∝ 1/m and binding energy ∝ m.

60IBP3H-E1-20r13-07The Bohr model Paper 1AHard1 mark

In the Bohr model of hydrogen, the electron orbit radius is proportional to n², and the electron's angular momentum is nh/2π. How does the orbital speed of the electron in the n = 3 orbit compare with its speed in the n = 1 orbit?

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From mvr = nh/2π, v = nh/(2πmr). With r ∝ n², v ∝ n/n² = 1/n, so v3 = v1/3. One ninth follows the radius or kinetic-energy scaling (1/n²), three times assumes v ∝ n, and 1/√3 comes from v ∝ 1/√n, which does not follow from the quantisation condition.✓ 1Answer B

Answer: B

Common error: Assuming speed increases in higher orbits. In the Bohr model the electron is slower the further it is from the nucleus.

61IBP3H-E1-20r33-07Nuclear radius and closest approach Paper 1AHard1 mark

An α particle with kinetic energy E is fired directly at a stationary nucleus of charge +Ze. It stops momentarily at a distance r0 from the centre of the nucleus. A proton with the same kinetic energy E is fired directly at a stationary nucleus of charge +2Ze. Ignore any motion of the nuclei. What is the closest approach of the proton to the centre of its target nucleus?

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At closest approach all the kinetic energy has become electric potential energy, so r = kq1q2/E. For the α particle q1q2 = (2e)(Ze) = 2Ze2, and for the proton it is (e)(2Ze) = 2Ze2. The products and the energies are the same, so the distance is r0. The answer r0/2 inverts the ratio of the charges. The answer 2r0 forgets that the α particle has charge +2e. The answer 4r0 brings in the mass ratio of the particles, but the mass does not enter this energy balance.✓ 1Answer B

Answer: B

Exam tip: For head-on closest approach set the initial kinetic energy equal to kq₁q₂/r; the masses of the particles do not appear.

62E-1B-03Nuclear radius and density Paper 1BMedium7 marks

Beams of high-energy electrons are scattered by thin targets of six different elements. From the angle of the first minimum of each diffraction pattern, the radius R of the target nuclei is deduced. A student wishes to test the hypothesis that R = R0An, where A is the nucleon number and R0 and n are constants, and so plots ln(R / fm) against ln A. The graph shows the data and the line of best fit, which has been drawn only across the range of the data.

(1 fm = 1.0 × 10−15 m; 1 u = 1.66 × 10−27 kg)

ElementAR / fmln Aln(R / fm)
carbon122.782.481.022
aluminium273.553.301.267
cobalt594.724.081.552
yttrium895.314.491.670
tin1206.004.791.792
lead2087.075.341.956
0123456ln A0.00.20.40.60.81.01.21.41.61.82.02.2ln(R / fm)
ln(R / fm) against ln A (graph drawn to scale)
(a)
(i)

Determine n.

(2)
(b)
(i)

Determine R0 by extrapolating the line of best fit.

(2)
(c)
(i)

Assume that n = 1/3 exactly. Use your answer to (b) to estimate the density of nuclear matter.

(2)
(d)
(i)

The uncertainty in R0 is ±0.05 fm. Determine the absolute uncertainty in your answer to (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ln R = ln R0 + n ln A, so n is the gradient of the graph✓ 1
Gradient from well-separated points on the line, e.g. (2.48, 1.01) and (5.34, 1.96): n = 0.33 ≈ 1/3✓ 1Accept 0.31–0.35.
Part (b)(i)
Intercept on the ln(R / fm) axis at ln A = 0: 0.19✓ 1Accept 0.15–0.23. ALT: intercept = ln R − n ln A for a point on the line, using n from (a); Allow ECF from (a).
R0 = e0.19 = 1.21 fm✓ 1Accept 1.16–1.26 fm.
Part (c)(i)
ρ = Au/(4/3 πR0³A) = 3u/(4πR0³): A cancels✓ 1Mass of a nucleus ≈ Au.
ρ = 3 × 1.66 × 10−27/(4π × (1.21 × 10−15)³) = 2.2 × 1017 kg m−3✓ 1Allow ECF from (b). With R0 = 1.20 fm: 2.3 × 1017 kg m−3. Accept 1.9–2.5 × 1017 kg m−3.
Part (d)(i)
%ρ = 3 × (0.05/1.21) × 100 = 12 %, so Δρ = ±3 × 1016 kg m−3✓ 1Allow ECF from (b) and (c). Answer to 1 s.f. required. Using the percentage in R0 once instead of three times: 0.

Answers: (a)(i) 0.33  ·  (b)(i) 1.21 fm  ·  (c)(i) 2.2 × 1017 kg m−3  ·  (d)(i) ±3 × 1016 kg m−3 (the remaining parts are explanations — see the table above)

63E-1B-04Rutherford scattering and nuclear charge Paper 1BMedium7 marks

A beam of alpha particles, all with the same kinetic energy, is directed in turn at thin foils of six different metals in an evacuated chamber. Every foil contains the same number of atoms per unit area. A detector at a fixed scattering angle counts the number N of alpha particles that arrive in 30 minutes. Background counts are negligible. Rutherford's model predicts that N ∝ Z², where Z is the proton number of the nuclei in the foil.

The graph shows ln N against ln Z with the line of best fit. The error bars show the random uncertainty ±√N in each count.

FoilZNln Zln N
aluminium134232.5656.047
titanium2212203.0917.107
copper2920903.3677.645
silver4754563.8508.604
tin5062143.9128.735
gold79154794.3699.647
2.42.62.83.03.23.43.63.84.04.24.44.6ln Z5.56.06.57.07.58.08.59.09.510.0ln N
ln N against ln Z (graph drawn to scale)
(a)
(i)

Determine the gradient of the line and state whether the data support the prediction N ∝ Z².

(2)
(b)
(i)

Show that the error bar on ln N for aluminium is about ±0.05.

(2)
(ii)

Suggest how, using the same apparatus, the percentage uncertainty in the count for aluminium could be reduced to 2 %.

(1)
(c)
(i)

A foil of an unknown metal, with the same number of atoms per unit area, gives 13200 counts in 30 minutes. Determine the proton number of the metal.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (2.50, 5.92) and (4.50, 9.90): (9.90 − 5.92)/2.00 = 1.99✓ 1Accept 1.9–2.1. Points must be read from the line.
ln N = 2 ln Z + constant if N ∝ Z²; the gradient is 2 within the uncertainty (and the points lie on a straight line), so the data support the prediction✓ 1Allow ECF from the gradient. The conclusion must agree with the candidate's gradient.
Part (b)(i)
Percentage uncertainty in N = √423/423 × 100 = 4.9 %✓ 1Or ΔN = √423 = 21 counts.
Δ(ln N) = ΔN/N = 0.049 ≈ ±0.05✓ 1The uncertainty in ln N equals the fractional uncertainty in N. Full working or an answer to at least 2 s.f. must be seen.
Part (b)(ii)
Need √N/N ≤ 0.02, i.e. N ≥ 2500 counts: count for about 6 times as long (about 3 h) with the same foil and detector✓ 1Accept "repeat the 30-minute count about six times and add the counts". Do not accept "use a stronger source" or "a better detector".
Part (c)(i)
ln 13200 = 9.49; from the line ln Z = 4.29✓ 1Allow ECF from the line. ALT: Z = 79 × √(13200/15479) = 73.0 using gold, since N ∝ Z².
Z = e4.29 = 73✓ 1Accept 72–74. An integer is expected; 73.0 accepted.

Answers: (a)(i) 2.0  ·  (b)(i) ±0.05  ·  (c)(i) 73 (the remaining parts are explanations — see the table above)

64E-1B-12Rutherford scattering Paper 1BMedium6 marks

In a teaching version of the Geiger–Marsden–Rutherford experiment, a narrow beam of alpha particles strikes a thin gold foil in an evacuated chamber. A detector on a rotating arm counts the alpha particles N arriving in 20 minutes at each angle reading θ on the arm's scale. Before the foil was inserted, the student found that the unscattered beam gave its maximum count at a scale reading of +2.0°. Rutherford's model predicts N = k(sin(θ/2))−n with n = 4, where k is a constant.

The graph shows ln N against ln(sin(θ/2)), calculated from the scale readings, with the line of best fit.

θ (scale reading) / °Nln(sin(θ/2))ln N
2087699−1.7511.38
3014999−1.359.62
452925−0.967.98
60932−0.696.84
90226−0.355.42
12095−0.144.55
15062−0.034.13
-2.00-1.75-1.50-1.25-1.00-0.75-0.50-0.250.00ln(sin(θ/2))3456789101112ln N
ln N against ln(sin(θ/2)) using the scale readings (graph drawn to scale)
(a)
(i)

Determine n from the graph.

(2)
(b)
(i)

Explain how the zero error accounts for the difference between your answer to (a) and the predicted value of 4.

(2)
(c)
(i)

Assume n = 4. Use the count at the scale reading of 90°, with corrected angles, to predict the count at the scale reading of 30°, and compare it with the measured value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ln N = ln k − n ln(sin(θ/2)); gradient from well-separated points on the line, e.g. (−1.75, 11.3) and (−0.03, 4.1)✓ 1
Gradient = −4.2, so n = 4.2✓ 1Accept 4.1–4.35.
Part (b)(i)
The true angles are 2.0° smaller than the readings; this changes sin(θ/2) by a much larger fraction at small angles (≈ 10 % at a reading of 20°) than at large angles (< 1 % at 150°)✓ 1Allow ECF from (a).
Corrected, the left-hand points move further left while the right-hand points hardly move, so the line becomes less steep and n moves down towards 4 (≈ 4.0)✓ 1The direction of the change must be consistent with the candidate's answer to (a).
Part (c)(i)
Corrected angles 88° and 28°: N30 = 226 × (sin 44°/sin 14°)⁴ = 226 × 68.0 = 1.5 × 104✓ 1Allow ECF from (b). Using the uncorrected readings gives 1.3 × 104: [1 max].
Agrees with the measured 14999 to within 2.4 %, supporting n = 4 once the zero error is removed✓ 1The comparison must follow from the candidate's prediction.

Answers: (a)(i) 4.2  ·  (c)(i) 1.5 × 104 (the remaining parts are explanations — see the table above)

65E-1B-22Excitation energy from electron collisions Paper 1BMedium7 marks

In a tube containing sodium vapour at low pressure, electrons from a hot cathode are accelerated from rest through a variable potential difference V towards a wire grid. Electrons that pass the grid must overcome a small retarding potential difference to reach a collecting plate. As V is increased, the collector current rises and falls repeatedly, because an electron that has gained exactly enough energy can give the first excitation energy E of a sodium atom to an atom in an inelastic collision.

A simple model predicts that the k-th current maximum occurs at Vk = kE/e. The table gives V at the first six maxima; each value is uncertain by ±0.10 V. The graph shows V against k with the line of best fit, extended to k = 0. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

k123456
V / V3.595.607.789.9111.9114.07
01234567k0246810121416V / V
Potential difference at each current maximum against k (graph drawn to scale)
(a)
(i)

Determine, using the gradient of the line, the first excitation energy E of sodium, in eV.

(2)
(b)
(i)

According to the simple model the line should pass through the origin. Determine the intercept on the V axis and state, with a reason, whether it affects your answer to (a).

(2)
(c)
(i)

Using only the first and last readings in the table, determine the absolute uncertainty in E.

(1)
(d)
(i)

Sodium vapour lamps emit yellow light of wavelength 589 nm. Deduce whether this light could be produced by sodium atoms excited in this experiment.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient = ΔV/Δk from two well-separated points on the line = 2.10 V✓ 1Accept 2.05–2.15 V.
E = e × gradient = 2.1 eV✓ 1Accept 2.05–2.15 eV. Using a single reading, e.g. V1 = 3.59 V as E/e: 0 for this mark.
Part (b)(i)
Intercept = 1.5 V✓ 1Accept 1.3–1.6 V.
It is a systematic error that adds the same amount to every V, so it does not change the gradient or E✓ 1Accept "offset" or "zero error". Reason needed.
Part (c)(i)
ΔE = (0.10 + 0.10)/5 = ±0.04 eV✓ 1Both readings' uncertainties must be added and divided by the 5 intervals. ±0.02 eV (one reading only) scores 0.
Part (d)(i)
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/(2.10 × 1.60 × 10−19) = 592 nm✓ 1Allow ECF from (a).
With ΔE = ±0.04 eV the range is about 581–604 nm (±11 nm), which includes 589 nm, so yes: the de-excitation of the excited atoms can produce this light✓ 1Allow ECF from (c). The conclusion must agree with the candidate's own range. A bare "yes" with no comparison: 0.

