Question

At a given instant, say t = 0, two radioactive substance A and B have equal activities. The ratio \(\frac{R_B}{R_A}\) of their activities after time t itself decays with time t as \(e^{-3t}\). If the half-life of A is \(\ln 2\), the half-life of B is :

• A. \(4\ln 2\)
• B. \(\frac{\ln 2}{2}\)
• C. \(\frac{\ln 2}{4}\)
• D. \(2\ln 2\)

Hint:

With equal initial activities, R_B/R_A = exp(-(lambda_B - lambda_A) t). Compare with e^(-3t) to get lambda_B, then T_half = ln2/lambda.

Answer 1

The Correct Option is C

Activity \(R = R_0 e^{-\lambda t}\) and \(R_{0A} = R_{0B}\) (equal at t = 0), so \(\frac{R_B}{R_A} = e^{-(\lambda_B - \lambda_A)t}\).
Given this equals \(e^{-3t}\): \(\lambda_B - \lambda_A = 3\).
Half-life of A \(= \frac{\ln 2}{\lambda_A} = \ln 2 \Rightarrow \lambda_A = 1\).
So \(\lambda_B = 4\) and the half-life of B \(= \frac{\ln 2}{\lambda_B} = \frac{\ln 2}{4}\).
Answer: (3) \(\frac{\ln 2}{4}\)