OxfordAQA International AS and A-level Physics: Nuclear radius, mass-energy and binding energy (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.12.1 (Radius of the nucleus) and 3.12.2 (Mass and energy) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. Section 3.12 is International A-level only. Take e = 1.60 × 10⁻¹⁹ C, 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 u = 1.66 × 10⁻²⁷ kg = 931.5 MeV, 1 MeV = 1.60 × 10⁻¹³ J, N_A = 6.02 × 10²³ mol⁻¹, m_p = 1.007276 u and m_n = 1.008665 u.

See the study guide, revision notes, the OxfordAQA Physics hub, the printable checklist and the free diagnostics.

Questions

1. (a) State a typical value for the radius of a nucleus. [1] (b) Give one reason why a closest-approach calculation gives only an upper limit for the radius. [1] (c) Explain why electrons used to find nuclear radii by diffraction need energies of the order of hundreds of MeV. [1]

2. An alpha particle of kinetic energy 7.2 MeV is fired head-on at a tin nucleus (proton number 50). (a) Calculate the distance of closest approach. [3] (b) A tin nucleus has A = 120. Use R = R₀A^(1/3) with R₀ = 1.2 fm to compare your answer with its radius, and comment. [2]

3. A nucleus with nucleon number 125 has a measured radius of 6.0 fm. (a) Calculate R₀ in R = R₀A^(1/3). [2] (b) Calculate the nucleon number of a nucleus of radius 4.2 fm, taking R₀ = 1.2 fm. [2] (c) A graph of lg R against lg A is plotted for many nuclei. State the gradient and the intercept. [2]

4. (a) Starting from the formula for the volume of a sphere, prove that every nucleus has the same density whatever its nucleon number. [3] (b) Calculate this density, taking R₀ = 1.30 fm and the mass of a nucleon as 1.67 × 10⁻²⁷ kg. [2]

5. Electrons of energy 500 MeV are diffracted by a nucleus. The first minimum in the intensity is at 34.0° to the incident direction. Treat the electron energy as pc. (a) Calculate the electron wavelength. [2] (b) Use sin θ ≈ 1.22λ/D, with D the diameter of the nucleus, to find the nuclear radius. [3] (c) Describe the graph of intensity against angle. [2]

6. The nuclear mass of lithium-7 is 7.014357 u. Calculate its mass defect, its binding energy in MeV and its binding energy per nucleon. [4]

7. (a) A nuclear process releases 3.20 × 10⁻¹¹ J. Calculate the decrease in mass in kg. [2] (b) Express the mass decrease in u. [1] (c) A chemical reaction transfers 4.0 × 10⁵ J of energy from 1.0 kg of fuel. Calculate the mass change and explain why it is not noticed. [2]

8. Two carbon-12 nuclei (7.68 MeV per nucleon) fuse to form magnesium-24 (8.26 MeV per nucleon). (a) Calculate the energy released in MeV. [3] (b) State which region of the binding energy per nucleon graph this lies in, and explain why energy is released. [2]

9. A nucleus with A = 236 and binding energy per nucleon 7.59 MeV splits into two identical fragments, each with A = 118 and 8.52 MeV per nucleon. (a) Calculate the energy released in MeV. [3] (b) Express this in joules. [1]

10. Two helium-3 nuclei fuse: ³He + ³He → ⁴He + 2p. Nuclear masses: helium-3 3.014932 u, helium-4 4.001506 u; the proton mass is given above. (a) Calculate the energy released per reaction in MeV. [3] (b) Calculate how many reactions per second are needed for a power of 1.0 MW. [2] (c) Calculate the mass of helium-3 consumed each second. [2]

11. Describe the shape of the curve of mean binding energy per nucleon plotted against A, giving approximate values at two labelled points, and say which parts correspond to fusion and to fission. [4]

