Nuclei - Result Question 70
71. The activity of a radioactive sample is measured as $N_0$ counts per minute at $t=0$ and $N_0 / e$ counts per minute at $t=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
(a) $\log _{e} 2 / 5$
(b) $\frac{5}{\log _{e} 2}$
(c) $5 \log _{10} 2$
(d) $5 \log _{e} 2$
[2010]
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Answer:
Correct Answer: 71. (d)
Solution:
- (d) $N=N_0 e^{-\lambda t}$
Here, $t=5$ minutes
$\frac{N_0}{e}=N_0 \cdot e^{-5 \lambda}$
$\Rightarrow 5 \lambda=1$,
$\lambda=\frac{1}{5}$,
Now, $T _{1 / 2}=\frac{\ln 2}{\lambda}=5 \ln 2$
