Physics - Modern Physics Question with Solution | TestHub

$\because \frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
$N$= no. of un-decayed nuclei
$N_0$= initial no. of nuclei
Also$t=n\times T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
$\Rightarrow \frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}$
$\Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{7.5}{15}}$
$\Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{1}{2}}\Rightarrow N=\frac{N_0}{\sqrt{2}}$
$\therefore$Number of nuclei decayed
$=\left(N_0-\frac{N_0}{\sqrt{2}}\right)=N_0\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
$=\frac{0.414}{1.414}{N}_0$
$=0.29N_0$
$\because N_0=\frac{1}{24}\times 6.023\times {10}^{23}$
$\Rightarrow$No. of nuclei decayed$=0.29\times \frac{1}{24}\times 6.023\times {10}^{23}$
$=0.0727\times {10}^{23}$
$=7.5\times {10}^{21}$