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A2 · Practice questions · Radioactive decay

Random nuclei, a certain law.

Six original Cambridge-style questions on random and spontaneous decay, activity A = λN, the decay constant and half-life linked by λt½ = ln 2, and the exponential law N = N₀e^(−λt).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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The rate is proportional to the number left.

01
Analysis
[2 marks]

Radioactive decay is described as both random and spontaneous. Explain the meaning of each term.

  • Random: it cannot be predicted which nucleus will decay next, or when a particular nucleus will decay; only the probability of decay per unit time is fixed. ✓
  • Spontaneous: the decay is not caused or affected by external conditions such as temperature, pressure or chemical state, or by other nuclei. ✓
02
Calculation
[3 marks]

Cobalt-60 has a half-life of 5.27 years. A source contains 1.0 µg of cobalt-60. Calculate the decay constant of cobalt-60 in s⁻¹, the number of cobalt-60 nuclei in the source, and its activity. (1 year = 3.15 × 10⁷ s; mass of a cobalt-60 nucleus = 60 u, 1 u = 1.66 × 10⁻²⁷ kg.)

  • t½ = 5.27 × 3.15 × 10⁷ = 1.66 × 10⁸ s, so λ = ln 2 / t½ = 0.693 / 1.66 × 10⁸ = 4.2 × 10⁻⁹ s⁻¹. ✓
  • N = 1.0 × 10⁻⁹ / (60 × 1.66 × 10⁻²⁷) = 1.0 × 10¹⁹ nuclei. ✓
  • A = λN = 4.17 × 10⁻⁹ × 1.0 × 10¹⁹ = 4.2 × 10¹⁰ Bq. ✓
03
Analysis
[3 marks]

Define the decay constant and the half-life of a radioactive nuclide. Starting from N = N₀e^(−λt), show that the half-life is given by t½ = ln 2 / λ.

  • The decay constant λ is the probability per unit time that an individual nucleus will decay (so that A = λN). ✓
  • The half-life is the time taken for the number of undecayed nuclei in a sample (and so its activity) to fall to half the initial value. ✓
  • Put N = N₀/2 at t = t½: ½ = e^(−λt½), so taking natural logs ln(½) = −λt½, giving λt½ = ln 2 and t½ = ln 2 / λ. ✓
04
Calculation
[3 marks]

Technetium-99m, used in medical imaging, has a half-life of 6.0 hours. A patient is injected with a sample of activity 400 MBq. Calculate the decay constant in hour⁻¹, the activity 9.0 hours after the injection, and the time after injection at which the activity has fallen to 25 MBq.

  • λ = ln 2 / t½ = 0.693 / 6.0 = 0.116 hour⁻¹. ✓
  • A = A₀e^(−λt) = 400 × e^(−0.1155 × 9.0) = 400 × 0.354 = 140 MBq. ✓
  • 400 / 25 = 16 = 2⁴, so four half-lives: t = 4 × 6.0 = 24 hours (or t = ln 16 / 0.1155 = 24 hours). ✓
05
Calculation
[3 marks]

The activity of a freshly prepared source is 6.0 × 10⁴ Bq. After 20 minutes it has fallen to 3.6 × 10⁴ Bq. Calculate the decay constant of the nuclide in min⁻¹, its half-life, and the number of undecayed nuclei that were present when the source was prepared.

  • A = A₀e^(−λt), so λ = (1/t) ln(A₀/A) = (1/20) ln(6.0/3.6) = 0.511 / 20 = 0.0255 min⁻¹. ✓
  • t½ = ln 2 / λ = 0.693 / 0.0255 = 27 minutes. ✓
  • In s⁻¹, λ = 0.0255 / 60 = 4.26 × 10⁻⁴ s⁻¹, so N₀ = A₀/λ = 6.0 × 10⁴ / 4.26 × 10⁻⁴ = 1.4 × 10⁸ nuclei. ✓
06
Calculation
[4 marks]

A domestic smoke detector contains americium-241, which has a half-life of 432 years, with an activity of 37 kBq. (1 year = 3.15 × 10⁷ s; mass of an americium-241 nucleus = 241 u, 1 u = 1.66 × 10⁻²⁷ kg.)

(a) Calculate the decay constant of americium-241 in s⁻¹. (b) Calculate the number of americium-241 nuclei in the detector and hence their mass. (c) Explain, with a calculation, why the detector does not need replacing because of decay during its ten-year working life.

  • (a) t½ = 432 × 3.15 × 10⁷ = 1.36 × 10¹⁰ s, so λ = 0.693 / 1.36 × 10¹⁰ = 5.1 × 10⁻¹¹ s⁻¹. ✓
  • (b) N = A / λ = 3.7 × 10⁴ / 5.09 × 10⁻¹¹ = 7.3 × 10¹⁴ nuclei. ✓
  • Mass = 7.27 × 10¹⁴ × 241 × 1.66 × 10⁻²⁷ = 2.9 × 10⁻¹⁰ kg (about 0.3 µg). ✓
  • (c) In 10 years, A/A₀ = e^(−0.693 × 10 / 432) = e^(−0.016) = 0.98, so the activity falls by only about 2%, far too little to matter. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.