E.1 Structure of the atom: IB Physics HL exam-style questions
The nuclear model of the atom comes from the Geiger–Marsden–Rutherford scattering experiment. Both levels use nuclear notation, emission and absorption spectra, and transitions between atomic energy levels with E = hf.
HL adds the distance of closest approach in head-on scattering, the nuclear radius R = R₀A^(1/3) and the near-constant density of nuclear matter, and the Bohr model of hydrogen with quantised angular momentum and energy levels that vary as 1/n².
46 questions
199 marks
Paper 1A: 28
Paper 1B: 8
Paper 2: 10
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21 practice questions on E.1 Structure of the atom
1E-1A-01
Energy levels and spectra·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark4 steps to full marksDeduce
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.
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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.
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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.
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Step 4So I and III only.
✓ 1
Answer 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.
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV (HL) Command term: Deduce
2E-1A-02
The Bohr model·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With r ∝ n², v ∝ n/n² = 1/n.
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Step 3v2/v1 = 1/2.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π Command term: Deduce
3E-1A-03
Nuclear radius and density·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Nucleus X has nucleon number 27 and nucleus Y has nucleon number 216.
Step 3So the density ratio is 1: nuclear matter has the same density whatever the nucleus.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities Command term: Deduce
4E-1A-17
Energy levels and spectra·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
Step 3λ32 = λ31λ21/(λ21 − λ31) — positive because λ21 > λ31. (For hydrogen: 102.6 nm and 121.6 nm give 656 nm.)
✓ 1
Answer 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.
Syllabus understandingE.1 — that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; that photons are emitted and absorbed during atomic transitions Command term: Deduce
5E-1A-22
Counting spectral lines·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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.
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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.
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Step 3Total = 3 + 2 + 1 = 6 (in general n(n − 1)/2).
✓ 1
Answer 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.
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions Command term: Determine
6E-1A-25
The Bohr model·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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).
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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.
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Step 3With r = n²r1: λ = 2πnr1.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π; E.2 (HL) — the de Broglie wavelength λ = h/pCommand term: Deduce
7E-1A-27
Transitions on an energy level diagram·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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?
Some energy levels of the hydrogen atom (not to scale) with four emission transitions marked.Show mark scheme
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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.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3C has the smallest energy (0.66 eV), giving λ ≈ 1.9 μm (infrared).
✓ 1
Answer 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.
Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hfCommand term: Deduce
8E-1A-30
Rutherford scattering·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
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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.
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Step 3So the distance of closest approach at the threshold energy (about 25 MeV) is an estimate of the nuclear radius.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — deviations from Rutherford scattering at high energies; the distance of closest approach in head-on scattering experiments; E.3 — the existence of the strong nuclear force, a short-range, attractive force between nucleons Command term: Deduce
9E-1A-31
Rutherford scattering·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
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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.
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Step 3With the same number of nuclei in the beam's path, more alpha particles are turned back through more than 90°.
✓ 1
Answer 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".
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus Command term: Deduce
10E-1A-32
Emission and absorption spectra·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksExplain
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.
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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.
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Step 3Therefore sodium is present in the cooler gas the light passes through: absorption spectra reveal chemical composition.
✓ 1
Answer 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.
Syllabus understandingE.1 — that emission and absorption spectra provide information on the chemical composition; that photons are emitted and absorbed during atomic transitions Command term: Explain
11E-1A-33
Rutherford scattering·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
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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).
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Step 3The two factors cancel: the distance of closest approach is again d.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — the distance of closest approach in head-on scattering experiments (Rutherford's energy-conservation argument); D.2 (HL) — the electric potential energy Ep = kq1q2/rCommand term: Deduce
12E-1A-34
Nuclear radius and closest approach·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With R = R0A1/3: Edev ∝ Z/A1/3.
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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.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; deviations from Rutherford scattering at high energies; the distance of closest approach in head-on scattering experiments; D.2 (HL) — Ep = kq1q2/rCommand term: Deduce
13E-1A-60
Nuclear notation and isotopes·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
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.
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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.
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Step 3Only 3919X has Z = 19: it has 19 protons and 39 − 19 = 20 neutrons, so it is an isotope of P.
✓ 1
Answer 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.
Syllabus understandingE.1 — nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; E.3 — isotopes Command term: Identify
14E-1A-61
Emission spectra and chemical composition·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
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
Emission line spectra of three elements and of the lamp (wavelength scale drawn to scale).Show mark scheme
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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.
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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.
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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.
✓ 1
Answer 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.
Syllabus understandingE.1 — that emission and absorption spectra provide information on the chemical composition; that emission spectra provide evidence for discrete atomic energy levels Command term: Deduce
15E-1A-62
The Bohr model·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Step 1From L = nh/2π, the quantum number is n = 2πL/h.
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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.
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Step 3Check: for L = 3h/2π this gives 13.6/9 = 1.51 eV, the ionisation energy from n = 3.
✓ 1
Answer 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.
Syllabus understandingE.1 (HL) — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV; that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model as given by mvr = nh/2π Command term: Determine
16E-1A-72
Nuclear notation for an ion·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An ion of the nuclide 5726X has a charge of +3e.
Which row gives the number of electrons and the number of neutrons in the ion?
ElectronsNeutrons
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Step 1Z = 26 is the proton number and A = 57 the nucleon number, so the nucleus has 57 − 26 = 31 neutrons.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The neutral atom has 26 electrons, one for each proton.
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Step 3A charge of +3e means that three electrons have been removed: 26 − 3 = 23 electrons.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: 23 electrons (three fewer than the 26 protons) and 57 − 26 = 31 neutrons.
BThis takes the nucleon number, 57, as the number of neutrons, forgetting that the nucleons include the 26 protons.
CThis ignores the charge of the ion and gives the 26 electrons of the neutral atom.
DThis adds three electrons instead of removing them: an ion with 29 electrons and 26 protons would have a charge of −3e.
Syllabus understandingE.1 — nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; E.3 — isotopes Command term: Deduce
17E-1A-73
An alpha particle approaching a nucleus·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An alpha particle moves directly towards a gold nucleus, which remains at rest. Only the electric force acts.
Which statements are correct while the alpha particle is approaching the nucleus?
I. The kinetic energy of the alpha particle decreases.
II. The electric potential energy of the alpha particle and the nucleus increases.
III. The magnitude of the force on the alpha particle increases.
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Step 1Both the alpha particle and the nucleus are positive, so the force on the alpha particle is a repulsion directed away from the nucleus, opposite to its motion: it slows down (I is correct).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Work is done against the repulsion, so the electric potential energy kq1q2/r of the system rises as r falls; it gains what the kinetic energy loses (II is correct).
