Question

A radioactive substance of half life $138.6$ days is placed in a box. After $n$ days only $20 \%$ of the substance is present then the value of $n$ is
$[\ln (5)=1.61]$

• A. 693
• B. 138.6
• C. 277.2
• D. 322

Answer 1

Answer: (D)

Half life of a radioactive substance,
$n_{1 / 2}=138.6$ days
If $N_{0}$ be the initial amount of radioactive substance,
then remaining amount after $n$ days is given by
$N=20 \% \text { of } N_{0}=\frac{20}{100} \times N_{0}=\frac{N_{0}}{5}$
By radioative decay's law,
$N=N_{0}\left(\frac{1}{2}\right)^{\frac{n}{n_{1 / 2}}} \Rightarrow \frac{N_{0}}{5}=N_{0}\left(\frac{1}{2}\right)^{\frac{n}{138.6}} \Rightarrow \frac{1}{5}=\left(\frac{1}{2}\right)^{\frac{n}{138.6}}$
Taking log on the both sides, we get
$\ln \frac{1}{5} =\ln \left(\frac{1}{2}\right)^{\frac{n}{138.6}} \Rightarrow \ln \frac{1}{5}=\frac{n}{138.6} \ln \left(\frac{1}{2}\right) $
$\ln 5 =\frac{n}{138.6} \ln 2$
$\Rightarrow n =138.6 \times \frac{\ln 5}{\ln 2} \begin{bmatrix}
\because \ln 5=1.61 \\
\ln 2=0.693
\end{bmatrix}$
$n=138.6 \times \frac{1.61}{0.693} \Rightarrow n=322 $ days