A2 · Paper 4 practice · Nuclear physics

Mass, binding, and decay.

Ten original structured questions in the style of Paper 4, covering the whole of Nuclear physics: mass defect and binding energy, E = mc², fusion and fission energy, and the random exponential law of radioactive decay. The later questions tie a decay's mass change to its energy. Each is tagged with its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Data: c = 3.00 × 10⁸ m s⁻¹ · 1 u = 1.66 × 10⁻²⁷ kg · 1 u ≡ 931 MeV · 1 eV = 1.60 × 10⁻¹⁹ J · ln2 = 0.693 · Nₐ = 6.02 × 10²³ mol⁻¹
Keep these straight

Defect, energy, decay.

01
Paper 4
[9 marks]

A helium-4 nucleus is made of 2 protons and 2 neutrons. Use: proton mass = 1.00728 u, neutron mass = 1.00867 u, helium-4 nucleus mass = 4.00151 u.

(a) State what is meant by the mass defect of a nucleus. [1]

(b) Calculate the mass defect of the helium-4 nucleus, in u. [2]

(c) Calculate the binding energy, in MeV. [2]

(d) Calculate the binding energy per nucleon, in MeV. [2]

(e) State what the binding energy per nucleon tells you about the nucleus. [2]

  • (a)The difference between the total mass of the separate nucleons and the (smaller) mass of the nucleus ✓
  • (b)Δm = 2(1.00728) + 2(1.00867) − 4.00151 ✓ = 4.03190 − 4.00151 = 0.03039 u ✓
  • (c)BE = 0.03039 × 931 ✓ = 28.3 MeV ✓
  • (d)28.3 / 4 ✓ = 7.07 MeV per nucleon ✓
  • (e)It measures how tightly bound, and so how stable, the nucleus is ✓; a higher value means more energy is needed per nucleon to pull it apart ✓
02
Paper 4
[6 marks]

Show how the conversion 1 u ≡ 931 MeV comes from E = mc².

(a) Calculate the energy equivalent of a mass of 1 u, in joules. [2]

(b) Convert this energy to MeV, and compare with 931 MeV. [2]

(c) State why mass and energy are treated as equivalent in nuclear physics. [2]

  • (a)E = mc² = 1.66 × 10⁻²⁷ × (3.00 × 10⁸)² ✓ = 1.49 × 10⁻¹⁰ J ✓
  • (b)E = 1.49 × 10⁻¹⁰ / 1.60 × 10⁻¹³ ✓ = 933 MeV ≈ 931 MeV ✓
  • (c)By E = mc² a change in mass corresponds to a change in energy ✓; in nuclear reactions the "missing" mass appears as released energy ✓
03
Paper 4
[8 marks]

The binding energy per nucleon rises steeply for light nuclei, peaks near iron (A ≈ 56), then falls slowly toward the heaviest nuclei.

(a) State which nuclei are the most stable, and why. [2]

(b) Explain, using the curve, why energy is released when a heavy nucleus undergoes fission. [3]

(c) Explain why energy is released when two light nuclei undergo fusion. [2]

(d) State one reason fusion is hard to achieve on Earth. [1]

  • (a)Nuclei near the peak (around iron) ✓; they have the greatest binding energy per nucleon, so are the most tightly bound ✓
  • (b)A heavy nucleus lies to the right of the peak ✓; its fragments lie nearer the peak with a greater binding energy per nucleon ✓; the increase in binding energy is released as energy ✓
  • (c)Light nuclei lie to the left; fusing them gives a nucleus nearer the peak with higher binding energy per nucleon ✓, so energy is released ✓
  • (d)The nuclei must overcome strong electrostatic repulsion, needing very high temperatures and pressures ✓
04
Paper 4
[8 marks]

A deuterium and a tritium nucleus fuse: ²₁H + ³₁H → ⁴₂He + ¹₀n. Masses (u): ²₁H = 2.01410, ³₁H = 3.01605, ⁴₂He = 4.00260, n = 1.00867.

(a) Calculate the change in mass for the reaction, in u. [2]

(b) Calculate the energy released, in MeV. [2]

(c) Calculate the energy released, in joules. [2]

(d) State what happens to this energy. [2]

  • (a)Δm = (2.01410 + 3.01605) − (4.00260 + 1.00867) ✓ = 5.03015 − 5.01127 = 0.01888 u ✓
  • (b)E = 0.01888 × 931 ✓ = 17.6 MeV ✓
  • (c)E = 17.6 × 10⁶ × 1.60 × 10⁻¹⁹ ✓ = 2.81 × 10⁻¹² J ✓
  • (d)It appears as kinetic energy of the products (the helium nucleus and the neutron) ✓✓
05
Paper 4
[6 marks]

The Sun radiates energy at a power of 3.8 × 10²⁶ W, all of it coming from a loss of mass.

(a) State the relation between the power radiated and the rate of mass loss. [1]

(b) Calculate the mass lost by the Sun each second. [3]

(c) Suggest why this enormous rate barely changes the Sun's mass over a human lifetime. [2]

  • (a)P = (Δm/Δt) c², so the rate of mass loss = P / c² ✓
  • (b)Δm/Δt = 3.8 × 10²⁶ / (3.00 × 10⁸)² ✓ = 3.8 × 10²⁶ / 9.0 × 10¹⁶ ✓ = 4.2 × 10⁹ kg s⁻¹ ✓
  • (c)The Sun's mass (about 2 × 10³⁰ kg) is vastly larger ✓; even over decades the fraction lost is negligible ✓
06
Paper 4
[8 marks]

A radioactive source contains 5.0 × 10²₀ undecayed nuclei of a nuclide whose decay constant is 2.0 × 10⁻⁸ s⁻¹.