Answers: (a)(i) 2.1 eV  ·  (b)(i) 1.5 V  ·  (c)(i) ±0.04 eV  ·  (d)(i) 592 ± 11 nm; yes (the remaining parts are explanations — see the table above)

66E-1B-11The hydrogen spectrum Paper 1BHard7 marks

Light from a hydrogen discharge tube is passed through a diffraction grating with 600 lines per mm mounted on a spectrometer. The student measures the first-order angle of diffraction θ for five visible lines, each produced by a transition from an upper level n to the same lower level nf. Modelling the energy levels as En = −EI/n², she predicts that 1/λ = (EI/hc)(1/nf² − 1/n²) and plots 1/λ against 1/n².

The graph shows four of the points with the line of best fit. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

nθ / °1/n²1/λ / 106 m−1
323.280.11111.518
416.990.06252.053
515.150.0400
614.280.02782.432
713.840.02042.508
0.000.020.040.060.080.100.121/n²1.41.61.82.02.22.42.62.81/λ / 10⁶ m⁻¹
1/λ against 1/n² (graph drawn to scale; the point for n = 5 is not plotted)
(a)
(i)

Determine 1/λ for the line with n = 5.

(2)
(b)
(i)

State whether your answer to (a) lies on the line of best fit.

(1)
(c)
(i)

Determine EI in eV.

(2)
(d)
(i)

Deduce nf by extrapolating the line.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Grating spacing = 1/600 mm = 1.667 × 10−6 m; λ = d sin θ = 1.667 × 10−6 × sin 15.15° = 4.36 × 10−7 m✓ 1
1/λ = 2.296 × 106 m−1✓ 1Accept 2.29–2.31 × 106 m−1.
Part (b)(i)
At 1/n² = 0.0400 the line gives 2.30 × 106 m−1, equal to the value from (a): the point lies on the line✓ 1Allow ECF from (a). The comparison must use a value read from the line.
Part (c)(i)
Gradient = −1.09 × 107 m−1 from well-separated points on the line✓ 1Accept −1.06 to −1.12 × 107 m−1.
EI = |gradient| × hc/e = 13.6 eV✓ 1Allow ECF from the gradient. Accept 13.2–14.0 eV.
Part (d)(i)
Intercept at 1/n² = 0: 2.73 × 106 m−1 = EI/(hcnf²) = |gradient|/nf²✓ 1Accept 2.70–2.76 × 106 m−1.
1/nf² = 2.73/10.93 = 0.250, so nf = 2✓ 1Allow ECF from (c). An integer must be stated.

Answers: (a)(i) 2.30 × 106 m−1  ·  (c)(i) 13.6 eV  ·  (d)(i) 2 (the remaining parts are explanations — see the table above)

67IBP2H-036-00-09Nuclear notation and isotopes Paper 2Easy5 marks

Iodine-131, 13153I, is used in medicine to treat thyroid conditions. The stable form of iodine found in food is iodine-127.

(a)

State the number of protons and the number of neutrons in a nucleus of iodine-131.

(2)
(b)

Outline why iodine-131 and iodine-127 are described as isotopes of the same element.

(1)
(c)

Nuclear radius is given by R = R0A1/3, where R0 = 1.20 × 10−15 m. Calculate the radius of an iodine-131 nucleus.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
53 protons ; 131 − 53 = 78 neutrons✓ 2
Part (b)
Both have the same number of protons (Z = 53) but different numbers of neutrons, so they have different nucleon numbers✓ 1
Part (c)
R = 1.20 × 10−15 × 1311/3 ; R = 6.09 × 10−15 m✓ 2

Common error: In R = R₀A^(1/3), use the nucleon number A (131), not the proton number Z.

68IBP2H-036-01-09Excitation of atoms in a fluorescent lamp Paper 2Easy5 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)

Outline how an atom of sodium vapour in the lamp emits light.

(2)
(b)

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

(2)
(c)

State why a mercury lamp emits light of different colours from a sodium lamp.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Electrons in the atom are excited to higher energy levels, for example by collisions with electrons in the discharge . When an electron falls to a lower level, it emits a photon whose energy equals the difference between the two levels .✓ 2
Part (b)
E = hc/λ = (6.63×10−34 × 3.00×108)/(589×10−9) = 3.38×10−19 J . Dividing by 1.60×10−19 gives E = 2.11 eV .✓ 2
Part (c)
Each element has its own unique set of energy levels, so the differences between levels, and hence the photon wavelengths, are different .✓ 1

Exam tip: Convert nm to m before using hc/λ, then divide by e to get eV. Carry at least three significant figures through the calculation.

69IBP2S-035-02-12Rutherford scattering and the nuclear radius Paper 2Easy5 marks

Household smoke detectors contain americium-241, which emits alpha particles with kinetic energy 5.49 MeV. A student considers one of these alpha particles travelling directly towards a tin nucleus, \(^{120}_{50}\text{Sn}\), in the solder of the detector's circuit board. Assume the tin nucleus stays at rest.
k = 8.99×109 N m2 C−2, e = 1.60×10−19 C.

(a)

Outline the energy changes as the alpha particle approaches the nucleus until it stops momentarily.

(2)
(b)

Calculate the distance of closest approach.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The kinetic energy of the alpha particle decreases as it is transferred to electric potential energy . At the point of closest approach the kinetic energy is zero, so all of the initial kinetic energy is stored as electric potential energy .✓ 2
Part (b)
Ek = 5.49×106 × 1.60×10−19 = 8.78×10−13 J .✓ 1
Using Ek = k(2e)(50e)/d gives d = 8.99×109 × 100 × (1.60×10−19)2 / 8.78×10−13 .✓ 1
d = 2.6×10−14 m .✓ 1

Common error: Use 2e for the alpha particle and Ze for the nucleus. Using e for the alpha particle, or the nucleon number instead of the proton number, are both common slips.

70IBP3H-E1-20r13-09Excitation of atoms in a fluorescent lamp Paper 2Easy6 marks

In a fluorescent lamp, a low-pressure mercury vapour is excited by electrons that travel along the tube. The excited mercury atoms emit ultraviolet radiation of wavelength 254 nm. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

Outline how collisions with free electrons cause mercury atoms to emit ultraviolet photons.

(2)
(b)

The inside of the tube has a coating that emits visible light when it absorbs the ultraviolet radiation. Explain why the photons emitted by the coating have less energy than the photons absorbed.

(2)
(c)

Calculate the energy, in eV, of a photon of wavelength 254 nm. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
an electron colliding with a mercury atom transfers energy, raising an atomic electron to a higher energy level✓ 1
the atom returns to a lower level and emits a photon with energy equal to the difference between the levels✓ 1
Part (b)
the coating atoms absorb UV photons and are excited, then return to the ground state in more than one step / lose some energy as thermal energy✓ 1
so each photon emitted has less energy than the absorbed photon (hence longer wavelength, visible)✓ 1
Part (c)
M1 \(E=hc/\lambda=\frac{6.63\times10^{-34}\times3.00\times10^{8}}{254\times10^{-9}}=7.83\times10^{-19}\) J✓ 1
A1 divide by \(1.60\times10^{-19}\): 4.89 eV✓ 1

Common error: Leaving a photon energy in joules when the question asks for eV, or the reverse. Convert with 1 eV = 1.60 × 10⁻¹⁹ J.

71IBP3H-E1-20r13-10Energy levels and spectra Paper 2Easy6 marks

A research greenhouse uses a vapour lamp. Atoms in the vapour have the following energy levels.
n = 1: −10.0 eV; n = 2: −4.5 eV; n = 3: −2.0 eV; n = 4: −1.0 eV. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

energyn = 1−10.0 eVn = 2−4.5 eVn = 3−2.0 eVn = 4−1.0 eV
Energy levels not to scale
(a)

State the ionisation energy of an atom in the ground state.

(1)
(b)

A photon of energy 6.0 eV is incident on atoms in the ground state. Explain why it is not absorbed.

(2)
(c)

Determine the longest wavelength that can be emitted by these atoms, in μm. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
10.0 eV✓ 1
Part (b)
the possible gaps from n = 1 are 5.5, 8.0 and 9.0 eV, and 6.0 eV matches none of them✓ 1
6.0 eV is below the ionisation energy of 10.0 eV so the atom cannot be ionised either✓ 1
Part (c)
M1 the smallest gap is between n = 4 and n = 3: \(-1.0-(-2.0)=1.0\) eV✓ 1
M1 \(\lambda=hc/\Delta E=\frac{6.63\times10^{-34}\times3.00\times10^{8}}{1.0\times1.60\times10^{-19}}\)✓ 1
A1 \(1.24\times10^{-6}\) m = 1.24 μm✓ 1

Exam tip: The longest wavelength always comes from the smallest energy gap, which is usually between the two highest levels.

72IBP3H-E1-20r13-16Emission and absorption spectra Paper 2Easy6 marks

Aircraft obstruction beacons on a hill use neon lamps that emit red light of wavelength 640 nm. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

State what is meant by a line emission spectrum.

(1)
(b)

Explain why a neon lamp gives a line spectrum whereas a hot filament lamp gives a continuous spectrum.

(2)
(c)

Calculate the frequency of the light. Give your answer to 3 significant figures.

(1)
(d)

Calculate the energy difference, in eV, between the two levels involved. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
a spectrum consisting of discrete (bright) lines at particular wavelengths on a dark background✓ 1
Part (b)
in a neon gas the atoms have discrete energy levels, so photons are emitted only with energies equal to the gaps between levels✓ 1
in a hot solid the interacting atoms emit a continuous range of photon energies✓ 1
Part (c)
A1 \(f=c/\lambda=4.69\times10^{14}\) Hz✓ 1
Part (d)
M1 \(E=hf\) or \(hc/\lambda=3.11\times10^{-19}\) J✓ 1
A1 1.94 eV✓ 1

Common error: Giving the wavelength in nanometres to the equation without converting to metres. Use 640 × 10⁻⁹ m.

73IBP3H-E1-20r14-09Excitation of atoms in a fluorescent lamp Paper 2Easy7 marks

A mercury-vapour lamp is used to disinfect a water supply. One strong emission line from the vapour has wavelength 254 nm. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.

(a)

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

(3)
(b)

The photon is emitted when an atom makes a transition from the energy level at −1.60 eV to a lower level. Determine the energy of the lower level.

(2)
(c)

Explain why the emission spectrum of the vapour contains only certain discrete wavelengths.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = hc/λ = 6.63×10−34 × 3.00×108 / 2.54×10−7✓ 1
7.83×10−19 J✓ 1
4.89 eV✓ 1
Part (b)
lower level = −1.60 − 4.89 (ECF from (a))✓ 1
−6.49 eV✓ 1
Part (c)
the atoms have discrete energy levels✓ 1
each photon's energy equals the difference between two levels, so only certain photon energies (wavelengths) occur✓ 1

Exam tip: Convert joules to electronvolts by dividing by 1.60×10⁻¹⁹; keep the unrounded value for later steps.

74IBP3H-E1-20r14-10Emission and absorption spectra Paper 2Easy7 marks

In a pharmacy laboratory the sodium content of a rehydration solution is checked using atomic absorption. A vapourised sample is illuminated with white light and the transmitted light shows a dark line at 589 nm. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.

(a)

Outline how a dark line arises in the spectrum of the transmitted light.

(3)
(b)

Calculate the energy gap, in eV, corresponding to the dark line.

(2)
(c)

The emission spectrum of sodium vapour contains a bright line at the same wavelength. Explain this.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
atoms absorb photons whose energy equals the gap between two energy levels (electrons are excited)✓ 1
the excited atoms re-emit photons in random directions✓ 1
so the intensity of that wavelength in the original direction is reduced✓ 1
Part (b)
E = hc/λ = 3.38×10−19 J✓ 1
2.11 eV✓ 1
Part (c)
electrons de-excite from the higher level to the lower level✓ 1
the photon emitted has energy equal to the same gap (2.11 eV), so the same wavelength✓ 1

Common error: A dark line in an absorption spectrum sits at exactly the same wavelength as a bright line in the emission spectrum of the same element.

