Nuclear physics (radioactive decay, nuclear radius, binding energy) – A-level Physics exam questions with mark scheme

1Core3 marks

(a) Radium-226 ($^{226}_{88}$Ra) decays by alpha emission to form radon (Rn). Write the equation for this decay. [1]

(b) Carbon-14 ($^{14}_{6}$C) decays by β$^-$ emission to form nitrogen (N). Write the equation for this decay. [2]

(a) $^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{Rn} +\ ^{4}_{2}\alpha$ (accept He)(1)
(b) $^{14}_{6}\text{C} \rightarrow\ ^{14}_{7}\text{N} +\ ^{0}_{-1}\beta$ (accept e)(1)
(b) $+\ \bar{\nu}_e$ (electron antineutrino) included (neutrino without bar scores 0 for this mark)(1)
Answer: (a) Ra-226 → Rn-222 + α (b) C-14 → N-14 + β⁻ + antineutrino
Examiner tip: β⁻ decay always produces an electron antineutrino — leaving it out, or writing a neutrino, loses a mark.

2Standard4 marks

A sample of iodine-131 has an initial activity of 5.0 MBq. The half-life of iodine-131 is 8.0 days.

(a) Calculate the decay constant of iodine-131 in s$^{-1}$. [1]

(b) Calculate the initial number of iodine-131 nuclei in the sample. [1]

(a) $\lambda = \frac{\ln 2}{8.0 \times 24 \times 3600}$ = 1.0 × 10$^{-6}$ s$^{-1}$(1)
(b) $N = \frac{A}{\lambda} = \frac{5.0\times10^6}{1.0\times10^{-6}}$ = 5.0 × 10$^{12}$(1)
(c) $A = A_0e^{-\lambda t}$ with consistent units, e.g. $5.0\,e^{-(\ln 2/8.0)\times 20}$(1)
(c) $A$ = 0.88 MBq (8.8 × 10$^5$ Bq)(1)
Answer: (a) 1.0 × 10$^{-6}$ s$^{-1}$ (b) 5.0 × 10$^{12}$ (c) 0.88 MBq
Examiner tip: λ and t must be in matching units; for A = λN, λ must be in s⁻¹ because activity is in Bq.

3Standard4 marks

A gamma source is placed 0.20 m from a Geiger–Müller tube. The measured count rate is 480 counts per minute. The background count rate is 30 counts per minute.

(a) Calculate the count rate you would expect to measure when the tube is 0.50 m from the source. Assume no absorption of gamma radiation in the air. [3]

(b) Suggest why the distance from the source to the tube may be subject to a systematic error. [1]

(a) corrected count rate at 0.20 m = 480 − 30 = 450 counts min$^{-1}$(1)
(a) inverse square law: $450 \times \left(\frac{0.20}{0.50}\right)^2$ = 72 counts min$^{-1}$(1)
(a) measured count rate = 72 + 30 = 102 counts min$^{-1}$ (allow 100)(1)
(b) the position of the source inside its sealed holder/capsule is not known exactly, so the measured distance is not the true distance from the source (accept: radiation is detected at an unknown point inside the GM tube, not at its window; ruler zero/end not aligned with source or detector)(1)
Answer: (a) 102 counts min$^{-1}$
Examiner tip: Only the source's count rate obeys the inverse-square law — subtract background first, then add it back for a 'measured' value.

4Hard6 marks

The radius $R$ of a nucleus with nucleon number $A$ is given by $R = r_0A^{1/3}$, where $r_0$ = 1.2 fm.

(a) Calculate the radius of a gold-197 nucleus. [1]

(b) Show that the density of nuclear matter does not depend on $A$, and calculate its value. Take the mass of a nucleon as 1.661 × 10$^{-27}$ kg. [3]

(c) In a scattering experiment, alpha particles of kinetic energy 5.0 MeV are directed at gold nuclei (proton number 79). Calculate the distance of closest approach of an alpha particle travelling head-on towards a gold nucleus. [2]

(a) $R = 1.2 \times 197^{1/3}$ = 7.0 fm (6.98 × 10$^{-15}$ m)(1)
(b) $\rho = \frac{Am}{\frac{4}{3}\pi r_0^3 A}$ — volume ∝ $R^3$ ∝ $A$ and mass ∝ $A$(1)
(b) so $A$ cancels: $\rho = \frac{3m}{4\pi r_0^3}$, independent of $A$(1)
(b) $\rho$ = 2.3 × 10$^{17}$ kg m$^{-3}$(1)
(c) KE = electric PE at closest approach: $5.0\times1.60\times10^{-13} = \frac{(79e)(2e)}{4\pi\varepsilon_0 r}$(1)
(c) $r$ = 4.5 × 10$^{-14}$ m (4.55)(1)
Answer: (a) 7.0 fm (b) 2.3 × 10$^{17}$ kg m$^{-3}$ (c) 4.5 × 10$^{-14}$ m
Examiner tip: Convert MeV to joules (× 1.60 × 10⁻¹³) and use charge 2e for the alpha particle.