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Step 3The force kq1q2/r² grows as the separation r decreases (III is correct).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the electric force as constant, independent of the separation; Coulomb's law gives a force proportional to 1/r².
BThis takes the potential energy to fall as the charges approach, as it does for attracting charges or masses; for two like charges it rises.
CThis misses where the stored potential energy comes from: the alpha particle does work against the repulsion, so it slows down and its kinetic energy falls.
DCorrect: the repulsion grows as the separation falls, slows the alpha particle and stores its kinetic energy as electric potential energy.
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; the distance of closest approach in head-on scattering experiments; D.2 — Coulomb's law and the electric potential energy Ep = kq1q2/rCommand term: Deduce
18E-1A-74
Emission and absorption spectra·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
White light passes through a cool gas G and is analysed by a spectrometer. The graph shows the intensity of the transmitted light against wavelength.
Gas G is then heated in a discharge tube so that it glows, and its light is analysed with the same spectrometer. Which describes the spectrum of the light from the glowing gas?
Intensity of white light transmitted through the cool gas G against wavelength (drawn to scale).Show mark scheme
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Step 1The dips occur because atoms of G absorb photons whose energies hf equal the differences between their energy levels.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2When the atoms are excited in the discharge tube they make downward transitions between the same levels and emit photons of the same energies, so bright lines appear at 450 nm, 520 nm and 610 nm.
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Step 3The glowing gas produces only these discrete wavelengths (an emission line spectrum), with darkness between them; it may also emit other lines from transitions between excited levels, hence "including".
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis repeats the absorption spectrum: a glowing gas with no white light behind it has no continuous background to absorb from.
BThis treats emission lines as the "missing" wavelengths of the absorption spectrum; in fact the same energy differences give both, so the wavelengths coincide.
CCorrect: the same pairs of energy levels are involved in absorption and emission, so the emission lines include the absorption wavelengths.
DThis treats a low-pressure glowing gas like a hot dense solid; isolated atoms can emit only photons whose energies match their discrete level differences.
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; that emission and absorption spectra provide information on the chemical composition Command term: Deduce
19E-1A-75
Alpha-particle paths near a nucleus·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
An alpha particle travels towards a gold nucleus N along a line that does not pass through N. The diagram shows four suggested paths P, Q, R and S that the alpha particle might follow.
Which path could be followed by the alpha particle?
Four suggested paths P, Q, R and S for an alpha particle passing a nucleus N (not to scale).Show mark scheme
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Step 1The alpha particle and the nucleus are both positive, so the electric force on the alpha particle is a repulsion directed away from N.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The force acts at every point of the path and grows smoothly as the alpha particle approaches, so the path bends gradually, with no sudden change of direction.
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Step 3The alpha particle is therefore deflected smoothly away from N: path R.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis path bends towards N, as if the force between the alpha particle and the nucleus were attractive.
BThis ignores the force altogether, as if the alpha particle were deflected only when it struck the nucleus.
CCorrect: a repulsive force acting continuously along the path bends it smoothly away from the nucleus.
DThis treats the deflection as a collision at a single point, like a contact force, instead of a force that acts all along the path.
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; D.2 — Coulomb's law Command term: Identify
20E-1A-76
Photon frequency from an energy-level diagram·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The diagram shows some energy levels of an atom. The arrow shows a transition in which a photon is emitted.
What is the frequency of the photon?
Some energy levels of an atom (not to scale). The arrow shows a transition in which a photon is emitted.Show mark scheme
Marking point
Mark
Notes
Step 1The photon carries away the difference between the two levels: ΔE = −1.60 − (−4.70) = 3.10 eV.
—
All 3 steps must be completed — there is no mark for a part-answer.
AThis uses the energy of the upper level alone (1.60 eV) instead of the difference between the levels.
BCorrect: ΔE = 3.10 eV, so f = ΔE/h = 7.5 × 1014 Hz.
CThis uses the energy of the lower level alone (4.70 eV) instead of the difference between the levels.
DThis adds the magnitudes of the two levels (6.30 eV) instead of subtracting them.
Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hfCommand term: Determine
21E-1A-77
Nuclear radius·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The radius of a nucleus is 6.0 fm. Take R0 = 1.2 fm.
What is the nucleon number of the nucleus?
Show mark scheme
Marking point
Mark
Notes
Step 1R = R0A1/3, so A1/3 = 6.0/1.2 = 5.0.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2A = 5.0³ = 125.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis stops at R/R0 = 5, which is A1/3, not A.
BThis treats the power 1/3 as a division by 3 and so multiplies 5 by 3.
CThis squares 5 instead of cubing it, as if R ∝ A1/2.
DCorrect: A = (R/R0)³ = 5³ = 125.
Syllabus understandingE.1 — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol Command term: Determine
22E-1A-78
Wavelengths in the hydrogen spectrum·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The energy levels of hydrogen are given by E = −13.6/n² eV. A sample of hydrogen contains atoms that have all been excited to the n = 3 level.
Which wavelength could be emitted by these atoms as they return to the ground state?
Show mark scheme
Marking point
Mark
Notes
Step 1From n = 3 the possible transitions are 3 → 2, 3 → 1 and 2 → 1, with photon energies equal to the differences between the levels.
—
All 3 steps must be completed — there is no mark for a part-answer.
DThis uses the magnitude of the n = 3 level alone, 1.51 eV, as the photon energy, instead of a difference between two levels.
Syllabus understandingE.1 — that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; that photons are emitted and absorbed during atomic transitions; E.1 (HL) — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV Command term: Deduce
23E-1A-79
Closest approach after acceleration·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A particle of charge q and mass m is accelerated from rest through a potential difference V. It is then directed head-on at a fixed nucleus of proton number Z. Only the electric force acts, and the particle starts far from the nucleus.
What is the distance of closest approach? (k = 1/4πε0)
Show mark scheme
Marking point
Mark
Notes
Step 1The kinetic energy gained in the acceleration is qV.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At closest approach all of it has become electric potential energy: qV = kqZe/d.
—
Step 3d = kZe/V: the charge and mass of the particle cancel.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: qV = kqZe/d gives d = kZe/V, independent of q and m.
BThis takes the kinetic energy as ½qV, confusing qV with ½mv², so d comes out twice as large.
CThis takes the kinetic energy as V instead of qV, so the charge q fails to cancel.
DThis equates the Coulomb force kqZe/d² to the energy qV instead of equating two energies.
Syllabus understandingE.1 — the distance of closest approach in head-on scattering experiments; Guidance: Rutherford's simple energy conservation considerations can be used to determine the distance of closest approach; D.2 — the work done on a charge moved through a potential difference, W = qV, and Ep = kq1q2/rCommand term: Determine
24E-1A-80
Nuclear density·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Nuclear matter has the same density in every nucleus. The mass of the Earth, 6.0 × 1024 kg, is imagined compressed into a sphere of nuclear matter.