12. In one fission, uranium-235 absorbs a neutron: ²³⁵U + n → ¹³⁹Xe + ⁹⁵Sr + 2n. Atomic masses: uranium-235 235.043930 u, xenon-139 138.918787 u, strontium-95 94.919358 u; neutron 1.008665 u. (a) Calculate the energy released in one fission, in MeV. [4] (b) Calculate the energy released by 1.0 kg of uranium-235 undergoing this fission, in J, taking its molar mass as 0.235 kg mol⁻¹. [3] (c) A reactor has a thermal power of 3.0 GW. Calculate the mass of uranium-235 fissioned per day. [2]

Answers

1. (a) About 10⁻¹⁵ m (1 fm), anywhere from about 1 fm to 7 fm [1] (b) It assumes a head-on approach, a stationary nucleus and only the Coulomb force [1] (c) The wavelength must be comparable with, or smaller than, the nuclear size (a few fm); λ = hc/E, so this needs E of hundreds of MeV [1] [3] Examiner insight: In (a) give a number and unit; in (c) link wavelength to size, then to energy.

2. (a) E_k = 7.2 × 1.60 × 10⁻¹³ = 1.152 × 10⁻¹² J [1] r = (8.99 × 10⁹ × 2 × 50 × (1.60 × 10⁻¹⁹)²)/(1.152 × 10⁻¹²) [1] r = 2.0 × 10⁻¹⁴ m [1] (b) R = 1.2 × 120^(1/3) = 5.9 fm [1] The closest approach is about 3.4 times the radius, so the alpha particle stays outside the nucleus; the radius is below this limit [1] [5] Examiner insight: Check the charge factors (2e and 50e) and convert MeV to joules before substituting.

3. (a) A^(1/3) = 125^(1/3) = 5 [1] R₀ = 6.0/5 = 1.2 fm [1] (b) A = (R/R₀)³ = (4.2/1.2)³ = (3.5)³ = 42.9 [1] A = 43 (nearest whole number) [1] (c) Gradient 1/3 [1]; intercept lg R₀ [1] [6] Examiner insight: Round A to a whole number at the end; cubing and cube-rooting are easy to swap.

4. (a) Volume of nucleus = (4/3)πR³ = (4/3)πR₀³A [1] Mass ≈ A × (mass of one nucleon) [1] Density = 3m/(4πR₀³); A cancels, so density is the same for all nuclei [1] (b) ρ = (3 × 1.67 × 10⁻²⁷)/(4π × (1.30 × 10⁻¹⁵)³) [1] ρ = 1.8 × 10¹⁷ kg m⁻³ [1] [5] Examiner insight: On a proof question, write the volume and mass expressions before cancelling A.

5. (a) E = 500 × 1.60 × 10⁻¹³ = 8.0 × 10⁻¹¹ J [1] λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(8.0 × 10⁻¹¹) = 2.5 × 10⁻¹⁵ m [1] (b) D = 1.22λ/sin 34.0° [1] D = 5.4 × 10⁻¹⁵ m [1] R = D/2 = 2.7 × 10⁻¹⁵ m [1] (c) Central maximum at 0° falling to a first minimum, then weaker maxima separated by further minima [1] Each maximum is lower than the last, and the minima are shallow [1] [11] Examiner insight: Halve the diameter to get the radius; in the graph description, mention the falling heights.

6. Mass of separate nucleons = 3 × 1.007276 + 4 × 1.008665 = 7.056488 u [1] Mass defect = 7.056488 − 7.014357 = 0.042131 u [1] Binding energy = 0.042131 × 931.5 = 39.2 MeV [1] Per nucleon = 39.2/7 = 5.61 MeV [1] Examiner insight: State the mass defect in u before converting.

7. (a) Δm = ΔE/c² = 3.20 × 10⁻¹¹/(3.00 × 10⁸)² [1] Δm = 3.56 × 10⁻²⁸ kg [1] (b) 3.56 × 10⁻²⁸/1.66 × 10⁻²⁷ = 0.214 u [1] (c) Δm = 4.0 × 10⁵/(3.00 × 10⁸)² = 4.4 × 10⁻¹² kg [1] About 4 parts in 10¹² of the fuel mass, too small to measure [1] [5] Examiner insight: Part (c) tests that E = mc² applies to chemical changes too; “no mass change” is wrong.

