Activity of a radioactive sample is measured as N_0 counts per minute at t = 0 and N_0/e counts per

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13.Nuclei

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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

$5\,{\log _e}\,2$

Solution

$N=N_{0} \mathrm{e}^{-\lambda t}$

Here, $t=5$ $minutes$

$\frac{N_{0}}{e}=N_{0} \cdot e^{-5 \lambda}$

$\Rightarrow \quad 5 \lambda=1,$ or $\quad \lambda=\frac{1}{5},$

Now, ${T_{1/2}} = \frac{{\ell n2}}{\lambda } = 5\ell n2$

Standard 12

Physics

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