(a) State what is meant by the decay constant. [1]

(b) Calculate the activity of the source, in Bq. [2]

(c) Calculate the half-life of the nuclide, in seconds. [2]

(d) State and explain how the activity changes as time passes. [3]

  • (a)The probability per unit time that a given nucleus decays ✓
  • (b)A = λN = 2.0 × 10⁻⁸ × 5.0 × 10²₀ ✓ = 1.0 × 10¹³ Bq ✓
  • (c)t½ = ln2 / λ = 0.693 / 2.0 × 10⁻⁸ ✓ = 3.5 × 10⁷ s ✓
  • (d)The activity falls ✓; as nuclei decay N decreases, and since A = λN the activity decreases too ✓; it does so exponentially, halving each half-life ✓
07
Paper 4
[8 marks]

A radioactive nuclide has a half-life of 8.0 days.

(a) Calculate its decay constant, in day⁻¹. [2]

(b) State the fraction of the original nuclei remaining after 24 days. [2]

(c) Calculate the fraction remaining after 20 days. [3]

(d) State whether the half-life depends on how much of the sample is left. [1]

  • (a)λ = 0.693 / 8.0 ✓ = 0.0866 day⁻¹ ✓
  • (b)24 days = 3 half-lives ✓; fraction = (½)³ = 1/8 ✓
  • (c)N/N₀ = e⁻λᵗ = e⁻⁽⁰.₀₈⁶₆ × ²₀⁾ ✓ = e⁻¹.⁷³ ✓ = 0.18 (about 18%) ✓
  • (d)No, the half-life is constant for the nuclide, independent of the amount remaining ✓
08
Paper 4
[8 marks]

A sample of a nuclide of half-life 15 hours has an initial activity A₀.

(a) Calculate the time for the activity to fall to A₀/16. [2]

(b) Calculate the decay constant, in hour⁻¹. [2]

(c) Calculate the time for the activity to fall to 10% of A₀. [3]

(d) State why a measured count rate is usually less than the activity. [1]

  • (a)A₀/16 = (½)⁴ A₀, so 4 half-lives ✓ = 4 × 15 = 60 hours ✓
  • (b)λ = 0.693 / 15 ✓ = 0.0462 hour⁻¹ ✓
  • (c)t = (1/λ) ln(A₀/A) = (1/0.0462) ln(10) ✓ = 21.6 × 2.303 ✓ = 50 hours ✓
  • (d)The detector intercepts only a fraction of the radiation, and some is absorbed or misses it ✓
09
Paper 4
[8 marks]

Carbon-14 has a half-life of 5730 years. A living sample has a C-14 activity of 0.25 Bq per gram of carbon. An old wooden artefact reads 0.0625 Bq per gram.

(a) State the assumption that makes carbon dating possible. [1]

(b) Calculate the age of the artefact using the number of half-lives. [3]

(c) Confirm the age using N = N₀e⁻λᵗ. [3]

(d) Suggest one reason very old samples cannot be dated this way. [1]

  • (a)The C-14 activity per gram in living matter has stayed constant over time ✓
  • (b)0.0625 / 0.25 = 1/4 = (½)² ✓, so 2 half-lives ✓ = 2 × 5730 = 11460 years ✓
  • (c)λ = 0.693/5730; t = (1/λ) ln(4) = 5730 × (ln4/ln2) ✓ = 5730 × 2 ✓ = 11460 years ✓
  • (d)After many half-lives the remaining activity is too small to measure above background ✓
10
Paper 4
[11 marks]

Radium-226 decays by alpha emission to radon: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He. Masses (u): Ra-226 = 226.0254, Rn-222 = 222.0176, He-4 = 4.0026. The half-life of radium-226 is 1600 years.

(a) Show that the nucleon and proton numbers balance in the equation. [2]

(b) Calculate the change in mass, in u. [2]

(c) Calculate the energy released, in MeV and in joules. [3]

(d) Calculate the decay constant of radium-226, in s⁻¹ (1 year = 3.15 × 10⁷ s). [2]

(e) State whether the energy released per decay changes as the sample ages, and why. [2]

  • (a)Nucleon: 226 = 222 + 4 ✓; proton: 88 = 86 + 2 ✓
  • (b)Δm = 226.0254 − (222.0176 + 4.0026) ✓ = 226.0254 − 226.0202 = 0.0052 u ✓
  • (c)E = 0.0052 × 931 ✓ = 4.8 MeV ✓ = 4.8 × 10⁶ × 1.60 × 10⁻¹⁹ = 7.7 × 10⁻¹³ J ✓
  • (d)t½ = 1600 × 3.15 × 10⁷ = 5.04 × 10¹⁰ s; λ = 0.693 / 5.04 × 10¹⁰ ✓ = 1.4 × 10⁻¹¹ s⁻¹ ✓
  • (e)No ✓; each decay involves the same nuclei and the same mass change, so the energy per decay is fixed; only the number of decays per second falls ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.