8. (a) Before: 24 × 7.68 = 184.32 MeV [1] After: 24 × 8.26 = 198.24 MeV [1] Energy released = 13.9 MeV [1] (b) Fusion region (light nuclei, A below about 56, on the rising part of the curve) [1] Magnesium-24 sits higher on the curve, so it is more tightly bound and weighs less than two carbon-12 nuclei; that difference is released [1] [5] Examiner insight: Multiply each per-nucleon value by A before subtracting.

9. (a) Before: 236 × 7.59 = 1791.2 MeV [1] After: 236 × 8.52 = 2010.7 MeV [1] Energy released = 219 MeV [1] (b) 219 × 1.60 × 10⁻¹³ = 3.50 × 10⁻¹¹ J [1] [4] Examiner insight: Graph values have three figures, so quote no more than three.

10. (a) Mass before = 2 × 3.014932 = 6.029864 u [1] Mass after = 4.001506 + 2 × 1.007276 = 6.016058 u, difference 0.013806 u [1] Energy = 0.013806 × 931.5 = 12.9 MeV [1] (b) Energy per reaction = 12.86 × 1.60 × 10⁻¹³ = 2.058 × 10⁻¹² J [1] Reactions per second = 1.0 × 10⁶/2.058 × 10⁻¹² = 4.9 × 10¹⁷ s⁻¹ [1] (c) Two nuclei per reaction: 2 × 3.014932 × 1.66 × 10⁻²⁷ = 1.00 × 10⁻²⁶ kg [1] Mass per second = 4.86 × 10¹⁷ × 1.00 × 10⁻²⁶ = 4.9 × 10⁻⁹ kg [1] [11] Examiner insight: Subtract in the right order and carry unrounded values forward.

11. Rises steeply from low values for the lightest nuclei, for example about 1.1 MeV for hydrogen-2 [1] Peaks near iron-56 at about 8.8 MeV [1] Then falls slowly to about 7.6 MeV for uranium-235 [1] Fusion is to the left of the peak (light nuclei); fission is to the right (heavy nuclei) [1] Examiner insight: A labelled sketch is clearer than prose.

12. (a) Mass before = 235.043930 + 1.008665 = 236.052595 u [1] Mass after = 138.918787 + 94.919358 + 2 × 1.008665 = 235.855475 u [1] Difference = 0.197120 u [1] Energy = 0.197120 × 931.5 = 183.6 MeV [1] (b) Energy per fission = 183.6 × 1.60 × 10⁻¹³ = 2.94 × 10⁻¹¹ J [1] Nuclei in 1.0 kg = 6.02 × 10²³/0.235 = 2.56 × 10²⁴ [1] Energy = 2.94 × 10⁻¹¹ × 2.56 × 10²⁴ = 7.5 × 10¹³ J [1] (c) Fissions per second = 3.0 × 10⁹/2.94 × 10⁻¹¹ = 1.02 × 10²⁰; mass per day = 1.02 × 10²⁰ × 86400/2.56 × 10²⁴ [1] Mass ≈ 3.4 kg per day [1] [9] Examiner insight: Check that the proton numbers balance (92 = 54 + 38) before using atomic masses, and show each conversion on its own line.

Where marks are usually lost

• Using e instead of 2e for the alpha particle, or leaving MeV in the Coulomb equation.
• Calling the closest approach “the radius” without comment.
• Cubing or cube-rooting the wrong way in R = R₀A^(1/3).
• Not halving the diameter in the diffraction question.
• Subtracting masses the wrong way round.
• Forgetting that atomic masses include electrons.

Next steps

• Re-read the study guide and the revision notes.
• Use the OxfordAQA Physics hub and tick off topics on the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.12.1 and 3.12.2.