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If $10\%$ of a radioactive material decays in $5\, days$ then the amount of the original material left after $20\, days$ is approximately .......... $\%$
$60$
$65$
$70$
$75$
Solution
$10 \%$ decays in $5$ days $\Rightarrow \frac{\mathrm{N}^{1}}{\mathrm{N}_{0}}=\frac{10}{100}$
Thus active fraction in $5$ days $=\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{5 / T_{\mathrm{H}}}$
$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{90}{100}=\left(\frac{1}{2}\right)^{5 / T_{\mathrm{H}}}$
Thus amount of left material after $20$ days
$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{20 / \mathrm{T}_{\mathrm{H}}}=\left(\frac{1}{2}\right)^{5 / T_{\mathrm{H}} \times 4}$
$=\left(\frac{9 \not 0}{10 \not 0}\right)^{4}=\left(\frac{9}{10}\right)^{4}=0.6561$
in percentage approx $=65.61 \%$
