Radioactive decay and nuclear instability – 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.

3Hard5 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.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session