Question

A radioactive sample \(S_{1}\) having activity \(A_{1}\) has twice the number of nuclei as another sample of \(S_{2}\) of activity \(A_{2}\). If \(A_{2}=2A_{1}\) then the ratio of half life of \(S_{1}\) to the half life of \(S_{2}\) is

• A. \(4\)
• B. \(2\)
• C. \(0.25\)
• D. \(0.75\)

Hint:

Activity equals decay constant times the number of nuclei, and half life is inversely related to the decay constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Activity is \(A=\lambda N\) and half life is \(T=\dfrac{0.693}{\lambda}\).

Step 2: Given Data
\(N_1=2N_2\) and \(A_2=2A_1\).

Step 3: Decay Constants
\(\lambda_1=\dfrac{A_1}{N_1}\) and \(\lambda_2=\dfrac{A_2}{N_2}=\dfrac{2A_1}{N_1/2}=\dfrac{4A_1}{N_1}\).
So \(\dfrac{\lambda_1}{\lambda_2}=\dfrac{1}{4}\).

Step 4: Half Life Ratio
\(\dfrac{T_1}{T_2}=\dfrac{\lambda_2}{\lambda_1}=4\), option (A).

Final Answer:
The half life of the first sample is 4 times that of the second. \[ \boxed{\text{(A) }4} \]