A-Level Physics May/June 2025 Q10(b)(ii): Determine, in terms of T, the time at which the two samples will have equal activities.

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  The activity $A$ of a sample at time $t$ is given by the equation for radioactive decay: $A = A_0 e^{-\lambda t}$

  Where $A_0$ is the initial activity and $\lambda$ is the decay constant. The decay constant is related to the half-life $T_{1/2}$ by: $\lambda = \frac{\ln 2}{T_{1/2}}$

  For sample X, $A_{X,0} = 4A$ and $T_{1/2, X} = 2T$. So, $\lambda_X = \frac{\ln 2}{2T}$. Its activity at time $t$ is: $A_X(t) = 4A \exp\left(-\frac{t \ln 2}{2T}\right)$

  For sample Y, $A_{Y,0} = A$ and $T_{1/2, Y} = 3T$. So, $\lambda_Y = \frac{\ln 2}{3T}$. Its activity at time $t$ is: $A_Y(t) = A \exp\left(-\frac{t \ln 2}{3T}\right)$

  We need to find the time $t$ when the activities are equal, so we set $A_X(t) = A_Y(t)$: $4A \exp\left(-\frac{t \ln 2}{2T}\right) = A \exp\left(-\frac{t \ln 2}{3T}\right)$

  Divide both sides by $A$ and rearrange the exponential terms: $4 = \frac{\exp\left(-\frac{t \ln 2}{3T}\right)}{\exp\left(-\frac{t \ln 2}{2T}\right)}$

  Using the rule $\frac{e^a}{e^b} = e^{a-b}$: $4 = \exp\left(-\frac{t \ln 2}{3T} - \left(-\frac{t \ln 2}{2T}\right)\right)$ $4 = \exp\left(\frac{t \ln 2}{2T} - \frac{t \ln 2}{3T}\right)$

  Factor out the common terms: $4 = \exp\left(t \ln 2 \left(\frac{1}{2T} - \frac{1}{3T}\right)\right)$ $4 = \exp\left(t \ln 2 \left(\frac{3 - 2}{6T}\right)\right)$ $4 = \exp\left(\frac{t \ln 2}{6T}\right)$

  Take the natural logarithm (ln) of both sides: $\ln(4) = \ln\left(\exp\left(\frac{t \ln 2}{6T}\right)\right)$ $\ln(2^2) = \frac{t \ln 2}{6T}$ $2 \ln 2 = \frac{t \ln 2}{6T}$

  Cancel $\ln 2$ from both sides: $2 = \frac{t}{6T}$

  Solve for $t$: $t = 2 \times 6T$ $t = 12T$

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  How the marks are awarded

  • C1 — Correctly writing the activity equation for at least one of the samples, for example, $A_X(t) = 4A \exp(-\frac{t \ln 2}{2T})$, which substitutes the initial activity and the decay constant in terms of half-life.
  • C1 — Equating the two expressions for the activities of sample X and sample Y, as shown in the line: $4A \exp(-\frac{t \ln 2}{2T}) = A \exp(-\frac{t \ln 2}{3T})$.
  • A1 — Correctly solving the exponential equation through algebraic manipulation and use of logarithms to arrive at the final, correct answer of $t = 12T$.

  Common mistakes

  • Algebraic errors when combining exponents, for example adding the magnitudes of the negative exponents instead of subtracting, leading to an incorrect term in the final exponential.
  • Mistakes in applying logarithms, such as incorrectly simplifying ln(4) or being unable to cancel the ln(2) term.
  • Using an incorrect formula for decay, such as $A = A_0 e^{-t/T_{1/2}}$ instead of $A = A_0 e^{-(\ln 2 / T_{1/2})t}$. This leads to an answer of $t = 12T \ln 2$.
  • Swapping the initial activities or half-lives between the two samples, which leads to a non-physical negative time or an incorrect positive time if the negative sign is ignored.

  Examiner tip: Practice rearranging equations involving natural logarithms and exponents, as this is a key mathematical skill for solving problems in radioactive decay, capacitor discharge, and other exponential processes.

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