75E-2-09The hydrogen spectrum and a diffraction grating Paper 2Medium10 marks

Light from a hydrogen discharge tube is incident normally on a diffraction grating that has 600 lines per millimetre. The diagram shows some of the energy levels of the hydrogen atom, En = −13.6/n² eV.

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

n = ∞0 eVn = 6−0.38 eVn = 5−0.54 eVn = 4−0.85 eVn = 3−1.51 eVn = 2−3.40 eVn = 1−13.6 eV(gap not to scale)
Some energy levels of the hydrogen atom (not to scale)
(a)
(i)

A violet line is observed in the first order at an angle of 15.1° to the straight-through direction. Determine its wavelength.

(2)
(ii)

Identify the transition that produces this line.

(2)
(b)
(i)

Determine the highest order in which the red line produced by the transition n = 3 → n = 2 can be observed.

(3)
(ii)

The lines produced by transitions that end on n = 2 crowd together towards a limiting wavelength. Determine the angle at which this limit occurs in the first order.

(2)
(c)
(i)

Explain why hydrogen gas at room temperature does not absorb light of the wavelength found in (a)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
d = 1/600 mm = 1.67 × 10−6 m✓ 1
λ = d sin 15.1° = 4.34 × 10−7 m✓ 1Accept 434 nm.
Part (a)(ii)
E = hc/λ = 2.86 eV✓ 1Allow ECF from (a)(i).
This matches n = 5 → n = 2: −0.54 − (−3.40) = 2.86 eV✓ 1The direction (downwards, ending on n = 2) is required.
Part (b)(i)
ΔE = 3.40 − 1.51 = 1.89 eV, so λ = hc/ΔE = 658 nm✓ 1Accept 656–658 nm.
sin θ ≤ 1, so n ≤ d/λ = 2.53✓ 1Allow ECF from (a)(i) for d.
The highest order is 2✓ 1MP3 needs the integer below d/λ; "3" or "2.5" scores [2] max.
Part (b)(ii)
Limit: n = ∞ → n = 2, ΔE = 3.40 eV, λ = 366 nm✓ 1
θ = sin−1(3.66 × 10−7/1.67 × 10−6) = 12.7°✓ 1Allow ECF from (a)(i). Accept 12.6–12.7°.
Part (c)(i)
Almost all the atoms are in the ground state (n = 1); a 2.86 eV photon could only be absorbed by an atom already in n = 2, as no transition from n = 1 has this energy✓ 1Allow ECF from (a)(ii).

Answers: (a)(i) 434 nm  ·  (a)(ii) n = 5 → n = 2  ·  (b)(i) 2  ·  (b)(ii) 12.7° (the remaining parts are explanations — see the table above)

76E-2-17Excitation of atoms in a fluorescent lamp Paper 2Medium9 marks

In a fluorescent lamp, electrons accelerated by an electric field collide with atoms in low-pressure mercury vapour. The diagram shows some of the energy levels of the mercury atom.

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

0 eV (ionisation)−2.71 eV−4.98 eV−5.55 eV−5.77 eV−10.44 eV (ground state)(gap not to scale)
Some energy levels of the mercury atom (not to scale)
(a)
(i)

Show that the photon emitted in the transition from the −5.55 eV level to the ground state has a wavelength of about 254 nm.

(1)
(ii)

The lamp also emits green light of wavelength 546 nm. Identify, using the diagram, the transition responsible.

(2)
(b)
(i)

Determine the minimum distance that an electron, starting from rest, must travel along an electric field of strength 600 V m−1 before it can excite a mercury atom in its ground state.

(2)
(ii)

An electron with kinetic energy 5.2 eV can excite a ground-state mercury atom to the −5.55 eV level, but a photon of energy 5.2 eV cannot. Explain this difference.

(2)
(c)
(i)

The inside of the lamp is coated with a phosphor that absorbs the 254 nm photons and emits visible photons. Deduce the maximum efficiency with which ultraviolet energy is converted into light if each 254 nm photon produces one photon of wavelength 546 nm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/(4.89 × 1.60 × 10−19) = 2.54 × 10−7 m✓ 1Must see 4.89 eV and 254.2 nm (or the full substitution).
Part (a)(ii)
E = hc/λ = 3.64 × 10−19 J = 2.28 eV✓ 1
From −2.71 eV to −4.98 eV (difference 2.27 eV)✓ 1Direction required.
Part (b)(i)
It must gain at least the energy to reach the lowest excited level: 10.44 − 5.77 = 4.67 eV✓ 1Using 4.89 eV (8.2 mm) scores [1] max.
eEd = 4.67 eV, so d = 4.67/600 = 7.8 × 10−3 m✓ 1
Part (b)(ii)
A photon is absorbed completely, so its energy must exactly equal the difference between two levels; 5.2 eV is not such a difference✓ 1
The electron can transfer just 4.89 eV in the collision and keep the remaining 0.31 eV as kinetic energy✓ 1Allow ECF from (a)(i).
Part (c)(i)
Photon energies: 4.89 eV in and 2.27 eV out✓ 1Allow ECF from (a)(i) and (a)(ii).
Efficiency = 2.27/4.89 = 0.46 (46 %)✓ 1ALT: 254/546.

Answers: (a)(ii) −2.71 eV → −4.98 eV  ·  (b)(i) 7.8 × 10−3 m  ·  (c)(i) 46 % (the remaining parts are explanations — see the table above)

77IBP2H-036-00-10The Bohr model Paper 2Medium6 marks

Hydrogen atoms in a stellar atmosphere are modelled using the Bohr model. The radius of the ground-state orbit is 5.29 × 10−11 m. Consider an electron in the n = 3 orbit.
Electron mass = 9.11 × 10−31 kg; h = 6.63 × 10−34 J s.

(a)

State the Bohr condition for the angular momentum of the electron.

(1)
(b)

Calculate the radius of the n = 3 orbit.

(1)
(c)

Determine the speed of the electron in the n = 3 orbit.

(2)
(d)

Calculate the time taken for the electron to complete one orbit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angular momentum is quantised: mvr = nh/2π, where n is a positive integer✓ 1
Part (b)
r = 32 × 5.29 × 10−11 = 4.76 × 10−10 m✓ 1
Part (c)
v = nh/(2πmr) = 3 × 6.63 × 10−34 / (2π × 9.11 × 10−31 × 4.76 × 10−10) ; v = 7.30 × 105 m s−1 . ECF from (b)✓ 2
Part (d)
T = 2πr/v = 2π × 4.76 × 10−10 / 7.30 × 105 ; T = 4.10 × 10−15 s . ECF from (b) and (c)✓ 2

Exam tip: Keep unrounded values from earlier parts when you calculate later ones, to avoid rounding errors in the final answer.

78IBP2H-036-00-11Transitions on an energy level diagram Paper 2Medium8 marks

An engineer is choosing a gas for a new display lamp. The four lowest energy levels of the gas atoms are given in the table.

Level | Energy / eV
E1 | −9.80
E2 | −5.10
E3 | −3.20
E4 | −1.40

Visible light has photon energies from 1.77 eV to 3.10 eV.

energyE₁−9.80 eVE₂−5.10 eVE₃−3.20 eVE₄−1.40 eV
Energy levels not to scale
(a)

State the number of different photon energies that can be emitted by atoms excited to E4.

(1)
(b)

Determine which of these transitions emit visible light.

(3)
(c)

Explain why an atom in its ground state cannot absorb a photon of energy 3.00 eV.

(2)
(d)

Calculate the shortest wavelength emitted by atoms excited to E4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
6✓ 1
Part (b)
Gaps calculated: 4.70, 6.60, 8.40, 1.90, 3.70, 1.80 eV ; E3 → E2 (1.90 eV) is visible ; E4 → E3 (1.80 eV) is visible; all others lie outside 1.77–3.10 eV✓ 3
Part (c)
A photon is absorbed only if its whole energy equals the energy difference between two levels ; no level lies 3.00 eV above E1, because the smallest gap from the ground state is 4.70 eV✓ 2
Part (d)
ΔE = 8.40 eV = 1.344 × 10−18 J ; λ = hc/ΔE = 1.48 × 10−7 m✓ 2

Common error: Convert eV to joules (× 1.60 × 10⁻¹⁹) before using λ = hc/E. Otherwise the answer is wrong by a factor of about 10¹⁹.

79IBP2H-036-00-14Energy levels and spectra Paper 2Medium8 marks

A solar telescope uses a narrow filter centred on the hydrogen-alpha line at 656 nm to image the Sun's outer layers. The energy levels of hydrogen are En = −13.6/n2 eV.
h = 6.63 × 10−34 J s; c = 3.00 × 108 m s−1; e = 1.60 × 10−19 C.

(a)

Outline how dark absorption lines form in the spectrum of the Sun.

(2)
(b)

Calculate the energy, in eV, of a photon of wavelength 656 nm.

(2)
(c)

Explain why hydrogen gas in which almost all atoms are in the ground state shows no hydrogen-alpha absorption line.

(2)
(d)

Show that the hydrogen-alpha line corresponds to a transition between n = 3 and n = 2.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Cooler gas in the outer layers absorbs photons from the continuous spectrum, but only photons whose energy equals an energy-level difference ; the excited atoms re-emit in random directions, so less light at those wavelengths reaches the observer✓ 2
Part (b)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108 / (656 × 10−9) = 3.03 × 10−19 J ; E = 3.03 × 10−19 / 1.60 × 10−19 = 1.90 eV (1.895 eV; accept 1.89 eV from a rounded intermediate value)✓ 2
Part (c)
Absorption of the hydrogen-alpha line requires atoms that are already in the n = 2 level ; ground-state atoms can absorb only photons of at least 10.2 eV, which are ultraviolet✓ 2
Part (d)
ΔE = 13.6 × (1/22 − 1/32) ; ΔE = 1.889 eV ≈ 1.89 eV, which agrees with the photon energy of about 1.90 eV calculated earlier, within rounding✓ 2

Exam tip: In a 'show that' question, give the calculated value to one more significant figure than the target. Then compare it explicitly with the earlier result.

80IBP2H-036-00-16Nuclear radius and density Paper 2Medium6 marks

A neutron star can be roughly modelled as a single giant nucleus with the same density as ordinary nuclear matter. Nuclear radius is R = R0A1/3, with R0 = 1.20 × 10−15 m. The mass of a nucleon is 1.67 × 10−27 kg.

(a)

Show that the density of a nucleus does not depend on its nucleon number A.

(2)
(b)

Calculate the density of nuclear matter.

(2)
(c)

A neutron star has a mass of 2.8 × 1030 kg. Estimate its radius.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Mass = A × mn and volume = (4/3)πR03A ; ρ = Amn/((4/3)πR03A) = mn/((4/3)πR03), in which A cancels✓ 2
Part (b)
ρ = 1.67 × 10−27 / ((4/3)π × (1.20 × 10−15)3) ; ρ = 2.31 × 1017 kg m−3 (accept 2.3 × 1017)✓ 2
Part (c)
V = M/ρ = 2.8 × 1030 / 2.31 × 1017 = 1.2 × 1013 m3 ; r = (3V/4π)1/3 ≈ 1.4 × 104 m, about 14 km . Allow error carried forward from the density calculated in the previous part.✓ 2

Exam tip: For an 'estimate', give an answer to one or two significant figures. A radius of about 14 km is consistent with what is known about real neutron stars.

81IBP2H-036-01-13Excitation of atoms in a fluorescent lamp Paper 2Medium7 marks

A gas X is being tested for use in a new ultraviolet sterilisation lamp. Its lowest four energy levels are:
n = 1 (ground state): −8.20 eV
n = 2: −4.10 eV
n = 3: −2.60 eV
n = 4: −1.30 eV
Use h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.

energyn = 1−8.20 eVn = 2−4.10 eVn = 3−2.60 eVn = 4−1.30 eV
Energy levels not to scale
(a)

Calculate the ionisation energy of X in joules.

(1)
(b)

State the number of different photon wavelengths that can be emitted by atoms excited to n = 4.

(1)
(c)

The lamp emits a strong line at 302 nm. Determine the transition responsible.

