Modern Physics 3 Question 12
12. The half-life of ${ }^{215} At$ is $100 \mu$ s. The time taken for the activity
of a sample of ${ }^{215}$ At to decay to $\frac{1}{16}$ th of its initial value is $\underset{(2002,2 M)}{ }$
(a) $400 \mu s$
(b) $63 \mu s$
(c) $40 \mu s$
(d) $300 \mu s$
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Solution:
$$ R=R _0 \quad \frac{1}{2}{ }^{n} $$
Here $R$ = activity of radioactive substance after $n$ half-lives
$$ =\frac{R _0}{16} $$
Substituting in Eq. (i), we get $n=4$
$$ \therefore \quad t=(n) t _{1 / 2}=(4)(100 \mu s)=400 \mu s $$