What is the radius of this sphere? (R0 = 1.20 fm; 1 u = 1.661 × 10−27 kg)
Show mark scheme
Marking point
Mark
Notes
Step 1The sphere behaves like one giant nucleus with A = 6.0 × 1024/1.661 × 10−27 = 3.6 × 1051 nucleons.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3R = 1.8 × 102 m. (ALT: ρ = 3u/(4πR0³) = 2.3 × 1017 kg m−3, V = m/ρ, then r = (3V/4π)1/3.)
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses the mass in kilograms, 6.0 × 1024, as the nucleon number instead of dividing by the mass of one nucleon.
BCorrect: R = R0A1/3 with A = 3.6 × 1051 gives 1.8 × 102 m.
CThis finds the volume (2.6 × 107 m³) correctly and then takes its cube root as the radius, omitting the factor 4π/3.
DThis converts 1.20 fm to 1.20 × 10−12 m, confusing femto (10−15) with pico (10−12).
Syllabus understandingE.1 — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities Command term: Determine
25E-1A-81
What is the same in every nucleus·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The radius of a nucleus of nucleon number A is R = R0A1/3.
Which quantities are approximately the same for all nuclei?
I. the number of nucleons per unit volume
II. the electric charge per unit volume
III. R/A1/3
Show mark scheme
Marking point
Mark
Notes
Step 1Volume ∝ R³ ∝ A, so the number of nucleons per unit volume, A/volume, is the same for all nuclei (I).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2R/A1/3 = R0, a constant (III).
—
Step 3Charge per unit volume ∝ Z/A. Heavy nuclei contain proportionally more neutrons: Z/A = 0.50 for carbon-12 but 0.39 for uranium-238, so II is not the same.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: nucleon density and R0 are constant; the charge density falls because Z/A falls in heavy nuclei.
BThis misses that R/A1/3 is simply the constant R0.
CThis assumes larger nuclei are more tightly packed, so that the number of nucleons per unit volume rises with A; the cube-root law means it is constant.
DThis assumes the charge density is constant like the nucleon density, forgetting that the proportion of neutrons grows with A.
Syllabus understandingE.1 — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; E.3 (HL) — the neutron–proton (N–Z) ratio and the stability of nuclei Command term: Deduce
26E-1A-82
Orbit radius and angular momentum in the Bohr model·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In the Bohr model of hydrogen, the angular momentum of the electron is L = mvr = nh/2π and the radius of orbit n is n2r1, where r1 is a constant.
Which graph shows how the orbital radius r varies with the angular momentum L for the allowed orbits?
Sketch graphs of orbital radius r against angular momentum L (options A to D).Show mark scheme
Marking point
Mark
Notes
Step 1L = nh/2π, so L ∝ n: n = 2πL/h.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2r = n²r1 = (2πL/h)²r1, so r ∝ L².
—
Step 3The graph is a parabola through the origin that becomes steeper as L increases: D.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes r ∝ L, i.e. r ∝ n instead of n².
BThis inverts the relation, taking L ∝ r² (so r ∝ √L).
CThis confuses the radius with the magnitude of the energy, which is proportional to 1/n² and hence to 1/L².
DCorrect: L ∝ n and r ∝ n², so r ∝ L².
Syllabus understandingE.1 — that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π; the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV Command term: Deduce
27E-1A-83
Energies of the electron in the Bohr model·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In the Bohr model of hydrogen the electron moves in a circular orbit, the centripetal force being provided by the electric force of the proton. The total energy of the atom in level n is −13.6/n² eV.
Which row gives the kinetic energy of the electron and the electric potential energy of the atom when n = 2?
Kinetic energy / eVElectric potential energy / eV
Show mark scheme
Marking point
Mark
Notes
Step 1Electric force = centripetal force: ke²/r² = mv²/r, so Ek = ½mv² = ke²/(2r).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ep = −ke²/r = −2Ek, so the total energy is Ek + Ep = −Ek.
—
Step 3For n = 2 the total energy is −3.40 eV, so Ek = +3.40 eV and Ep = −6.80 eV.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats −13.6/n² eV as the potential energy instead of the total energy, then takes Ek = ½|Ep|.
BThis takes Ek as half the magnitude of the total energy; the total is still −3.40 eV but Ep is no longer −2Ek.
CCorrect: Ep = −2Ek and the total is −Ek = −3.40 eV, so Ek = 3.40 eV and Ep = −6.80 eV.
DThis swaps the roles: it takes Ek equal to |Ep| = 6.80 eV, so the potential energy must then be −10.2 eV.
Syllabus understandingE.1 — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV; that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π; D.2 — Ep = kq1q2/r; A.2 — centripetal force Command term: Deduce
28E-1A-84
Speed during a head-on approach·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An alpha particle, far from a fixed nucleus, moves directly towards it with speed v0. Its distance of closest approach is d. Only the electric force acts.
What is the speed of the alpha particle when it is at a distance 2d from the nucleus?
Show mark scheme
Marking point
Mark
Notes
Step 1At closest approach all the initial kinetic energy Ek0 has become electric potential energy: kqQ/d = Ek0.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At 2d the potential energy is kqQ/(2d) = ½Ek0, so the kinetic energy is ½Ek0.
—
Step 3Speed ∝ √Ek: v = v0/√2 = 0.71v0.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis finds that the kinetic energy has halved but then halves the speed as well, forgetting that v ∝ √Ek.
BCorrect: half the kinetic energy remains at 2d, so v = v0/√2.
CThis takes the potential energy ∝ 1/r² (the force law), so that ¾ of the kinetic energy remains, and then takes the speed ∝ Ek.
DThis takes the potential energy ∝ 1/r² (the force law), so that ¾ of the kinetic energy remains: v = √0.75 v0.
Syllabus understandingE.1 — the distance of closest approach in head-on scattering experiments; Guidance: Rutherford's simple energy conservation considerations can be used to determine the distance of closest approach; D.2 — Ep = kq1q2/rCommand term: Deduce
29E-1B-03
Nuclear radius and density·E.1 Structure of the atom
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
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)
Element
A
R / fm
ln A
ln(R / fm)
carbon
12
2.78
2.48
1.022
aluminium
27
3.55
3.30
1.267
cobalt
59
4.72
4.08
1.552
yttrium
89
5.31
4.49
1.670
tin
120
6.00
4.79
1.792
lead
208
7.07
5.34
1.956
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 point
Mark
Notes
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
✓ 1
Accept 0.31–0.35.
Part (b)(i)
Intercept on the ln(R / fm) axis at ln A = 0: 0.19
✓ 1
Accept 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).