(3)
(d)

An electron of kinetic energy 5.00 eV collides with an atom of X in its ground state. Determine the possible kinetic energies of the electron after the collision.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Ionisation energy = 8.20 × 1.60×10−19 = 1.31×10−18 J .✓ 1
Part (b)
6 .✓ 1
Part (c)
E = hc/λ = (6.63×10−34 × 3.00×108)/(302×10−9) = 6.59×10−19 J . This is 6.59×10−19/1.60×10−19 = 4.12 eV . It matches the 4.10 eV gap between n = 2 and n = 1, so the transition is n = 2 → n = 1 .✓ 3
Part (d)
In an elastic collision the electron leaves with 5.00 eV . In an inelastic collision it excites the atom to n = 2 and leaves with 5.00 − 4.10 = 0.90 eV. It cannot excite the atom to n = 3 or higher, because that needs at least 5.60 eV .✓ 2

Exam tip: A colliding electron can give up only part of its kinetic energy, whereas a photon must be absorbed whole. List the elastic outcome and every level the electron has enough energy to reach.

82IBP2H-036-01-15Transitions on an energy level diagram Paper 2Medium9 marks

Images of star-forming nebulae are often taken through filters that pass the red H-alpha line of hydrogen. The hydrogen energy levels are En = −13.6/n2 eV.
h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.

(a)

H-alpha is the longest-wavelength line of the Balmer series (transitions ending on n = 2). Show that its wavelength is about 660 nm.

(2)
(b)

Calculate the shortest possible wavelength in the Balmer series.

(2)
(c)

Explain why the lines of the Balmer series get closer together as the wavelength decreases.

(2)
(d)

The shortest visible wavelength is 400 nm. Determine the highest initial level n that gives a visible Balmer line, and calculate the wavelength of that line.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = 13.6(1/4 − 1/9) = 1.89 eV = 3.02×10−19 J . λ = hc/ΔE = 1.989×10−25/3.02×10−19 = 6.6×10−7 m .✓ 2
Part (b)
This is the series limit, n = ∞ → 2, so ΔE = 3.40 eV = 5.44×10−19 J . λ = 1.989×10−25/5.44×10−19 = 3.66×10−7 m .✓ 2
Part (c)
The energy levels get closer together as n increases, because E ∝ −1/n2 . Successive photon energies therefore differ by smaller and smaller amounts as they approach the series limit of 3.40 eV .✓ 2
Part (d)
A 400 nm photon has energy 4.97×10−19 J = 3.11 eV. A visible line therefore needs 3.40 − 13.6/n2 ≤ 3.11, which gives n2 ≤ 47 . So n = 6 (the n = 7 line is at 398 nm, which is not visible) . For 6 → 2, ΔE = 3.40 − 0.378 = 3.02 eV, so λ = 1.989×10−25/(3.02 × 1.60×10−19) = 411 nm .✓ 3

Exam tip: In 'highest n' problems, write the condition as an inequality, solve it, and then check the integers on either side of the boundary.

83IBP2S-035-00-14Transitions on an energy level diagram Paper 2Medium7 marks

In a laboratory plasma experiment, some hydrogen atoms are in excited states. Infrared light of wavelength 1064 nm from a Nd:YAG laser, of the kind used in surgery, is shone on the gas. The Bohr model gives the energy levels of hydrogen as En = −13.6/n2 eV.
(h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)

(a)

Calculate the ionization energy of a hydrogen atom in the n = 3 level.

(1)
(b)

Determine whether a single laser photon can ionize a hydrogen atom in the n = 3 level.

(3)
(c)

Determine the lowest level from which a laser photon could ionize an atom, and the kinetic energy of the electron released from this level.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
13.6/32 = 1.51 eV✓ 1
Part (b)
E = hc/λ = 1.989×10−25/1064×10−9 = 1.87×10−19 J✓ 1
= 1.17 eV✓ 1
1.17 eV < 1.51 eV, so the photon cannot ionize the atom✓ 1
ECF from (a)—
Part (c)
13.6/n2 ≤ 1.17 gives n2 ≥ 11.6, so n = 4✓ 1
Ionization energy from n = 4 = 13.6/16 = 0.85 eV✓ 1
Kinetic energy = 1.17 − 0.85 = 0.32 eV✓ 1
ECF from (b)—

Common error: n must be a whole number. Round n up, not to the nearest integer, so that the photon has enough energy.

84IBP2S-035-02-13Rutherford scattering and the nuclear radius Paper 2Medium7 marks

Engineers designing a small ion-beam facility want to know the minimum kinetic energy an alpha particle needs to just make contact with a lead-208 nucleus, \(^{208}_{82}\text{Pb}\), in a head-on approach. Treat both the alpha particle and the lead nucleus as spheres. Contact occurs when the separation of their centres equals the sum of their radii. Assume the lead nucleus does not recoil.
R0 = 1.20 fm, k = 8.99×109 N m2 C−2, e = 1.60×10−19 C, 1 MeV = 1.60×10−13 J.

(a)

Calculate the radius of the lead-208 nucleus.

(1)
(b)

Show that the radius of an alpha particle is about 1.9 fm.

(1)
(c)

Determine, in MeV, the minimum kinetic energy of the alpha particle needed for contact.

(3)
(d)

Explain why alpha particles with kinetic energies greater than this value are not scattered as the Rutherford model predicts.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = 1.20 × 2081/3 = 7.11 fm .✓ 1
Part (b)
R = 1.20 × 41/3 = 1.20 × 1.587 = 1.90 fm, which is about 1.9 fm .✓ 1
Part (c)
Separation at contact: d = 7.11 + 1.90 = 9.01 fm = 9.01×10−15 m. Allow ECF from (a) and (b) .✓ 1
Equate the initial kinetic energy to the electric potential energy at contact: E = k(2e)(82e)/d = 8.99×109 × 164 × (1.60×10−19)2 / 9.01×10−15 = 4.19×10−12 J .✓ 1
E = 4.19×10−12 / 1.60×10−13 = 26.2 MeV (accept 26 MeV) .✓ 1
Part (d)
The alpha particles now reach the nucleus, where the short-range strong nuclear force acts in addition to the Coulomb force . The Rutherford model assumes that only the Coulomb force acts, so the observed scattering deviates from its prediction .✓ 2

Exam tip: Convert femtometres to metres (×10⁻¹⁵) before substituting into k q₁q₂/d. Convert joules to MeV only at the final step.

85IBP2S-035-03-14Nuclear radius and density Paper 2Medium7 marks

Fuel rods for nuclear power stations contain mostly 23892U. The radius R of a nucleus of nucleon number A is given by R = R0A1/3, where R0 = 1.20 × 10−15 m.
1 u = 1.661 × 10−27 kg.

(a)

State the nucleon number and the neutron number of 23892U.

(2)
(b)

Calculate the radius of a 23892U nucleus.

(2)
(c)

Determine the density of the 23892U nucleus.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Nucleon number = 238✓ 1
Neutron number = 238 − 92 = 146✓ 1
Part (b)
R = 1.20 × 10−15 × 2381/3✓ 1
R = 7.44 × 10−15 m✓ 1
Part (c)
Mass ≈ 238 u = 238 × 1.661 × 10−27 = 3.95 × 10−25 kg✓ 1
Volume = (4/3)πR3 = 1.72 × 10−42 m3; allow error carried forward from the radius calculated in the previous part✓ 1
Density = 3.95 × 10−25 / 1.72 × 10−42 = 2.3 × 1017 kg m−3✓ 1

Exam tip: Because R³ is proportional to A, nuclear density is roughly the same for every nucleus. An answer far from about 10¹⁷ kg m⁻³ signals an arithmetic slip.

86IBP3H-E1-20r13-11Transitions on an energy level diagram Paper 2Medium8 marks

A technician at a concert hall tests a hydrogen discharge tube for a lighting demonstration. The energy levels of hydrogen are given by \(E_n=-\frac{13.6}{n^2}\) eV. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

Show that the energy of the n = 3 level is −1.51 eV.

(1)
(b)

Calculate the wavelength, in nm, of the photon emitted in the transition n = 3 to n = 2. Give your answer to 3 significant figures.

(3)
(c)

Calculate the wavelength, in μm, of the photon emitted in the transition n = 4 to n = 3. Give your answer to 3 significant figures.

(2)
(d)

Determine the shortest wavelength, in nm, in the series of lines that end on n = 2. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(-13.6/9=-1.51\) eV✓ 1
Part (b)
M1 \(\Delta E=-1.51-(-3.40)=1.89\) eV✓ 1
M1 \(\lambda=hc/\Delta E\) with \(\Delta E=1.89\times1.60\times10^{-19}\) J✓ 1
A1 658 nm✓ 1
Part (c)
M1 \(\Delta E=13.6\left(\frac19-\frac1{16}\right)=0.661\) eV and \(\lambda=hc/\Delta E\)✓ 1
A1 1.88 μm✓ 1
Part (d)
M1 the shortest wavelength comes from a transition from \(n=\infty\) (E = 0) to n = 2, so \(\Delta E=3.40\) eV✓ 1
A1 \(\lambda=\frac{hc}{3.40\times1.60\times10^{-19}}=366\) nm✓ 1

Exam tip: The series limit (n = ∞) gives the largest photon energy and so the shortest wavelength in a series.

87IBP3H-E1-20r13-12Energy levels and spectra Paper 2Medium7 marks

An airborne instrument passes light from a lamp that emits all visible wavelengths through a cell of cool hydrogen gas. The transmitted light is analysed and dark lines are seen at certain wavelengths. The energy levels of hydrogen are given by \(E_n=-\frac{13.6}{n^2}\) eV. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

Explain why dark lines are seen in the transmitted spectrum.

(3)
(b)

Calculate the energy, in eV, absorbed by an atom in a transition from n = 2 to n = 5. Give your answer to 3 significant figures.

(2)
(c)

Calculate the wavelength, in nm, of the photon absorbed in this transition. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
photons whose energy equals the gap between two energy levels of hydrogen are absorbed✓ 1
the atoms are excited to a higher level✓ 1
the atoms re-emit photons in random directions, so the intensity of these wavelengths in the transmitted beam is reduced✓ 1
Part (b)
M1 \(\Delta E=13.6\left(\frac14-\frac1{25}\right)\) or \(3.40-0.544\)✓ 1
A1 2.86 eV✓ 1
Part (c)
M1 \(\lambda=hc/\Delta E\) with \(\Delta E=2.86\times1.60\times10^{-19}\) J✓ 1
A1 435 nm✓ 1

Common error: Treating dark lines as places where no photons exist at all. Photons are absorbed and then re-emitted in random directions, so the beam is only reduced at those wavelengths.

88IBP3H-E1-20r13-14The Bohr model Paper 2Medium9 marks

In the Bohr model of hydrogen, the radius of the nth orbit is \(r_n=n^2r_1\), where \(r_1=5.29\times10^{-11}\) m, and the angular momentum of the electron is \(\frac{nh}{2\pi}\). Consider the electron in the n = 3 orbit. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

Calculate the radius of the n = 3 orbit. Give your answer to 3 significant figures.

(2)
(b)

Calculate the orbital speed of the electron. Give your answer to 3 significant figures.

(3)
(c)

Calculate the de Broglie wavelength of the electron in this orbit, in nm. Give your answer to 3 significant figures.

(2)
(d)

Use your answers to show how the electron in the n = 3 orbit is consistent with a standing-wave picture.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r_3=3^2\times5.29\times10^{-11}\)✓ 1
A1 \(4.76\times10^{-10}\) m✓ 1
Part (b)
M1 use \(m_evr=\frac{nh}{2\pi}\)✓ 1
M1 \(v=\frac{3\times6.63\times10^{-34}}{2\pi\times9.11\times10^{-31}\times4.76\times10^{-10}}\)✓ 1
A1 \(7.30\times10^{5}\) m s−1✓ 1
Part (c)
M1 \(\lambda=h/(m_ev)=\frac{6.63\times10^{-34}}{9.11\times10^{-31}\times7.30\times10^{5}}\)✓ 1
A1 0.997 nm✓ 1
Part (d)
circumference \(2\pi r_3=2.99\times10^{-9}\) m✓ 1
this equals \(3\lambda\) (\(3\times0.997\) nm), so a whole number of wavelengths (n = 3) fits round the orbit✓ 1

Exam tip: Both the Bohr condition \(mvr=\frac{nh}{2\pi}\) and the standing-wave condition \(2\pi r=n\lambda\) say the same thing once \(\lambda=h/mv\) is used.

