A radioactive substance of half life 138.6 days is placed in a box. After \(n\) days only \(20 \%\) of the…
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 \([\ln (5)=1.61]\)
693
138.6
277.2
322
Solution
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
\(\begin{aligned}
\ln \frac{1}{5} & =\ln \left(\frac{1}{2}\right)^{\frac{n}{138.6}} \Rightarrow \ln \frac{1}{5}=\frac{n}{138.6} \ln \left(\frac{1}{2}\right) \\
\ln 5 & =\frac{n}{138.6} \ln 2 \\
n & =138.6 \times \frac{\ln 5}{\ln 2} \quad \left[\begin{array}{l}\because \ln 5=1.61 \\ \ln 2=0.693\end{array}\right]\\
n & =138.6 \times \frac{1.61}{0.693} \Rightarrow n=322 \text { days }
\end{aligned}\)