Allow 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
✓ 1
Allow 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)
Syllabus understandingE.1 (HL) — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; Tools 3 — linearising a power law with logarithms, extrapolation to an intercept, propagation of uncertainties Command term: Determine
30E-1B-04
Rutherford scattering and nuclear charge·E.1 Structure of the atom
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
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.
Foil
Z
N
ln Z
ln N
aluminium
13
423
2.565
6.047
titanium
22
1220
3.091
7.107
copper
29
2090
3.367
7.645
silver
47
5456
3.850
8.604
tin
50
6214
3.912
8.735
gold
79
15479
4.369
9.647
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 point
Mark
Notes
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
✓ 1
Accept 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
✓ 1
Allow 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 %
✓ 1
Or ΔN = √423 = 21 counts.
Δ(ln N) = ΔN/N = 0.049 ≈ ±0.05
✓ 1
The 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
✓ 1
Accept "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
✓ 1
Allow ECF from the line. ALT: Z = 79 × √(13200/15479) = 73.0 using gold, since N ∝ Z².
Z = e4.29 = 73
✓ 1
Accept 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)
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; E.3 — the random and spontaneous nature of radioactive decay (random uncertainty √N in a count); Tools 3 — linearize graphs, determine gradients, draw and interpret uncertainty bars; Inquiry 3 — explain realistic and relevant improvements to an investigation Command term: Determine
31E-1B-11
The hydrogen spectrum·E.1 Structure of the atom
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
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)
n
θ / °
1/n²
1/λ / 106 m−1
3
23.28
0.1111
1.518
4
16.99
0.0625
2.053
5
15.15
0.0400
6
14.28
0.0278
2.432
7
13.84
0.0204
2.508
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 point
Mark
Notes
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
✓ 1
Accept 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
✓ 1
Allow 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
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)
Syllabus understandingE.1 — emission spectra provide evidence for discrete atomic energy levels; the frequency of the photon depends on the difference in energy levels, E = hf; (HL) the discrete energy levels in the Bohr model for hydrogen; C.3 — the diffraction grating equation nλ = d sin θ; Tools 3 — gradient and intercept of a linearised graph, extrapolation Command term: Determine
32E-1B-12
Rutherford scattering·E.1 Structure of the atom
Paper 1BMedium6 marks
Data-based question5 steps to full marksDetermine
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) / °
N
ln(sin(θ/2))
ln N
20
87699
−1.75
11.38
30
14999
−1.35
9.62
45
2925
−0.96
7.98
60
932
−0.69
6.84
90
226
−0.35
5.42
120
95
−0.14
4.55
150
62
−0.03
4.13
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 point
Mark
Notes
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
✓ 1
Accept 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°)
✓ 1
Allow 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)
✓ 1
The direction of the change must be consistent with the candidate's answer to (a).
Allow 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
✓ 1
The 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)
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; Tools 3 — linearising a power law with logarithms; Inquiry 3 — identifying and correcting a systematic (zero) error read from the instrument Command term: Determine
33E-1B-22
Excitation energy from electron collisions·E.1 Structure of the atom
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
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)
k
1
2
3
4
5
6
V / V
3.59
5.60
7.78
9.91
11.91
14.07
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 point
Mark
Notes
Part (a)(i)
Gradient = ΔV/Δk from two well-separated points on the line = 2.10 V
✓ 1
Accept 2.05–2.15 V.
E = e × gradient = 2.1 eV
✓ 1
Accept 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
✓ 1
Accept 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
✓ 1
Accept "offset" or "zero error". Reason needed.
Part (c)(i)
ΔE = (0.10 + 0.10)/5 = ±0.04 eV
✓ 1
Both readings' uncertainties must be added and divided by the 5 intervals. ±0.02 eV (one reading only) scores 0.
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
✓ 1
Allow 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)
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; D.2 — work done on a charge accelerated through a potential difference; Tools 3 — gradient and intercept of a line with units, systematic error, propagation of uncertainties Command term: Determine
34E-1B-24
Scattering and foil thickness·E.1 Structure of the atom
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
In a version of the Geiger–Marsden–Rutherford experiment, a narrow beam of alpha particles from a sealed source passes through a stack of n identical layers of gold leaf in an evacuated chamber. A detector at a fixed angle of 40° to the beam counts the alpha particles that arrive in 10 minutes. The count N is recorded for stacks of 1 to 6 layers. The source, detector and counting time are not changed.
The graph shows N against n with the line of best fit. The error bars show the random uncertainty ±√N in each count.
n
1
2
3
4
5
6
N
345
675
967
1275
1561
1901
Count N in 10 minutes against the number of layers n (drawn to scale).
(a)
(i)
Determine the gradient and the intercept on the N axis of the line of best fit.
(2)
(ii)
Suggest the origin of the counts given by the intercept.
(1)
(b)
(i)
Rutherford's model predicts that, for thin foils, the number of alpha particles scattered towards the detector is proportional to the number of gold nuclei in the path of the beam. Deduce, using three points from the table, whether the data support this prediction.
(2)
(c)
(i)
Calculate the percentage uncertainty in the count for n = 2.
(1)
(ii)
The detector is moved to a larger angle and the experiment is repeated with the same counting time. State and explain the effect on the gradient of the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (0, 46) and (7, 2195): 307 counts per layer
✓ 1
Accept 295–320. Points must be read from the line, not from the table.
Intercept = 46 counts
✓ 1
Accept 30–60.
Part (a)(ii)
Background radiation (counts recorded with no foil in the beam), a constant addition to every count / systematic offset
✓ 1
Allow ECF from (a)(i). "Zero error of the detector" alone: [0]; the source of the counts must be stated.
Part (b)(i)
Subtract the intercept and divide by n for at least three points, e.g. n = 1: (345 − 46)/1 = 299; n = 3: (967 − 46)/3 = 307; n = 6: (1901 − 46)/6 = 309
✓ 1
Allow ECF from (a)(i) for the intercept. Without subtracting the background: [1 max] only if the conclusion is reasoned.
The values agree to within about 3 %, less than the 5 % random uncertainty of the smallest count, so (N − background) ∝ n and the data support the prediction
✓ 1
The conclusion must agree with the candidate's own ratios. ALT: the line passes within all the error bars and through the background intercept.
Part (c)(i)
√675/675 × 100 = 3.8 %
✓ 1
Accept 3.8 % or 4 %.
Part (c)(ii)
The gradient decreases, because fewer alpha particles are scattered through a larger angle: a larger deflection requires the alpha particle to pass closer to a nucleus, which is less likely
✓ 1
Both the effect and the reason are needed. Allow "the intercept is unchanged" as additional comment.