5Standard5 marks

(a) Define the binding energy of a nucleus. [1]

(b) Calculate the binding energy, in MeV, of a helium-4 nucleus.

mass of helium-4 nucleus = 4.00151 u; mass of proton = 1.00728 u; mass of neutron = 1.00867 u; 1 u is equivalent to 931.5 MeV. [3]

(a) the energy needed to separate a nucleus into its individual (separate) protons and neutrons (accept: energy released when the nucleus forms from its nucleons)(1)
(b) total mass of nucleons $= 2(1.00728) + 2(1.00867)$ = 4.03190 u(1)
(b) mass defect $= 4.03190 - 4.00151$ = 0.03039 u(1)
(b) binding energy $= 0.03039 \times 931.5$ = 28.3 MeV(1)
(c) $28.3 / 4$ = 7.1 MeV per nucleon (7.08)(1)
Answer: (b) 28.3 MeV (c) 7.08 MeV
Examiner tip: Keep all the decimal places of the masses until the subtraction — rounding early wipes out the mass defect.

6Hard6 marks

The graph shows how the binding energy per nucleon varies with nucleon number $A$.

(a) Use the graph to explain why energy is released both when light nuclei fuse and when very heavy nuclei undergo fission. [3]

(b) One possible fission reaction is

$$^{235}_{92}\text{U} +\ ^{1}_{0}\text{n} \rightarrow\ ^{141}_{56}\text{Ba} +\ ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$$

The binding energies per nucleon are: U-235 7.59 MeV, Ba-141 8.33 MeV, Kr-92 8.51 MeV. Calculate the energy released in this fission. [3]

(a) energy is released when the products have a greater (total) binding energy than the original nucleus / products have higher binding energy per nucleon(1)
(a) fusion: light nuclei (A below about 56) combine to make nuclei further up the curve, nearer the peak, with higher binding energy per nucleon(1)
(a) fission: heavy nuclei split into medium nuclei closer to the peak (iron/nickel region), which have higher binding energy per nucleon(1)
(b) binding energy of products $= (141\times8.33) + (92\times8.51) = 1174.5 + 782.9$ = 1957.5 MeV(1)
(b) binding energy of U-235 $= 235\times7.59$ = 1783.7 MeV(1)
(b) energy released = 1957.5 − 1783.7 = 174 MeV (allow 170–174 MeV; ignore free neutrons, which have no binding energy)(1)
Answer: (b) 174 MeV
Examiner tip: Multiply binding energy per nucleon by the nucleon number of each nucleus before subtracting — free neutrons have zero binding energy.

7Hard5 marks

Carbon-14 has a half-life of 5730 years. Living wood has a carbon-14 activity of 0.25 Bq per gram of carbon. A sample of wood from an archaeological site has an activity of 0.080 Bq per gram of carbon.

(a) Calculate the age of the sample in years. [3]

(b) Explain why carbon-14 dating would not give a reliable age for a sample about 100 000 years old. [2]

(a) $\lambda = \frac{\ln 2}{5730}$ = 1.21 × 10$^{-4}$ year$^{-1}$ (allow equivalent route: $\left(\tfrac{1}{2}\right)^n = \frac{0.080}{0.25}$ gives $n = 1.64$ half-lives for this mark)(1)
(a) $0.080 = 0.25\,e^{-\lambda t}$ or $t = \frac{1}{\lambda}\ln\frac{0.25}{0.080}$ (or $t = 1.64 \times 5730$)(1)
(a) $t$ = 9400 years (9420)(1)
(b) 100 000 years is about 17 half-lives, so the activity would be extremely small (≈ 1 × 10$^{-6}$ Bq per gram) / very few C-14 nuclei remain(1)
(b) this is far below the background count rate, so the (percentage) uncertainty in the measured activity, and hence the age, is very large (accept: count dominated by random fluctuations of background)(1)
Answer: (a) 9.4 × 10$^3$ years
Examiner tip: Take logs carefully: t = (1/λ) ln(A₀/A) — inverting the ratio gives a negative age.

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