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Q.
$ t_{1/2} $ for $Kr$ is $10.6 \,yr$. The time taken for $99.9\%$ decay of $Kr$ is
AMUAMU 2003
Solution:
$\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$
where $n=$ number of half-life periods
$N_{0}=$ initial concentration of substance
$N=$ amount of concentration left after $n$ half-life period.
Total time $(T)=n \times t_{1 / 2}$
Amount disintegrated $=99.9 \%=\frac{999}{1000}$
Amount left $=1-\frac{999}{1000}$
$=\frac{1}{1000} \approx \frac{1}{1024} $
$ \frac{N}{N_{0}} =\left(\frac{1}{2}\right)^{n} $
$ \frac{1}{1024} =\left(\frac{1}{2}\right)^{n} $
$\left(\frac{1}{2}\right)^{10} =\left(\frac{1}{2}\right)^{n} $
$ n =10 $
Total time $=n \times t_{1 / 2} $
Here, $ t_{1 / 2} =10.6 \,yr$
Total time $=10 \times 10.6 $
$=106\, yr $