Answers: (a)(i) gradient 307 per layer; intercept 46 · (b)(i) supported · (c)(i) 3.8 % (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; E.3 — the random and spontaneous nature of radioactive decay; background radiation; Tools 3 — gradient and intercept of a linear graph, uncertainty bars, testing a proportional relationship Command term: Determine
35E-1B-25
Radio lines from highly excited hydrogen·E.1 Structure of the atom
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
In clouds of ionised hydrogen, electrons recombine with protons into very high energy levels. As an atom then moves down the levels, transitions from level n to level n − 1 emit radio waves. A radio astronomer measures the frequency f of six of these lines. For large n, the Bohr model predicts that
f ≈ 2EI/(hn³)
where EI is the ionisation energy of hydrogen from the ground state. The graph shows ln(f / GHz) against ln n with the line of best fit. (h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C)
n
f / GHz
ln n
ln(f / GHz)
100
6.665
4.605
1.897
120
3.854
4.787
1.349
140
2.415
4.942
0.882
170
1.350
5.136
0.300
200
0.826
5.298
−0.191
240
0.478
5.481
−0.738
ln(f / GHz) against ln n for six radio lines of hydrogen (drawn to scale).
(a)
(i)
Determine the gradient of the line.
(2)
(ii)
State whether the data support the prediction of the Bohr model.
(1)
(b)
(i)
Assume the gradient is −3. Use the line at n = 150 to determine EI in eV.
(2)
(ii)
The receiver of the radio telescope can detect frequencies up to 35 GHz. Use your answer to (b)(i) to determine the smallest value of n for which the line from n to n − 1 can be detected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (4.60, 1.91) and (5.50, −0.80)
✓ 1
Points must be read from the line.
Gradient = −3.01
✓ 1
Accept −2.90 to −3.10.
Part (a)(ii)
ln f = ln(2EI/h) − 3 ln n, so the model predicts a straight line of gradient −3; the gradient is −3.0 (and the points lie on a straight line), so yes
✓ 1
Allow ECF from (a)(i): the conclusion must agree with the candidate's gradient.
Part (b)(i)
At ln 150 = 5.01 the line gives ln(f / GHz) = 0.68, so f = 1.97 GHz
Allow ECF from (b)(i). Using 13.6 eV gives the same result.
n must be a whole number: n = 58
✓ 1
Rounding down to 57: [1 max]; the line from 57 has a frequency above 35 GHz.
Answers: (a)(i) −3.01 · (a)(ii) yes · (b)(i) 13.7 eV · (b)(ii) 58 (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 (HL) — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV; E.1 — that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; that emission and absorption spectra provide information on the chemical composition; Tools 3 — linearising a power law with logarithms, determining a gradient Command term: Determine
36E-1B-26
Nuclear radius from deviation energies·E.1 Structure of the atom
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
Alpha particles from a particle accelerator are scattered through angles close to 180° by thin foils of six elements. For each element the beam energy is raised until the count first falls below the prediction of Rutherford's model, in which only the electric force acts. This energy, Edev, is uncertain by ±0.5 MeV.
A student models the onset of the deviation as the energy at which, in a head-on collision with a fixed nucleus, the surfaces of the alpha particle and the target nucleus just touch. Both radii are given by R = R0A1/3. She plots Edev against x = Z/(A1/3 + 41/3), where Z and A are the proton number and nucleon number of the target nucleus.
(k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)
Target
Z
A
x
Edev / MeV
aluminium
13
27
2.83
6.1
titanium
22
48
4.21
9.4
nickel
28
58
5.13
11.6
silver
47
107
7.42
16.2
tin
50
120
16.4
gold
79
197
10.67
23.6
Edev against x for the six targets with the line of best fit (drawn to scale). The value of x for tin is not given in the table.
(a)
(i)
Show that the model predicts Edev = (2ke²/R0)x.
(2)
(ii)
Calculate x for tin.
(1)
(b)
(i)
Determine R0 using the gradient of the line.
(2)
(ii)
Determine the absolute uncertainty in R0, using lines of maximum and minimum gradient that pass through all the error bars.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
At the onset all the kinetic energy has become electric potential energy at a separation equal to the sum of the radii: Edev = k(2e)(Ze)/[R0(A1/3 + 41/3)]
✓ 1
The separation must include the radius of the alpha particle (A = 4).
Allow ECF from the gradient. Accept 1.25–1.37 × 10−15 m. The gradient must be converted to joules.
Part (b)(ii)
Maximum and minimum gradients, e.g. 2.28 MeV and 2.11 MeV
✓ 1
Accept a maximum of 2.24–2.32 MeV and a minimum of 2.06–2.14 MeV. The lines need not pass through the origin but must pass through all six error bars; a line steeper than about 2.29 MeV or shallower than about 2.10 MeV cannot.
R0 from 1.26 fm to 1.36 fm, so ΔR0 = ±0.05 fm (±5 × 10−17 m)
✓ 1
Allow ECF from (b)(i) and from the candidate's gradients. Accept ±0.03 to ±0.08 fm. ALT: fractional uncertainty of the gradient = fractional uncertainty of R0.
Answers: (a)(ii) 7.67 · (b)(i) 1.31 × 10−15 m · (b)(ii) ±0.05 fm (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 (HL) — deviations from Rutherford scattering at high energies; the distance of closest approach in head-on scattering experiments; the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; Guidance: Rutherford's simple energy conservation considerations can be used to determine the distance of closest approach; D.2 — Ep = kq1q2/r; Tools 3 — gradient with units, lines of maximum and minimum gradient, uncertainty in a derived quantity Command term: Determine
37E-2-01
The Bohr model and muonic hydrogen·E.1 Structure of the atom
Paper 2Hard10 marks
Short answer & extended response6 steps to full marksDeduce
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 point
Mark
Notes
Part (a)(i)
The electric force provides the centripetal force: e²/(4πε0r²) = mv²/r
✓ 1
Both 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
✓ 1
Allow ECF from (a)(i).
En ∝ −1/rn ∝ m, so the ratio is 207 (the muonic level is 207 times deeper)
✓ 1
A 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
✓ 1
Allow ECF from (a)(ii).
ΔE = 2111 eV = 2.1 keV
✓ 1
Accept 2.11 keV.
Part (b)(ii)
λ = hc/ΔE = 5.9 × 10−10 m: X-ray
✓ 1
Allow 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
✓ 1
Allow ECF from (a)(i). ALT: 5.3 × 10−11 m/207.
About 310 times the radius of the proton
✓ 1
Accept 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
✓ 1
Allow 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)
Syllabus understandingE.1 (HL) — the discrete energy levels in the Bohr model for hydrogen, E = −13.6/n² eV; that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen, mvr = nh/2π; E.1 — photons emitted in atomic transitions, E = hf; D.2 — Coulomb's law; A.2 — centripetal force Command term: Deduce
38E-2-09
The hydrogen spectrum and a diffraction grating·E.1 Structure of the atom
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine
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)
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 point
Mark
Notes
Part (a)(i)
d = 1/600 mm = 1.67 × 10−6 m
✓ 1
λ = d sin 15.1° = 4.34 × 10−7 m
✓ 1
Accept 434 nm.
