Question

Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives \(T_{1/2}(A)\ :\ T_{1/2}(B)\ :\ T_{1/2}(C)\) are in the ratio :

• A. 3 : 2 : 1
• B. 2 : 1 : 1
• C. 4 : 3 : 1
• D. 2 : 1 : 3

Hint:

\(\ln R = \ln R_0 - \lambda t\), so the slope magnitude of each line is \(\lambda\) and \(T_{1/2} = \ln 2/\lambda\). Read the intercepts from the graph.

Answer 1

The Correct Option is D

\(R = R_0 e^{-\lambda t} \Rightarrow \ln R = \ln R_0 - \lambda t\); the slope of the line is \(-\lambda\).
From the graph (ln R vs t, t in years):
A: from 6 at t = 0 to 0 at t = 10, so \(\lambda_A = 6/10 = 0.6\).
B: from 6 to 0 at t = 5, so \(\lambda_B = 6/5 = 1.2\).
C: from 2 to 0 at t = 5, so \(\lambda_C = 2/5 = 0.4\).
\(T_{1/2} = \dfrac{\ln 2}{\lambda} \propto \dfrac1\lambda\).
\(T_A : T_B : T_C = \dfrac1{0.6} : \dfrac1{1.2} : \dfrac1{0.4} = \dfrac53 : \dfrac56 : \dfrac52\).
Multiplying by 6/5: \(2 : 1 : 3\).
Answer: (4) 2 : 1 : 3