89IBP3H-E1-20r14-11Nuclear radius and closest approach Paper 2Medium10 marks

Alpha particles of kinetic energy 7.7 MeV are directed head-on at a thin gold foil. A gold nucleus has atomic number 79 and mass number 197. Ignore recoil of the nucleus. k = 8.99×109 N m2 C−2, e = 1.60×10−19 C; the nuclear radius is R = R0A1/3 with R0 = 1.20×10−15 m.

(a)

Calculate the distance of closest approach of an alpha particle to a gold nucleus.

(3)
(b)

Calculate the radius of a gold nucleus.

(2)
(c)

Deduce whether the alpha particles reach the surface of the nucleus, and state what this implies about the scattering.

(2)
(d)

Determine the kinetic energy, in MeV, that an alpha particle would need to reach a distance of 7.0×10−15 m from the centre of the gold nucleus.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
all kinetic energy becomes electric potential energy: Ek = k(2e)(79e)/d✓ 1
Ek = 7.7×106 × 1.60×10−19 = 1.23×10−12 J✓ 1
d = 2.95×10−14 m✓ 1
Part (b)
R = 1.20×10−15 × 1971/3✓ 1
6.98×10−15 m✓ 1
Part (c)
the closest approach (about 4.2 times the radius) is much larger than the nuclear radius, so the alpha particles do not reach the nucleus✓ 1
scattering is due to the electrostatic (Coulomb) force alone, with no strong-force interaction✓ 1
Part (d)
Ek = k(2e)(79e)/(7.0×10−15)✓ 1
5.19×10−12 J✓ 1
32.5 MeV✓ 1

Exam tip: Use Ek (in joules) = k Q₁Q₂/d and convert MeV to J with 1 MeV = 1.60×10⁻¹³ J.

90IBP3H-E1-20r14-12Rutherford scattering and the nuclear radius Paper 2Medium8 marks

A copper nucleus has mass number 64 and atomic number 29. In a scattering experiment alpha particles are fired head-on at copper nuclei. k = 8.99×109 N m2 C−2, e = 1.60×10−19 C; R0 = 1.20×10−15 m.

(a)

Calculate the radius of the copper nucleus.

(2)
(b)

Calculate the kinetic energy, in MeV, for which the distance of closest approach of an alpha particle equals the radius found in (a).

(3)
(c)

Explain how measuring the scattering of alpha particles at increasing energies can give an estimate of the nuclear radius.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = R0A1/3 = 1.20×10−15 × 641/3✓ 1
4.80×10−15 m✓ 1
Part (b)
Ek = k(2e)(29e)/R (ECF from (a))✓ 1
2.78×10−12 J✓ 1
17.4 MeV✓ 1
Part (c)
at low energies the scattering follows the pattern predicted for the electrostatic force alone✓ 1
when the energy is high enough for the alpha particle to reach the nuclear surface, the strong force acts and the scattering deviates from this pattern✓ 1
the closest approach at the energy where deviation begins gives the (upper limit of the) nuclear radius✓ 1

Exam tip: Closest approach is inversely proportional to kinetic energy: doubling the energy halves the distance.

91IBP3H-E1-20r14-13Rutherford scattering and the nuclear radius Paper 2Medium8 marks

In a physics simulation game, the player must identify an unknown nucleus. An alpha particle of kinetic energy 6.50 MeV is fired head-on at the nucleus and the game reports a distance of closest approach of 2.21×10−14 m. The game ignores recoil. k = 8.99×109 N m2 C−2, e = 1.60×10−19 C.

(a)

Determine the atomic number of the nucleus.

(3)
(b)

The nuclei available in the game are listed as tin (Z = 50), indium (Z = 49) and cadmium (Z = 48). State which is most likely.

(1)
(c)

The player now triples the kinetic energy of the alpha particle. Calculate the new distance of closest approach.

(2)
(d)

The tin nucleus has mass number 119. Show whether the alpha particle at the tripled energy reaches the surface of the nucleus (R0 = 1.20×10−15 m).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = k(2e)(Ze)/Ek so Z = Ekd/(2ke2)✓ 1
Ek = 6.50×106 × 1.60×10−19 = 1.04×10−12 J✓ 1
Z = 49.9 (≈ 50)✓ 1
Part (b)
tin✓ 1
Part (c)
d is inversely proportional to Ek for fixed nucleus✓ 1
d = 2.21×10−14/3 = 7.37×10−15 m✓ 1
Part (d)
R = 1.20×10−15 × 1191/3 = 5.90×10−15 m✓ 1
7.37×10−15 m is greater than R, so the particle does not reach the surface and the electrostatic-only model remains valid✓ 1

Exam tip: Rearrange the formula symbolically for the unknown before substituting numbers.

92IBP3H-E1-20r14-14The Bohr model Paper 2Medium9 marks

In the Bohr model of hydrogen the energy of level n is −13.6/n2 eV, the orbital radius is n2r1 with r1 = 5.29×10−11 m, and the angular momentum of the electron is nh/2π. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C; electron mass 9.11×10−31 kg.

(a)

Calculate the energy of the photon emitted when the electron moves from n = 3 to n = 2, in eV.

(2)
(b)

Calculate the wavelength of this photon.

(2)
(c)

Calculate the angular momentum of the electron in the n = 3 level.

(2)
(d)

Determine the orbital speed of the electron in the n = 3 level. Use the unrounded angular momentum.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 13.6(1/22 − 1/32)✓ 1
1.89 eV✓ 1
Part (b)
λ = hc/E = 6.63×10−34 × 3.00×108 / (1.89 × 1.60×10−19)✓ 1
6.58×10−7 m✓ 1
Part (c)
L = nh/2π = 3 × 6.63×10−34/(2π)✓ 1
3.17×10−34 kg m2 s−1✓ 1
Part (d)
r = 9 × 5.29×10−11 = 4.76×10−10 m✓ 1
L = mvr so v = L/(mr)✓ 1
7.30×105 m s−1 (accept 7.31×105 if rounded L used)✓ 1

Common error: Use the radius of level n (n² times the first radius), not r₁, when finding the speed from mvr = nh/2π.

93IBP3H-E1-20r33-11Emission spectra and chemical composition Paper 2Medium7 marks

A food scientist heats three samples of seasoning in a flame and examines the light with a spectrometer. Reference lines for four elements are shown. Each element is identified here by the single line listed.

Sodium: 589 nm
Potassium: 767 nm
Lithium: 671 nm
Barium: 554 nm

One sample produces lines at 554 nm, 589 nm and 767 nm. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1 and 1 eV = 1.60 × 10−19 J.

(a)

Explain why the wavelengths of the lines in the spectrum can be used to identify the elements in the sample.

(2)
(b)

Identify the elements present in this sample.

(2)
(c)

Calculate the energy, in eV, of a photon of the sodium line. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Each element has its own unique set of energy levels✓ 1
So the photons emitted in transitions between levels have a unique set of energies (and wavelengths) for that element✓ 1
Part (b)
A1 sodium and potassium✓ 1
A1 barium (and no lithium)✓ 1
Part (c)
M1 use \(E=\dfrac{hc}{\lambda}\)✓ 1
A1 \(E=3.38\times10^{-19}\) J✓ 1
A1 \(2.11\) eV✓ 1

Common error: A line spectrum is a property of the element's energy levels, not of how hot the flame is; the brightness may change but the wavelengths do not.

94IBP3H-E1-20r33-14The hydrogen spectrum and a diffraction grating Paper 2Medium8 marks

A hydrogen discharge lamp is observed with a diffraction grating that has 500 lines per millimetre. Light from the lamp is incident normally on the grating. The energy levels of hydrogen are given by \(E_n=-\dfrac{13.6}{n^{2}}\) eV. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1 and 1 eV = 1.60 × 10−19 J.

lampgrating500 lines per mmθ = ?central maximumfirst-order maximum
Diagram not to scale
(a)

Calculate the wavelength of the light emitted in the transition from n = 3 to n = 2. Give your answer to 3 significant figures.

(3)
(b)

Determine the highest order of maximum that can be observed for this wavelength.

(2)
(c)

Calculate the angle between the central maximum and the first-order maximum for this wavelength. Give your answer to 3 significant figures.

(2)
(d)

State the appearance of the central (zero-order) maximum when light from the lamp contains several wavelengths.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\Delta E=13.6\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.89\) eV✓ 1
M1 convert to joules and use \(\lambda=\dfrac{hc}{\Delta E}\)✓ 1
A1 \(\lambda=6.58\times10^{-7}\) m✓ 1
Part (b)
M1 slit spacing \(d=\dfrac{1}{500\times10^{3}}=2.00\times10^{-6}\) m and \(n\lambda=d\sin\theta\) with \(\sin\theta\le1\), so \(n\le\dfrac{d}{\lambda}=3.04\)✓ 1
A1 highest order is 3✓ 1
Part (c)
M1 \(\sin\theta=\dfrac{\lambda}{d}=\dfrac{6.58\times10^{-7}}{2.00\times10^{-6}}\)✓ 1
A1 \(\theta=19.2^{\circ}\)✓ 1
Part (d)
All wavelengths are in the same direction, so the central maximum shows the combined colour of all the lines (not separated into colours)✓ 1

Common error: Slit spacing is the reciprocal of the number of lines per metre, so convert lines per millimetre to lines per metre first.

95E-2-01The Bohr model and muonic hydrogen Paper 2Hard10 marks

In the Bohr model of hydrogen, a particle of charge −e and mass m moves with speed v in a circular orbit of radius r about a stationary proton. Its angular momentum is quantized: mvr = nh/2π. The total energy of the particle in orbit n is En = −e²/(8πε0rn). For ordinary hydrogen E1 = −13.6 eV.

In muonic hydrogen the electron is replaced by a muon, a particle with the same charge as the electron and a mass of 207 me.

(h = 6.63 × 10−34 J s, ε0 = 8.85 × 10−12 C² N−1 m−2, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C, c = 3.00 × 108 m s−1)

(a)
(i)

Show that the radius of orbit n is rn = n²h²ε0/(πme²).

(2)
(ii)

Deduce, without substituting numerical values, the ratio E1(muonic hydrogen)/E1(hydrogen).

(2)
(b)
(i)

Determine the energy, in keV, of the photon emitted when muonic hydrogen makes the transition from n = 2 to n = 1.

(2)
(ii)

Identify the region of the electromagnetic spectrum in which this photon lies.

(1)
(c)
(i)

Using (a)(i), calculate the radius of the n = 1 orbit in muonic hydrogen and compare it with the radius of the proton, 8.4 × 10−16 m.

(2)
(ii)

Suggest why the energy levels of muonic hydrogen are much more sensitive to the size of the proton than those of ordinary hydrogen.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The electric force provides the centripetal force: e²/(4πε0r²) = mv²/r✓ 1Both steps must be seen: the answer is given.
Substituting v = nh/(2πmr): e²/(4πε0r²) = n²h²/(4π²mr³), so r = n²h²ε0/(πme²)✓ 1
Part (a)(ii)
rn ∝ 1/m, so the muon's orbit is 207 times smaller than the electron's✓ 1Allow ECF from (a)(i).
En ∝ −1/rn ∝ m, so the ratio is 207 (the muonic level is 207 times deeper)✓ 1A ratio of 1/207 scores [1] if MP1 is seen.
Part (b)(i)
En = −207 × 13.6/n² eV, so ΔE = 207 × 13.6 × (1 − ¼) eV✓ 1Allow ECF from (a)(ii).
ΔE = 2111 eV = 2.1 keV✓ 1Accept 2.11 keV.
Part (b)(ii)
λ = hc/ΔE = 5.9 × 10−10 m: X-ray✓ 1Allow ECF from (b)(i). The wavelength, or a comparison with visible photon energies of about 2–3 eV, must be seen.
Part (c)(i)
r1 = h²ε0/(π × 207me × e²) = 2.6 × 10−13 m✓ 1Allow ECF from (a)(i). ALT: 5.3 × 10−11 m/207.
About 310 times the radius of the proton✓ 1Accept 290–310.
Part (c)(ii)
The muon is on average 207 times closer to the proton, so the proton's size is a far larger fraction of the orbit; the departure of the force from the point-charge (kq1q2/r²) form near the proton therefore changes the muon's energy much more✓ 1Allow ECF from (c)(i). Accept any answer that links the much smaller orbit to a larger effect of the proton's finite size.