Part (a)(ii)
E = hc/λ = 2.86 eV
✓ 1
Allow ECF from (a)(i).
This matches n = 5 → n = 2: −0.54 − (−3.40) = 2.86 eV
✓ 1
The 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
✓ 1
Accept 656–658 nm.
sin θ ≤ 1, so n ≤ d/λ = 2.53
✓ 1
Allow ECF from (a)(i) for d.
The highest order is 2
✓ 1
MP3 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°
✓ 1
Allow 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
✓ 1
Allow 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)
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions, E = hf; E.1 (HL) — E = −13.6/n² eV; C.3 — the diffraction grating, nλ = d sin θCommand term: Determine
39E-2-17
Excitation of atoms in a fluorescent lamp·E.1 Structure of the atom
Paper 2Medium9 marks
Short answer & extended response6 steps to full marksExplain
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)
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.
Must 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)
✓ 1
Direction required.
Part (b)(i)
It must gain at least the energy to reach the lowest excited level: 10.44 − 5.77 = 4.67 eV
✓ 1
Using 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
✓ 1
Allow ECF from (a)(i).
Part (c)(i)
Photon energies: 4.89 eV in and 2.27 eV out
✓ 1
Allow ECF from (a)(i) and (a)(ii).
Efficiency = 2.27/4.89 = 0.46 (46 %)
✓ 1
ALT: 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)
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; the frequency of the photon depends on the difference in energy levels, E = hf; D.2 — work done on a charge in a uniform electric field; A.3 — efficiency Command term: Explain
40E-2-18
Rutherford scattering and the nuclear radius·E.1 Structure of the atom
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksDetermine
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)
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 point
Mark
Notes
Part (a)(i)
Kinetic energy = electric potential energy at closest approach: Ek = k(2e)(50e)/d
E0 ≈ 18.4 MeV (where the curve leaves the ratio 1)
✓ 1
Accept 18–19 MeV.
Part (b)(ii)
d ∝ 1/Ek: d = 1.44 × 10−14 × 10.0/18.4 = 7.8 × 10−15 m
✓ 1
Accept 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
✓ 1
ECF.
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
✓ 1
Accept 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)
Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; nuclear notation; (HL) the distance of closest approach in head-on scattering experiments; deviations from Rutherford scattering at high energies; R = R0A1/3 and implications for nuclear densities Command term: Determine
41E-2-21
The Bohr model·E.1 Structure of the atom
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksExplain
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 point
Mark
Notes
Part (a)(i)
ΔE = 13.6(1/109² − 1/110²) eV
✓ 1
= 13.6 × 1.52 × 10−6 = 2.07 × 10−5 eV
✓ 1
The difference must be shown to at least 3 s.f. before rounding.
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)
Syllabus understandingE.1 (HL) — the discrete energy levels in the Bohr model for hydrogen, E = −13.6/n² eV; the existence of quantized energy and orbits arise from the quantization of angular momentum, mvr = nh/2π; E.1 — E = hf for atomic transitions; spectra provide information on chemical composition Command term: Explain
42E-2-37
Identifying an element from its spectrum·E.1 Structure of the atom
Paper 2Easy11 marks
Short answer & extended response8 steps to full marksDetermine
A small sample of an unknown salt is heated in a hot flame. The light from the flame is analysed with a spectrometer, which shows a line spectrum. The strongest line has a wavelength of 671 nm.
The diagram shows some of the energy levels of the atoms of two elements, P and Q. One of them is present in the salt. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
Some energy levels of the atoms of elements P and Q, in eV (not to scale). 0 eV corresponds to ionisation.
(a)
(i)
Explain why the light emitted by the atoms in the flame forms a line spectrum rather than a continuous spectrum.
(2)
(ii)
Show that the energy of a photon of wavelength 671 nm is about 1.85 eV.
(1)
(b)
(i)
Deduce which element, P or Q, is present in the salt.
(2)
(c)
(i)
A weaker line of wavelength 610 nm is also observed. Identify the transition that produces it.
(2)
(d)
(i)
The light of wavelength 671 nm reaching the detector of the spectrometer has a power of 2.0 μW. Determine the number of these photons that reach the detector per second.
(2)
(e)
(i)
State the ionisation energy, in eV, of an atom of P in its ground state.
(1)
(ii)
Calculate the minimum frequency of a photon that can ionise an atom of P in its ground state.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The electrons in an atom can only occupy discrete energy levels (the atom has only certain allowed energies)
✓ 1
Vague "electrons jump" without discrete levels: [0].
A photon is emitted when the atom makes a transition from a higher to a lower level, with energy hf equal to the difference between the levels, so only certain frequencies/wavelengths are emitted
✓ 1
Part (a)(ii)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(671 × 10−9 × 1.60 × 10−19) = 1.853 eV
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
Element P: −3.54 − (−5.39) = 1.85 eV, equal to the photon energy, so P can emit this line
✓ 1
Allow ECF from (a)(ii).
No pair of levels of Q differs by 1.85 eV (the nearest differences are 1.65 eV and 2.10 eV), so the element is P
✓ 1
The check on Q (or the absence of a match) is required for the second mark.
Part (c)(i)
E = hc/λ = 2.04 eV
✓ 1
From the −1.51 eV level to the −3.54 eV level of P (difference 2.03 eV)
✓ 1
Allow ECF from (b). The direction (downwards, ending on −3.54 eV) is required. −1.56 → −3.54 eV (1.98 eV) does not match: [1 max].
Answers: (a)(ii) 1.853 eV · (b)(i) P · (c)(i) −1.51 eV → −3.54 eV · (d)(i) 6.7 × 1012 s−1 · (e)(i) 5.39 eV · (e)(ii) 1.3 × 1015 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; that emission and absorption spectra provide information on the chemical composition; A.3 — power as the rate of energy transfer Command term: Determine
43E-2-38
The recoil of an emitting atom·E.1 Structure of the atom
Paper 2Medium12 marks
Short answer & extended response7 steps to full marksDetermine
A hydrogen atom, initially at rest, makes a transition from n = 3 to n = 2 and emits a photon. The energy levels of hydrogen are given by E = −13.6/n² eV. The mass of a hydrogen atom is mH = 1.67 × 10−27 kg. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
(a)
(i)
Show that the energy of the photon is about 1.9 eV.
(1)
(ii)
Calculate the wavelength of the photon.
(2)
(b)
(i)
The photon carries momentum p = h/λ. Determine the speed with which the atom recoils.
