Question

The activity of radioactive sample is measured as \(\mathrm{N}_0\) counts per minute at time \(\mathrm{t}=0\), and \(\frac{\mathrm{N}_0}{\mathrm{e}}\) counts per minute at time \(\mathrm{t}=3\) minute, The activity reduces to half its value in time (in minute)

• A. \(\frac{1}{3}\mathrm{log}_{\mathrm{e}}2\)
• B. \(3\mathrm{log}_{\mathrm{e}}2\)
• C. \(3\mathrm{log}_{10}2\)
• D. \(\frac{3}{\mathrm{log}_{10}2}\)

Hint:

Find the decay constant from the drop to N0/e in 3 minutes, then use the half life formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Activity decays as \(A=A_0e^{-\lambda t}\).

Step 2: Find the decay constant
At \(t=3\) min, \(A=\dfrac{A_0}{e}\), so \(e^{-3\lambda}=e^{-1}\) and \(\lambda=\dfrac13\ \text{min}^{-1}\).

Step 3: Half life
\(T=\dfrac{\ln2}{\lambda}=\dfrac{\ln 2}{1/3}=3\log_e2\) minutes.

Final Answer:
The activity halves in \(3\log_e 2\) minutes. \[ \boxed{\text{(B) }3\mathrm{log}_{\mathrm{e}}2} \]