Answers: (a)(ii) 207  ·  (b)(i) 2.1 keV  ·  (b)(ii) X-ray (5.9 × 10−10 m)  ·  (c)(i) 2.6 × 10−13 m (the remaining parts are explanations — see the table above)

96E-2-18Rutherford scattering and the nuclear radius Paper 2Hard14 marks

Alpha particles (42He) from a particle accelerator are scattered through angles close to 180° by a thin foil of tin, 12050Sn. For each beam energy, the measured count is divided by the count predicted by Rutherford's model, in which only the electric force acts. The graph shows this ratio against the kinetic energy of the alpha particles. Treat each scattering as a head-on collision and ignore the recoil of the tin nucleus.

(k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, R0 = 1.20 × 10−15 m, 1 u = 1.66 × 10−27 kg)

1012141618202224262830alpha-particle kinetic energy / MeV0.00.20.40.60.81.01.2measured / RutherfordRutherford
Graph drawn to scale. The dashed line is the Rutherford prediction.
(a)

Closest approach.

(i)

Show that the distance of closest approach of a 10.0 MeV alpha particle to a tin nucleus is about 1.4 × 10−14 m.

(2)
(b)

Deviations from Rutherford scattering.

(i)

Estimate, from the graph, the energy E0 above which the results deviate from the Rutherford prediction.

(1)
(ii)

Determine the distance of closest approach at E0.

(1)
(iii)

Explain why the measured count falls below the Rutherford prediction at energies above E0.

(2)
(c)

Nuclear radius.

(i)

Calculate the radius of a tin-120 nucleus and of an alpha particle using R = R0A1/3.

(2)
(ii)

Compare your answers to (b)(ii) and (c)(i) and comment on what the comparison suggests.

(2)
(d)

Nuclear density.

(i)

The radius of a tin atom is about 1.4 × 10−10 m. Calculate the ratio (density of the tin nucleus)/(mean density of a tin atom).

(2)
(ii)

Hence explain why most alpha particles pass straight through a thin foil, while a very few are scattered through large angles.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Kinetic energy = electric potential energy at closest approach: Ek = k(2e)(50e)/d✓ 1
d = 8.99 × 109 × 100 × (1.60 × 10−19)²/(10.0 × 106 × 1.60 × 10−19) = 1.44 × 10−14 m✓ 1Energy must be converted to joules.
Part (b)(i)
E0 ≈ 18.4 MeV (where the curve leaves the ratio 1)✓ 1Accept 18–19 MeV.
Part (b)(ii)
d ∝ 1/Ek: d = 1.44 × 10−14 × 10.0/18.4 = 7.8 × 10−15 m✓ 1Accept 7.5–8.0 × 10−15 m (for 18–19 MeV); ECF from (b)(i).
Part (b)(iii)
Above E0 the alpha particle gets close enough to touch the nucleus✓ 1
The (attractive, short-range) strong nuclear force then acts as well as the electric force / alpha particles are absorbed in nuclear reactions, so fewer are scattered back than Rutherford's electric-only model predicts✓ 1
Part (c)(i)
RSn = 1.20 × 10−15 × 1201/3 = 5.9 × 10−15 m✓ 1
Rα = 1.20 × 10−15 × 41/3 = 1.9 × 10−15 m✓ 1
Part (c)(ii)
The sum of the radii, 7.8 × 10−15 m, agrees closely with the distance of closest approach at E0✓ 1ECF.
This supports R = R0A1/3: deviations begin when the surfaces of the two nuclei meet, so the onset of deviation measures nuclear size✓ 1
Part (d)(i)
The nucleus and the atom have (almost) the same mass, since the electrons' mass is negligible, so the density ratio = (ratom/RSn)³✓ 1
Ratio = (1.4 × 10−10/5.92 × 10−15)³ = 1.3 × 1013✓ 1Accept 1.2–1.4 × 1013; ECF from (c)(i).
Part (d)(ii)
Almost all of the mass is concentrated in a nucleus occupying only ~10−13 of the atom's volume; the rest is empty space containing only light electrons, so most alpha particles never pass close to a nucleus and are hardly deflected✓ 1
The rare alpha particle that passes very close to the dense, massive, positive nucleus experiences a very large repulsive force and is deflected through a large angle✓ 1

Answers: (a)(i) 1.4 × 10−14 m  ·  (b)(i) ≈ 18.4 MeV  ·  (b)(ii) 7.8 × 10−15 m  ·  (c)(i) 5.9 fm, 1.9 fm  ·  (d)(i) 1.3 × 1013 (the remaining parts are explanations — see the table above)

97E-2-21The Bohr model Paper 2Hard13 marks

Radio telescopes detect a spectral line of frequency close to 5.0 GHz from clouds of ionised hydrogen in the Milky Way. The line is emitted when a hydrogen atom, formed with its electron in a very high level, makes the transition from n = 110 to n = 109. (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)

(a)

The line.

(i)

Show, using the Bohr model, that the energy of the photon emitted is about 2.1 × 10−5 eV.

(2)
(ii)

Calculate the frequency and the wavelength of this photon.

(2)
(b)

A giant atom.

(i)

In the Bohr model the orbital radius is rn = n²r1, where r1 = 5.29 × 10−11 m. Calculate the radius of the n = 110 orbit.

(1)
(ii)

Use the Bohr condition for angular momentum to calculate the speed of the electron in the n = 110 orbit.

(2)
(iii)

Explain how, in the Bohr model, the quantization of angular momentum leads to discrete energy levels.

(2)
(c)

Where the line is seen.

(i)

The number density of atoms in the cloud is about 1 × 109 m−3. Estimate the mean spacing between atoms.

(1)
(ii)

In a laboratory hydrogen discharge tube the number density is about 1 × 1022 m−3. Suggest why this line is observed from the cloud but not from the discharge tube.

(2)
(d)

Composition.

(i)

State why detecting a line at exactly this frequency shows that the cloud contains hydrogen.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ΔE = 13.6(1/109² − 1/110²) eV✓ 1
= 13.6 × 1.52 × 10−6 = 2.07 × 10−5 eV✓ 1The difference must be shown to at least 3 s.f. before rounding.
Part (a)(ii)
f = ΔE/h = 2.07 × 10−5 × 1.60 × 10−19/6.63 × 10−34 = 5.0 × 109 Hz✓ 1
λ = c/f = 0.060 m (6.0 cm)✓ 1ECF.
Part (b)(i)
r = 110² × 5.29 × 10−11 = 6.4 × 10−7 m✓ 1About 12 000 times larger than a ground-state atom.
Part (b)(ii)
mvr = nh/2π, so v = nh/(2πmer)✓ 1
v = 110 × 6.63 × 10−34/(2π × 9.11 × 10−31 × 6.40 × 10−7) = 2.0 × 104 m s−1✓ 1ECF from (b)(i).
Part (b)(iii)
The angular momentum mvr can only equal whole-number multiples of h/2π; combined with the electric force providing the centripetal force, this allows only certain orbit radii and speeds✓ 1
The kinetic and electric potential energies are therefore fixed for each n, so the total energy can only take discrete values (E = −13.6/n² eV)✓ 1
Part (c)(i)
Spacing ≈ (1 × 109)−1/3 = 1 × 10−3 m✓ 1
Part (c)(ii)
In the tube the spacing is (1 × 1022)−1/3 ≈ 5 × 10−8 m, much smaller than the diameter of the n = 110 atom (≈ 1.3 × 10−6 m)✓ 1
Such a large, weakly bound atom would be disturbed/ionised by collisions with neighbours before it could make the transition; in the cloud the atoms are about 800 atomic diameters apart and are left undisturbed✓ 1
Part (d)(i)
The frequency matches exactly the difference between two hydrogen energy levels; each element has its own unique set of energy levels (and hence spectral lines)✓ 1

Answers: (a)(i) 2.07 × 10−5 eV  ·  (a)(ii) 5.0 × 109 Hz, 6.0 cm  ·  (b)(i) 6.4 × 10−7 m  ·  (b)(ii) 2.0 × 104 m s−1  ·  (c)(i) 1 × 10−3 m (the remaining parts are explanations — see the table above)

98IBP2H-036-00-12Rutherford scattering and the nuclear radius Paper 2Hard9 marks

Rutherford backscattering spectrometry is used in the semiconductor industry to check thin metal films. Alpha particles of kinetic energy 2.0 MeV are fired at a tungsten film. Tungsten nuclei have Z = 74 and A = 184. Ignore recoil of the nucleus.
k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C; R0 = 1.20 × 10−15 m.

(a)

Outline why only a very small fraction of the 2.0 MeV alpha particles are scattered through large angles.

(2)
(b)

Calculate the distance of closest approach for a head-on collision of a 2.0 MeV alpha particle with a tungsten nucleus.

(2)
(c)

Determine the minimum kinetic energy, in MeV, that an alpha particle needs to just reach the surface of a tungsten nucleus. Ignore the size of the alpha particle.

(3)
(d)

Alpha particles with the minimum kinetic energy found in the previous part are now fired at the tungsten film. Explain why their scattering pattern would differ from the pattern predicted by the Rutherford model.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The positive charge and almost all of the mass of the atom are concentrated in a tiny nucleus ; only the few alpha particles passing very close to a nucleus feel a repulsive force large enough to deflect them through large angles✓ 2
Part (b)
Use Ek = k(2e)(74e)/d, so d = 8.99 × 109 × 148 × (1.60 × 10−19)2 / (2.0 × 106 × 1.60 × 10−19) ; d = 1.1 × 10−13 m (1.06 × 10−13 m)✓ 2
Part (c)
R = 1.20 × 10−15 × 1841/3 = 6.8 × 10−15 m ; either E = k(2e)(74e)/R, or E ∝ 1/d giving E = 2.0 MeV × (d / R) using the distance of closest approach calculated above ; E ≈ 31 MeV, and any answer from 31 to 32 MeV is accepted, including values obtained by carrying forward a rounded or incorrect distance of closest approach✓ 3
Part (d)
At this energy the alpha particle reaches the nuclear surface and comes within range of the short-range strong nuclear force ; the interaction is then no longer purely electrostatic, so the Coulomb-only Rutherford prediction no longer applies✓ 2

Exam tip: The alpha particle's charge is 2e. Using e instead halves the calculated distance of closest approach, which is a frequent slip.

99IBP2H-036-00-13The Bohr model and muonic hydrogen Paper 2Hard9 marks

In muonic hydrogen, the electron is replaced by a muon. The muon has the same charge as the electron, but its mass is 1.88 × 10−28 kg. Muonic hydrogen is used to measure the size of the proton. Treat the proton as stationary and use the Bohr model.
k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C; h = 6.63 × 10−34 J s; c = 3.00 × 108 m s−1.

(a)

Show that the radius of orbit n is rn = n2h2/(4π2mke2), where m is the mass of the orbiting particle.

(3)
(b)

Calculate the ground-state radius of muonic hydrogen.

(2)
(c)

The total energy in orbit n is En = −ke2/(2rn). Determine the ground-state energy in eV.

(2)
(d)

Calculate the wavelength of the photon emitted in the n = 2 → n = 1 transition, and state the region of the electromagnetic spectrum it belongs to.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The electric force provides the centripetal force: ke2/r2 = mv2/r ; from the Bohr condition mvr = nh/(2π), so v = nh/(2πmr) ; substituting and rearranging gives r = n2h2/(4π2mke2)✓ 3
Part (b)
r1 = (6.63 × 10−34)2 / (4π2 × 1.88 × 10−28 × 8.99 × 109 × (1.60 × 10−19)2) ; r1 = 2.57 × 10−13 m✓ 2
Part (c)
E1 = −8.99 × 109 × (1.60 × 10−19)2 / (2 × 2.57 × 10−13) = −4.47 × 10−16 J ; E1 = −2.8 × 103 eV . Allow error carried forward from the calculated ground-state radius.✓ 2
Part (d)
ΔE = (1 − 1/4) × 4.47 × 10−16 = 3.35 × 10−16 J, so λ = hc/ΔE = 5.9 × 10−10 m ; the photon is an X-ray . Allow error carried forward from the calculated ground-state energy.✓ 2

Exam tip: Since r ∝ 1/m and E ∝ m, replacing the electron by a particle about 207 times heavier makes the orbit about 207 times smaller and the energies about 207 times larger.