(3)
(ii)
Calculate the kinetic energy of the recoiling atom, in eV, and hence comment on the statement that the photon energy equals the difference between the energy levels.
(2)
(c)
(i)
Show that, for a photon of energy E emitted by an atom of mass M initially at rest, the ratio (kinetic energy of the atom)/(energy of the photon) is E/(2Mc²).
(2)
(d)
(i)
In the Bohr model the speed of the electron in the n = 1 orbit is 2.19 × 106 m s−1. Determine the radius of this orbit. (me = 9.11 × 10−31 kg)
p = 6.63 × 10−34/6.58 × 10−7 = 1.01 × 10−27 kg m s−1
✓ 1
Allow ECF from (a)(ii).
The total momentum is zero before and after, so the atom has momentum of equal magnitude in the opposite direction to the photon
✓ 1
The conservation principle must be stated.
v = p/mH = 1.01 × 10−27/1.67 × 10−27 = 0.60 m s−1
✓ 1
Part (b)(ii)
Ek = ½ × 1.67 × 10−27 × 0.603² = 3.0 × 10−28 J = 1.9 × 10−9 eV
✓ 1
Allow ECF from (b)(i).
This is about 10−9 of the photon energy, so the photon carries the whole energy difference to a very good approximation (the statement is valid)
✓ 1
A comparison with the photon energy is required.
Part (c)(i)
p = h/λ = E/c for the photon, and the atom has a momentum of the same magnitude
✓ 1
Ek = p²/(2M) = E²/(2Mc²), so the ratio is E/(2Mc²)
✓ 1
Answer given: both steps must be seen.
Part (d)(i)
mvr = h/2π for n = 1, so r = h/(2πmev)
✓ 1
r = 6.63 × 10−34/(2π × 9.11 × 10−31 × 2.19 × 106) = 5.29 × 10−11 m
✓ 1
Accept 5.3 × 10−11 m.
Answers: (a)(i) 1.889 eV · (a)(ii) 6.58 × 10−7 m · (b)(i) 0.60 m s−1 · (b)(ii) 1.9 × 10−9 eV · (d)(i) 5.29 × 10−11 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; E.1 (HL) — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV; that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π; E.2 (HL) — the de Broglie relation λ = h/p applied to photons; A.2 — conservation of linear momentum Command term: Determine
44E-2-39
Separating the isotopes of neon·E.1 Structure of the atom
Paper 2Medium12 marks
Short answer & extended response8 steps to full marksDetermine
Natural neon contains the isotopes 2010Ne and 2210Ne. In a mass spectrometer, neon atoms are ionised so that each ion has a charge of +e. The ions are accelerated from rest through a potential difference of 2.00 kV and then enter a region of uniform magnetic field of flux density 0.250 T, directed at right angles to their velocity. Each ion moves in a semicircle and lands on a detector plate.
Take the mass of an ion of nucleon number A to be A × 1 u, where 1 u = 1.661 × 10−27 kg. (e = 1.60 × 10−19 C, R0 = 1.20 fm)
(a)
(i)
State the number of neutrons in a nucleus of 2210Ne.
(1)
(ii)
State the number of electrons in an ion of 2210Ne.
(1)
(b)
(i)
Show that the speed of a 2010Ne ion entering the magnetic field is about 1.4 × 105 m s−1.
(2)
(ii)
Calculate the radius of the path of a 2010Ne ion in the magnetic field.
(2)
(iii)
Deduce that r ∝ √m for ions accelerated through the same potential difference, and hence determine the distance between the points where the 2010Ne ions and the 2210Ne ions land.
(3)
(c)
(i)
Calculate the radius of a 2210Ne nucleus, and compare the percentage difference between the radii of the two nuclei with the percentage difference between their masses.
(2)
(d)
(i)
Natural neon also contains a small amount of 2110Ne. Determine the distance between the points where the 2010Ne and 2110Ne ions land.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
12
✓ 1
Part (a)(ii)
9 (the neutral atom has 10 electrons and one has been removed)
✓ 1
11 (an electron added): [0].
Part (b)(i)
qV = ½mv², with m = 20 × 1.661 × 10−27 = 3.32 × 10−26 kg
✓ 1
v = √(2 × 1.60 × 10−19 × 2000/3.32 × 10−26) = 1.39 × 105 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(ii)
qvB = mv²/r, so r = mv/(qB)
✓ 1
r = 3.32 × 10−26 × 1.39 × 105/(1.60 × 10−19 × 0.250) = 0.115 m
✓ 1
Allow ECF from (b)(i).
Part (b)(iii)
v = √(2qV/m), so r = mv/(qB) = √(2mV)/(B√q) ∝ √m
✓ 1
Symbolic working required.
r22 = 0.1153 × √(22/20) = 0.1209 m
✓ 1
Allow ECF from (b)(ii).
The ions land a diameter from the entry point, so the separation = 2(0.1209 − 0.1153) = 0.0113 m (11 mm)
✓ 1
Using the radii instead of the diameters: [2 max] (5.6 mm).
Part (c)(i)
R = 1.20 × 221/3 = 3.36 fm (2010Ne: 3.26 fm)
✓ 1
The radius is 3.2 % larger while the mass is 10 % larger, because R ∝ A1/3 (so the volume, like the mass, is 10 % larger)
✓ 1
Accept 3 %. A reason linking to the cube root is required.
Allow ECF from (b)(ii) and (b)(iii). Accept 5.6–5.7 mm. Half of the answer to (b)(iii): [0], since r ∝ √m is not linear.
Answers: (a)(i) 12 · (a)(ii) 9 · (b)(i) 1.39 × 105 m s−1 · (b)(ii) 0.115 m · (b)(iii) 11 mm · (c)(i) 3.36 fm · (d)(i) 5.7 mm (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 — nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; E.1 (HL) — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; E.3 — isotopes; D.2 — the work done on a charge accelerated through a potential difference; D.3 — the motion of a charged particle in a uniform magnetic field, F = qvBCommand term: Determine
45E-2-40
Absorption of ultraviolet photons by hydrogen·E.1 Structure of the atom
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksExplain
A beam of ultraviolet photons with a wide range of energies passes through a sample of cool hydrogen gas in which all the atoms are in the ground state. The graph shows the fraction of the photons that are absorbed against photon energy. The energy levels of hydrogen are given by E = −13.6/n² eV. (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)
Fraction of photons absorbed by cool hydrogen against photon energy (drawn to scale).
(a)
(i)
Explain why, below 13.6 eV, photons are absorbed only at certain energies.
(2)
(ii)
Identify, with a calculation, the transition responsible for the peak at 12.75 eV.
(2)
(b)
(i)
The peaks crowd together towards 13.6 eV. Determine how many peaks lie below 13.50 eV.
