A sample of radioactive element contains $8 \times 10^{16}$ active nuclei. The halt-life of the element is…
A sample of radioactive element contains $8 \times 10^{16}$ active nuclei. The halt-life of the element is 15 days. The number of nuclei decayed after 60 days is
$7.5 \times 10^{+16}$
$2.0 \times 10^{16}$
$0.5 \times 10^{16}$
$4.0 \times 10^{16}$
Solution
Half life, $\mathrm{T}=15$ days, time $\mathrm{t}=60$ days $=4 \mathrm{~T}$ The number of nuclei remaining is given by
$\begin{aligned}
& \mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{N}_0\left(\frac{1}{2}\right)^4=\frac{1}{16} \times 8 \times 10^{16} \\
& =0.5 \times 10^{16} \\
& \therefore \text { No. of nuclei decayed }=\mathrm{N}_0-\mathrm{N} \\
& =8 \times 10^{16}-0.5 \times 10^{16}=7.5 \times 10^{16}
\end{aligned}$