100IBP2H-036-00-15Excitation of atoms in a fluorescent lamp Paper 2Hard8 marks

In a low-pressure hydrogen plasma used to etch microchips, free electrons with kinetic energy 12.9 eV collide with hydrogen atoms in the ground state. The energy levels of hydrogen are En = −13.6/n2 eV.
h = 6.63 × 10−34 J s; c = 3.00 × 108 m s−1; e = 1.60 × 10−19 C.

(a)

Determine the highest energy level that the atoms can reach through these collisions.

(2)
(b)

State the number of different spectral lines that the excited atoms can then emit.

(1)
(c)

Calculate the kinetic energy of an electron after it has excited an atom to the level in (a).

(1)
(d)

Explain why photons of energy 12.9 eV, instead of electrons, would not excite the ground-state atoms at all.

(2)
(e)

Calculate the longest wavelength among the lines in (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E4 − E1 = 13.6 × (1 − 1/16) = 12.75 eV, which is less than 12.9 eV, while E5 − E1 = 13.06 eV, which is more than 12.9 eV ; so the highest level is n = 4✓ 2
Part (b)
6 . ECF from (a)✓ 1
Part (c)
12.9 − 12.75 = 0.15 eV . ECF from (a)✓ 1
Part (d)
A photon is absorbed completely or not at all, so its energy must exactly equal an energy-level difference ; 12.9 eV matches no transition from n = 1, whereas an electron can give up part of its kinetic energy and keep the rest✓ 2
Part (e)
The smallest gap is n = 4 → 3: ΔE = 13.6 × (1/9 − 1/16) = 0.661 eV = 1.06 × 10−19 J ; λ = hc/ΔE = 1.88 × 10−6 m (infrared)✓ 2

Common error: Electrons and photons excite atoms differently. An electron can transfer any amount of energy up to its kinetic energy, but a photon must match a level difference exactly.

101IBP2H-036-01-12The Bohr model Paper 2Hard6 marks

Radio telescopes detect 'recombination lines' from hydrogen in interstellar clouds. These lines are emitted when electrons in very highly excited atoms drop between neighbouring levels, such as n = 110 to n = 109. Use the Bohr model with En = −13.6/n2 eV and a ground-state radius of 5.29×10−11 m.
h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.

(a)

State Bohr's condition for the allowed orbits of the electron.

(1)
(b)

Calculate the radius of the n = 110 orbit.

(1)
(c)

Calculate, in eV, the energy of the photon emitted in the transition from n = 110 to n = 109.

(2)
(d)

Determine the wavelength of this photon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The angular momentum of the electron is quantised: mvr = nh/2π, where n is a positive integer .✓ 1
Part (b)
Since r ∝ n2, r = 1102 × 5.29×10−11 = 6.40×10−7 m .✓ 1
Part (c)
ΔE = 13.6(1/1092 − 1/1102) .✓ 1
ΔE = 2.07×10−5 eV .✓ 1
Part (d)
ΔE = 2.07×10−5 × 1.60×10−19 = 3.31×10−24 J .✓ 1
λ = hc/ΔE = 1.989×10−25/3.31×10−24 = 0.060 m .✓ 1
Allow error carried forward from the photon energy found in the previous part. A wavelength of about 6 cm lies in the radio/microwave region, but this remark is not required for the mark.—

Common error: The two energy terms are almost equal, so subtracting them after rounding can give a wildly wrong result. Keep full precision, or use (110² − 109²)/(109² × 110²).

102IBP2H-036-01-14Rutherford scattering and the nuclear radius Paper 2Hard10 marks

A research group uses alpha particles from a radioactive source, travelling at 1.50×107 m s−1, to probe nuclear sizes. Ignore nuclear recoil and relativistic effects.
Mass of alpha particle = 6.64×10−27 kg, k = 8.99×109 N m2 C−2, e = 1.60×10−19 C, R0 = 1.20×10−15 m, 1 MeV = 1.60×10−13 J.

(a)

Show that the kinetic energy of each alpha particle is about 4.7 MeV.

(2)
(b)

Calculate the distance of closest approach of these alpha particles to a silver nucleus (Z = 47).

(2)
(c)

When higher-energy alpha particles are fired at aluminium (Z = 13, A = 27), the scattering no longer agrees with the predictions of Rutherford's model above a certain energy. Explain this.

(2)
(d)

The disagreement begins at an alpha kinetic energy of 7.0 MeV. Determine the distance of closest approach at this energy.

(2)
(e)

By considering the sizes of both the aluminium nucleus and the alpha particle (A = 4), calculate the separation of their centres when their surfaces just touch. Comment on how this compares with the distance of closest approach at 7.0 MeV.

(2)
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Marking pointMarkNotes
Part (a)
Ek = ½ × 6.64×10−27 × (1.50×107)2 = 7.47×10−13 J . Dividing by 1.60×10−13 J MeV−1 gives 4.67 MeV, which is about 4.7 MeV .✓ 2
Part (b)
Set Ek equal to the electric potential energy at closest approach: d = k(2e)(47e)/Ek . d = 8.99×109 × 94 × (1.60×10−19)2 / 7.47×10−13 = 2.9×10−14 m . Allow a consistent value carried forward from the kinetic energy calculated above.✓ 2
Part (c)
At higher energy the alpha particle gets close enough to reach, or touch, the nucleus . The short-range strong nuclear force then acts in addition to the electrostatic repulsion, so the pure Coulomb prediction no longer applies .✓ 2
Part (d)
d = k × 2 × 13 × e2 / (7.0 × 1.60×10−13) . d = 5.3×10−15 m .✓ 2
Part (e)
Separation = R0(271/3 + 41/3) = 1.20×10−15 × (3.00 + 1.59) = 5.5×10−15 m . This agrees with the 7.0 MeV closest-approach distance to within about 4%, so deviation begins when the surfaces of the two nuclei come into contact . Allow a consistent comment based on the student's own value for the 7.0 MeV distance.✓ 2

Common error: The onset of deviation gives the distance between the centres when the surfaces touch. It is not the radius of the target alone, because the alpha particle has a size of its own.

103IBP2S-035-01-13The Bohr model Paper 2Hard8 marks

Radio telescopes detect radiation from hydrogen in low-density nebulae, emitted when electrons in very highly excited atoms fall from n = 110 to n = 109. Use the Bohr model, in which En = −13.6/n2 eV and the orbital radius is rn = n2a0, where a0 = 5.29×10−11 m.
(h = 6.63×10−34 J s, e = 1.60×10−19 C)

(a)

State the condition on angular momentum that Bohr imposed on the electron's orbit.

(1)
(b)

Calculate the orbital radius for n = 110.

(1)
(c)

Calculate, in eV, the energy of the photon emitted in the transition n = 110 → n = 109.

(2)
(d)

Calculate the frequency of the radiation detected.

(2)
(e)

Explain why this line is observed from nebulae but not from hydrogen at normal laboratory pressures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The angular momentum of the electron is quantised: mvr = nh/(2π), where n is a positive integer .✓ 1
Part (b)
r = 1102 × 5.29×10−11 = 6.40×10−7 m .✓ 1
Part (c)
ΔE = 13.6 × (1/1092 − 1/1102) .✓ 1
ΔE = 2.07×10−5 eV .✓ 1
Part (d)
ΔE = 2.07×10−5 × 1.60×10−19 = 3.31×10−24 J . Allow error carried forward from (c).✓ 1
f = ΔE/h = 3.31×10−24/6.63×10−34 = 5.00×109 Hz .✓ 1
Part (e)
An atom with n = 110 has a radius of about 0.64 μm, so it is about 1.3 μm across. Either size is hundreds of times larger than the typical spacing between particles in a gas at normal pressure, which is a few nm .✓ 1
In a dense gas, collisions would ionise or de-excite such an atom before it could emit. The very low density of a nebula makes collisions rare, so the radiative transition can occur .✓ 1

Common error: For large n, the two energies are very close. Keep at least five significant figures in 1/109² and 1/110², or the subtraction loses most of its accuracy.

104IBP2S-035-00-15The Bohr model Paper 2Hard8 marks

Radio astronomers observe a line from hydrogen atoms in interstellar gas as electrons fall from n = 110 to n = 109. In the Bohr model, the orbit radius is rn = n2a0, where a0 = 5.3×10−11 m, and the energy levels are En = −13.6/n2 eV.
(h = 6.63×10−34 J s, e = 1.60×10−19 C)

(a)

Calculate the radius of the electron orbit for n = 110.

(2)
(b)

Show that the energy of the photon emitted is about 2×10−5 eV.

(2)
(c)

Calculate the frequency of the emitted radiation.

(2)
(d)

Explain why atoms in such high levels are observed only in gas of extremely low density.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
r = 1102 × 5.3×10−11✓ 1
r = 6.4×10−7 m✓ 1
Part (b)
ΔE = 13.6 × (1/1092 − 1/1102)✓ 1
ΔE = 13.6 × (8.4168×10−5 − 8.2645×10−5) = 13.6 × 1.523×10−6 = 2.07×10−5 eV; accept 2.07–2.08×10−5 eV✓ 1
Part (c)
f = ΔE/h = (2.07×10−5 × 1.60×10−19)/6.63×10−34✓ 1
f = 5.0×109 Hz✓ 1
Allow error carried forward from (b).—
Part (d)
The electron is very weakly bound: the ionization energy from n = 110 is only about 13.6/1102 ≈ 1×10−3 eV. Alternatively, the atom is very large (radius ≈ 6×10−7 m), so it presents a large target for collisions. Either point earns this mark.✓ 1
In denser gas, frequent collisions would ionize or de-excite the atom before it could emit the photon. Only very low density gives a long enough time between collisions.✓ 1

Exam tip: For large n, keep enough significant figures in 1/n². The difference is small, and rounding too early can wipe it out or shift the answer.

105IBP2S-035-02-15Rutherford scattering and the nuclear radius Paper 2Hard9 marks

An astrophysics student uses alpha-particle scattering data to estimate the size of a neutron star. Alpha particles are fired head-on at tellurium-125 nuclei, \(^{125}_{52}\text{Te}\). Deviations from Rutherford scattering first appear when the alpha-particle kinetic energy exceeds 21.4 MeV. Assume that, at this energy, the distance of closest approach equals the radius of the tellurium nucleus. Neglect the size of the alpha particle and the recoil of the nucleus.
k = 8.99×109 N m2 C−2, e = 1.60×10−19 C, mu = 1.66×10−27 kg.

(a)

Explain why deviations from Rutherford scattering appear above a certain kinetic energy.

(2)
(b)

Calculate the distance of closest approach for an alpha particle of kinetic energy 21.4 MeV.

(2)
(c)

Determine a value for R0 in R = R0A1/3.

(1)
(d)

Calculate the density of nuclear matter using your value of R0.

(2)
(e)

A neutron star of mass 2.8×1030 kg has an average density equal to that of nuclear matter. Estimate its radius.

(2)
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Marking pointMarkNotes
Part (a)
At higher kinetic energy the alpha particle gets closer to the nucleus, and above this energy it reaches the nuclear surface . The strong nuclear force then acts in addition to the Coulomb force, so the Rutherford (Coulomb-only) prediction no longer holds .✓ 2
Part (b)
Using 21.4×106 × 1.60×10−19 = k(2e)(52e)/d gives d = 8.99×109 × 104 × (1.60×10−19)2 / 3.42×10−12 .✓ 1
d = 7.0×10−15 m .✓ 1
Part (c)
R0 = 7.0×10−15 / 1251/3 = 7.0×10−15 / 5 = 1.4×10−15 m. ECF from (b) .✓ 1
Part (d)
ρ = A mu / (4/3 π R03A) = 3mu / (4πR03), which is independent of A .✓ 1
ρ = 3 × 1.66×10−27 / (4π × (1.40×10−15)3) ≈ 1.4×1017 kg m−3. Accept 1.4–1.5×1017 kg m−3. ECF from (c) .✓ 1
Part (e)
V = M/ρ = 2.8×1030 / 1.45×1017 ≈ 1.9×1013 m3, and r = (3V/4π)1/3. Equivalently, r = R0(M/mu)1/3 .✓ 1
r ≈ 1.7×104 m, about 17 km. ECF from (d) .✓ 1

Exam tip: Carry unrounded values from earlier parts through the working. Rounding R₀ too early and then cubing it noticeably shifts the density and the final radius.