(3)
(c)
(i)
Above 13.6 eV the absorption is continuous. A photon of energy 15.0 eV is absorbed by an atom in the ground state. Determine the speed of the electron that is released. Ignore the recoil of the proton.
(2)
(ii)
Calculate the de Broglie wavelength of this electron.
(1)
(d)
(i)
Deduce, using the Bohr model, the angular momentum of the electron immediately after the atom absorbs a photon of energy 12.75 eV.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A photon is absorbed only if its whole energy raises the atom from the ground state to one of the discrete higher energy levels
✓ 1
Reference to discrete levels needed.
So the photon energy must equal −13.6/n² − (−13.6) eV for a whole number n; other photons pass through
✓ 1
Accept "equal to the difference between two energy levels".
Part (a)(ii)
13.6(1 − 1/n²) = 12.75 gives 1/n² = 0.0625, so n = 4
✓ 1
Absorption from n = 1 to n = 4
✓ 1
Direction (upwards, from the ground state) required.
Part (b)(i)
Peaks at 13.6(1 − 1/n²) for n = 2, 3, 4, …; the levels approach 0 eV as n increases, so the peaks approach 13.6 eV
✓ 1
13.6(1 − 1/n²) < 13.50 requires 13.6/n² > 0.10, so n² < 136 and n ≤ 11 (n = 11: 13.488 eV; n = 12: 13.506 eV)
Answers: (a)(ii) n = 1 → n = 4 · (b)(i) 10 · (c)(i) 7.0 × 105 m s−1 · (c)(ii) 1.0 × 10−9 m · (d)(i) 4.2 × 10−34 J s (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; E.1 (HL) — the discrete energy levels in the Bohr model for hydrogen as given by E = −13.6/n² eV; that the existence of quantized energy and orbits arise from the quantization of angular momentum in the Bohr model for hydrogen as given by mvr = nh/2π; E.2 (HL) — λ = h/p; A.3 — kinetic energy Command term: Explain
46E-2-41
Closest approach to a light nucleus·E.1 Structure of the atom
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksDetermine
Alpha particles of kinetic energy 4.0 MeV are directed at a thin target of carbon, 126C. Consider an alpha particle that moves directly towards a carbon nucleus. Only the electric force acts until the surfaces of the two nuclei touch. (k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J, R0 = 1.20 fm)
(a)
Assume first that the carbon nucleus remains at rest.
(i)
Show that the distance of closest approach is about 4.3 fm.
(2)
(ii)
Calculate the sum of the radii of the alpha particle and the carbon nucleus.
(1)
(iii)
Hence state what this model predicts about the scattering of these alpha particles.
(1)
(b)
In fact the carbon nucleus is free to move. Take the mass of each nucleus as proportional to its nucleon number.
(i)
Explain why, at the instant of closest approach, the alpha particle and the carbon nucleus move with the same velocity, and show that this velocity is one quarter of the initial velocity of the alpha particle.
(2)
(ii)
Determine the distance of closest approach when the recoil of the carbon nucleus is included.
(3)
(iii)
Determine the minimum initial kinetic energy of the alpha particles for which the model predicts deviations from Rutherford scattering.
(3)
(c)
(i)
The same 4.0 MeV alpha particles are directed at a silver foil, 10747Ag. Explain why the recoil of the target nucleus could be ignored in this case, giving the fraction of the initial kinetic energy that can become electric potential energy.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
4.0 MeV = 6.4 × 10−13 J = kqαqC/d with qα = 2e and qC = 6e
✓ 1
The charge of the carbon nucleus must come from the proton number 6.
d = 8.99 × 109 × 12 × (1.60 × 10−19)²/6.4 × 10−13 = 4.32 × 10−15 m
✓ 1
An answer to at least 3 s.f. or full substitution is required.
Part (a)(ii)
1.20 × (121/3 + 41/3) = 2.75 + 1.90 = 4.65 fm
✓ 1
Radius of carbon alone (2.75 fm): [0].
Part (a)(iii)
The closest approach is less than the sum of the radii, so the nuclei would touch: the strong nuclear force would act and the scattering would deviate from Rutherford's (electric-only) prediction
✓ 1
Allow ECF from (a)(i) and (a)(ii); the prediction must follow from the comparison.
Part (b)(i)
The separation stops decreasing (and is about to increase) when the relative velocity is zero, i.e. both move with the same velocity
✓ 1
Momentum is conserved: 4uv0 = (4 + 12)uvc, so vc = v0/4
✓ 1
Any consistent mass unit. Answer given: the equation must be seen.
Part (b)(ii)
Kinetic energy remaining at closest approach = ½ × 16u × (v0/4)² = ¼ of the initial kinetic energy = 1.0 MeV
✓ 1
Allow ECF from (b)(i).
Energy converted to electric potential energy = 4.0 − 1.0 = 3.0 MeV
✓ 1
d = 4.32 × 10−15 × 4.0/3.0 = 5.75 × 10−15 m (5.75 fm)
✓ 1
Allow ECF from (a)(i). ALT: direct substitution of 3.0 MeV into kqαqC/d.
Part (b)(iii)
At contact the electric potential energy = 8.99 × 109 × 12 × (1.60 × 10−19)²/4.65 × 10−15 = 3.71 MeV
✓ 1
Allow ECF from (a)(ii).
Only ¾ of the initial kinetic energy can become potential energy, so Ekmin = 3.71 × 4/3
✓ 1
Allow ECF from (b)(i). Omitting the recoil factor (3.7 MeV): [1 max].
Ekmin = 4.95 MeV, so the 4.0 MeV alpha particles do not in fact reach the nucleus
✓ 1
Accept 4.9–5.0 MeV. The conclusion about 4.0 MeV must agree with the candidate's value.
Part (c)(i)
Fraction = M/(M + m) = 107/111 = 0.96
✓ 1
Allow ECF from (b)(i) for the method.
The silver nucleus is about 27 times more massive than the alpha particle, so it moves very little and almost all (96 %) of the kinetic energy becomes potential energy, compared with 75 % for carbon
✓ 1
A comparison of masses is required.
Answers: (a)(i) 4.32 × 10−15 m · (a)(ii) 4.65 fm · (b)(ii) 5.75 fm · (b)(iii) 4.95 MeV · (c)(i) 0.96 (the remaining parts are explanations — see the table above)
Syllabus understandingE.1 (HL) — the distance of closest approach in head-on scattering experiments; Guidance: Rutherford's simple energy conservation considerations can be used to determine the distance of closest approach; deviations from Rutherford scattering at high energies; the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3 and implications for nuclear densities; nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; D.2 — Ep = kq1q2/r; A.2 — conservation of linear momentum Command term: Determine
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