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9702 · 23.2

Radioactive decay — practice questions

Practice and worked examples for 9702 Radioactive decay. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A radioactive isotope has a decay constant of $2.5 \times 10^{-3}$ s^{-1}. Calculate its half-life in seconds. Then, if a sample initially has an activity of 480 Bq, what will its activity be after 5.0 minutes?

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  1. Calculate the half-life (t1/2t_{1/2}): Using the formula t1/2=ln⁡2λt_{1/2} = \frac{\ln 2}{\lambda} t1/2=0.6932.5×10−3 s−1t_{1/2} = \frac{0.693}{2.5 \times 10^{-3} \text{ s}^{-1}} t1/2=277.2t_{1/2} = 277.2 s So, the half-life is approximately 277 s (to 3 s.f.).
  2. Convert time to seconds: Time t=5.0 minutes=5.0×60 s=300 st = 5.0 \text{ minutes} = 5.0 \times 60 \text{ s} = 300 \text{ s}
  3. Calculate the activity (A) after 300 s: Using the exponential decay formula for activity: A=A0e−λtA = A_0 e^{-\lambda t} A=480 Bq×e(−2.5×10−3s−1×300 s)A = 480 \text{ Bq} \times e^{(-2.5 \times 10^{-3} \mathrm{s}^{-1} \times 300 \text{ s})} A=480×e(−0.75)A = 480 \times e^{(-0.75)} A=480×0.47236...A = 480 \times 0.47236... A=226.73 BqA = 226.73 \text{ Bq} Therefore, the activity after 5.0 minutes will be approximately 227 Bq (to 3 s.f.).

Worked example 2

An ancient wooden artifact is found to have a carbon-14 activity of 0.180 Bq per gram of carbon. A modern, living sample of wood has an activity of 0.250 Bq per gram of carbon. Given that the half-life of carbon-14 is 5730 years, calculate the age of the artifact.

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  1. Calculate the decay constant (\lambda): The half-life is given in years, so we should calculate \lambda in units of year^{-1}. t1/2=ln⁡2λ  ⟹  λ=ln⁡2t1/2t_{1/2} = \frac{\ln 2}{\lambda} \implies \lambda = \frac{\ln 2}{t_{1/2}} λ=0.6935730 years=1.2096×10−4 year−1\lambda = \frac{0.693}{5730 \text{ years}} = 1.2096 \times 10^{-4} \text{ year}^{-1}
  2. Use the exponential decay formula for activity: The formula is A=A0e−λtA = A_0 e^{-\lambda t}, where AA is the current activity and A0A_0 is the initial activity. A=0.180A = 0.180 Bq A0=0.250A_0 = 0.250 Bq $0.180 = 0.250 \times e^{(-1.2096 \times 10^{-4} \times t)}$
  3. Solve for time (t): First, isolate the exponential term. 0.1800.250=e(−1.2096×10−4×t)\frac{0.180}{0.250} = e^{(-1.2096 \times 10^{-4} \times t)} $0.72 = e^{(-1.2096 \times 10^{-4} \times t)}$ Take the natural logarithm of both sides to remove the exponential. ln⁡(0.72)=−1.2096×10−4×t\ln(0.72) = -1.2096 \times 10^{-4} \times t −0.3285=−1.2096×10−4×t-0.3285 = -1.2096 \times 10^{-4} \times t Now, solve for t. t=−0.3285−1.2096×10−4=2715.8 yearst = \frac{-0.3285}{-1.2096 \times 10^{-4}} = 2715.8 \text{ years}
  4. Final Answer: The age of the artifact is approximately 2720 years (to 3 significant figures).
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