106IBP2S-035-02-16Rutherford scattering and the nuclear radius Paper 2Hard9 marks

A materials engineer uses ion-beam analysis to measure a carbon coating on a medical implant. Alpha particles of kinetic energy 2.0 MeV are directed at the coating. Consider one alpha particle moving directly towards a carbon-12 nucleus, \(^{12}_{6}\text{C}\), which is initially at rest and free to recoil. Take the masses of the alpha particle and the carbon nucleus as 4u and 12u.
k = 8.99×109 N m2 C−2, e = 1.60×10−19 C.

(a)

Calculate the distance of closest approach, assuming the carbon nucleus stays fixed.

(2)
(b)

In reality the carbon nucleus recoils. Explain why, at the instant of closest approach, the alpha particle and the carbon nucleus move with the same velocity.

(2)
(c)

Show that three-quarters of the alpha particle's initial kinetic energy has been converted into electric potential energy at closest approach.

(3)
(d)

Determine the distance of closest approach when the recoil of the carbon nucleus is taken into account.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Set the initial kinetic energy equal to the electric potential energy: 2.0×106 × 1.60×10−19 = k(2e)(6e)/d .✓ 1
d = 8.99×109 × 12 × (1.60×10−19)2 / 3.2×10−13 = 8.6×10−15 m .✓ 1
Part (b)
The separation decreases while the alpha particle moves faster than the nucleus and increases once the nucleus moves faster, so at minimum separation the relative velocity is zero .✓ 1
Momentum is conserved and the initial momentum is non-zero, so both particles share the same non-zero velocity rather than being at rest .✓ 1
Part (c)
Conservation of momentum: 4u v0 = 16u v, so v = v0/4 .✓ 1
Kinetic energy remaining: ½(16u)(v0/4)2 = ½u v02, which is ¼ of the initial ½(4u)v02 = 2u v02 .✓ 1
By conservation of energy, the kinetic energy lost, 1 − ¼ = ¾ of the initial value, is stored as electric potential energy .✓ 1
Part (d)
Electric potential energy at closest approach = 0.75 × 2.0 MeV = 1.5 MeV = 2.4×10−13 J. Alternatively, note that d is inversely proportional to the electric potential energy, so d = 8.6×10−15/0.75 .✓ 1
d = 8.99×109 × 12 × (1.60×10−19)2 / 2.4×10−13 ≈ 1.15×10−14 m. Accept 1.1×10−14 m or 1.2×10−14 m. Allow error carried forward from the first or third part .✓ 1

Common error: With a light target, setting all the initial kinetic energy equal to the electric potential energy ignores momentum conservation and underestimates the closest approach.

107IBP2S-035-03-15Rutherford scattering and the nuclear radius Paper 2Hard8 marks

Rutherford backscattering is used in the semiconductor industry to analyse thin layers on silicon chips. A beam of alpha particles, each with kinetic energy 2.0 MeV, is directed at a silicon wafer containing 2814Si nuclei. Consider an alpha particle (42He) travelling directly towards a silicon nucleus.
Take k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, R = R0A1/3 with R0 = 1.20 × 10−15 m. Ignore the recoil of the silicon nucleus.

(a)

State the charge of a 2814Si nucleus in coulombs.

(1)
(b)

Calculate the distance of closest approach of the alpha particle to the centre of the silicon nucleus.

(3)
(c)

Calculate the separation of the centres of the alpha particle and the silicon nucleus when their surfaces just touch.

(2)
(d)

Determine the minimum initial kinetic energy, in MeV, that an alpha particle would need for its surface to just reach the surface of the silicon nucleus. Use this to comment on whether 2.0 MeV alpha particles are expected to show Rutherford scattering from silicon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
14 × 1.60 × 10−19 = 2.24 × 10−18 C✓ 1
Part (b)
Ek = 2.0 × 106 × 1.60 × 10−19 = 3.2 × 10−13 J✓ 1
Ek = k(2e)(14e)/d, so d = 8.99 × 109 × 3.20 × 10−19 × 2.24 × 10−18 / 3.2 × 10−13✓ 1
d = 2.0 × 10−14 m✓ 1
Part (c)
RSi = 1.20 × 10−15 × 281/3 = 3.64 × 10−15 m and RHe = 1.20 × 10−15 × 41/3 = 1.90 × 10−15 m✓ 1
Separation = 5.5 × 10−15 m✓ 1
Part (d)
E = k(2e)(14e)/(5.55 × 10−15) = 1.16 × 10−12 J ≈ 7.3 MeV. Accept 1.17 × 10−12 J if 5.5 × 10−15 m is used. ECF from (c)✓ 1
2.0 MeV is well below this, or equivalently the closest approach in (b) is greater than the separation in (c). So the alpha particle does not come within range of the nuclear force, and Rutherford scattering applies✓ 1

Exam tip: Write the charges as Ze using the proton numbers, 2 for the alpha particle and 14 for silicon. Nucleon numbers are needed only for the radii.

108IBP3H-E1-20r13-15The Bohr model and muonic hydrogen Paper 2Hard9 marks

Muonic hydrogen is a proton with a negative muon in orbit. The muon has the same charge as an electron and a mass of 207 times the electron mass. Treat the proton as stationary and use the Bohr model, for which the ground-state orbit radius of ordinary hydrogen is \(5.29\times10^{-11}\) m, the ground-state energy of ordinary hydrogen is −13.6 eV, and the radius is inversely proportional to the orbiting mass while the energy is proportional to it. The radius of the proton is \(8.4\times10^{-16}\) m. Use h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C (1 eV = 1.60 × 10−19 J) and me = 9.11 × 10−31 kg where needed.

(a)

Show that the ground-state orbit radius of muonic hydrogen is about \(2.56\times10^{-13}\) m.

(2)
(b)

Calculate the ground-state energy of muonic hydrogen in keV. Give your answer to 3 significant figures.

(2)
(c)

Calculate the wavelength, in pm, of the photon emitted when a muon falls from n = 2 to n = 1. Give your answer to 3 significant figures.

(3)
(d)

Suggest why the spectrum of muonic hydrogen is much more sensitive to the size of the proton than the spectrum of ordinary hydrogen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r=\frac{5.29\times10^{-11}}{207}\)✓ 1
A1 \(2.56\times10^{-13}\) m✓ 1
Part (b)
M1 \(13.6\times207=2815\) eV✓ 1
A1 −2.82 keV✓ 1
Part (c)
M1 \(E_2=E_1/4\), so \(\Delta E=\frac34\times2815=2111\) eV✓ 1
M1 \(\lambda=hc/\Delta E\) with \(\Delta E=2111\times1.60\times10^{-19}\) J✓ 1
A1 589 pm✓ 1
Part (d)
the muon's orbit radius is only about 300 times the proton radius (\(2.56\times10^{-13}/8.4\times10^{-16}\approx304\)), so the muon spends appreciable time very close to the proton✓ 1
the electron's orbit is about \(6\times10^{4}\) times the proton radius, so its energy levels are hardly affected by the proton's size✓ 1

Common error: Using the electron mass in the energy and radius scaling. Both depend on the mass of the orbiting particle, here the muon.

109IBP3H-E1-20r14-15Nuclear radius and density Paper 2Hard8 marks

The nuclear radius is R = R0A1/3 with R0 = 1.20×10−15 m, where A is the nucleon number. Take the mass of a nucleus to be A × 1.66×10−27 kg.

(a)

Show that the density of nuclear matter does not depend on the nucleon number A.

(3)
(b)

Calculate the density of nuclear matter.

(2)
(c)

A town uses 4.0×105 m3 of water each day, and water has density 1.0×103 kg m−3. Calculate the volume of nuclear matter that has the same mass as one day's water use. Use ρ = 2.29×1017 kg m−3.

(3)
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Marking pointMarkNotes
Part (a)
mass = A × 1.66×10−27 kg✓ 1
volume = (4/3)πR3 = (4/3)πR03A✓ 1
density = mass/volume, in which A cancels✓ 1
Part (b)
ρ = 1.66×10−27 / [(4/3)π(1.20×10−15)3]✓ 1
2.29×1017 kg m−3✓ 1
Part (c)
mass of water = 4.0×105 × 1.0×103 = 4.0×108 kg✓ 1
V = m/ρ = 4.0×108/2.29×1017✓ 1
1.75×10−9 m3✓ 1

Exam tip: In a "show that" question, carry the symbol A through every line so its cancellation is visible.

110IBP3H-E1-20r33-13The Bohr model Paper 2Hard10 marks

In the Bohr model of hydrogen, an electron of mass m moves in a circular orbit of radius r around a proton. Use k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, h = 6.63 × 10−34 J s and m = 9.11 × 10−31 kg. The electron's angular momentum in the nth orbit is \(mvr=\dfrac{nh}{2\pi}\).

(a)

Show that the speed of the electron in the nth orbit is \(v=\dfrac{2\pi ke^{2}}{nh}\).

(3)
(b)

Calculate the speed of the electron in the n = 3 orbit. Give your answer to 3 significant figures.

(2)
(c)

Using a speed of 7.27 × 105 m s−1, calculate the radius of the n = 3 orbit. Give your answer to 3 significant figures.

(3)
(d)

Deduce the ratio of the radius of the n = 3 orbit to the radius of the n = 1 orbit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 electric force provides the centripetal force: \(\dfrac{ke^{2}}{r^{2}}=\dfrac{mv^{2}}{r}\), so \(ke^{2}=mv^{2}r\)✓ 1
M1 write \(mv^{2}r=v\,(mvr)=v\dfrac{nh}{2\pi}\)✓ 1
A1 \(v=\dfrac{2\pi ke^{2}}{nh}\) (AG)✓ 1
Part (b)
M1 substitute: \(v=\dfrac{2\pi\times8.99\times10^{9}\times(1.60\times10^{-19})^{2}}{3\times6.63\times10^{-34}}\)✓ 1
A1 \(7.27\times10^{5}\) m s−1✓ 1
Part (c)
M1 use \(mvr=\dfrac{nh}{2\pi}\) so \(r=\dfrac{nh}{2\pi mv}\)✓ 1
M1 substitute: \(r=\dfrac{3\times6.63\times10^{-34}}{2\pi\times9.11\times10^{-31}\times7.27\times10^{5}}\)✓ 1
A1 \(r=4.78\times10^{-10}\) m✓ 1
Part (d)
M1 \(v\propto\dfrac{1}{n}\) and \(r=\dfrac{nh}{2\pi mv}\), so \(r\propto n^{2}\)✓ 1
A1 ratio \(=3^{2}=9\)✓ 1

Exam tip: In a 'show that' question, state the starting equations clearly and keep every algebraic step visible until you reach the given result.

111IBP3H-E1-20r33-15Rutherford scattering and the nuclear radius Paper 2Hard8 marks

In an experiment, α particles of kinetic energy 5.5 MeV are fired directly at a thin silver foil (atomic number 47). Use k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C and 1 MeV = 1.60 × 10−13 J. The charge of an α particle is +2e.

(a)

Calculate the distance of closest approach of an α particle that travels directly towards the centre of a silver nucleus. Give your answer to 3 significant figures.

(3)
(b)

The radius of a silver nucleus is about 7.0 × 10−15 m. Calculate the minimum kinetic energy, in MeV, that an α particle needs to reach the surface of the nucleus when fired directly at it. Give your answer to 3 significant figures.

(3)
(c)

Suggest why the scattering of α particles at this higher energy no longer follows the prediction based only on the electrostatic force.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 initial kinetic energy equals electric potential energy at closest approach✓ 1
M1 \(r=\dfrac{8.99\times10^{9}\times2\times47\times(1.60\times10^{-19})^{2}}{5.5\times1.60\times10^{-13}}\)✓ 1
A1 \(r=2.46\times10^{-14}\) m✓ 1
Part (b)
M1 kinetic energy equals potential energy at \(r=7.0\times10^{-15}\) m: \(E=\dfrac{k(2e)(47e)}{r}\)✓ 1
A1 \(E=3.09\times10^{-12}\) J✓ 1
A1 \(19.3\) MeV✓ 1
Part (c)
The α particles come close enough to the nucleus to be within range of the strong nuclear force✓ 1
So an extra force (attractive, not described by the inverse square law) acts or the α particles may be absorbed, and the scattering differs from the prediction✓ 1

Exam tip: Compare a distance of closest approach with the nuclear radius: if the two are similar, forces other than the electrostatic force must be considered.