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
$\ln \frac{1}{5}=\ln {\left(\frac{1}{2}\right)}^{\frac{n}{138.6}}\Rightarrow \ln \frac{1}{5}=\frac{n}{138.6}\ln \left(\frac{1}{2}\right)$
$\ln 5=\frac{n}{138.6}\ln 2$
$\Rightarrow n=138.6\times \frac{\ln 5}{\ln 2}\left[\begin{array}{c}<br/><br/>\because \ln 5=1.61\\ <br/><br/>\ln 2=0.693<br/><br/>\end{array}\right]$
$n=138.6\times \frac{1.61}{0.693}\Rightarrow n=322$ days