The half life of a radioactive substance is \( 20 \) minutes. The time taken between \( 50 \% \) decay and
\( 87.5 \% \) decay of the substance will be

Given, half-life of radioactive substance, \( T_{\frac{1}{2}}=20 \) minutes
Using
\[
\begin{array}{l}
N=N_{0} e^{-\lambda t} \Rightarrow \frac{N}{N_{0}}=e^{-\lambda t} \Rightarrow-\lambda t=\ln \left(\frac{N}{N_{0}}\right) \\
\Rightarrow t=-\frac{1}{\lambda} \ln \left(\frac{N}{N_{0}}\right)
\end{array}
\]
We know \( \frac{1}{\lambda}=\frac{T_{1 / 2}}{\ln 2} \)
\[
t=\frac{-\ln 2}{T_{1 / 2}} \ln \left(\frac{N}{N_{0}}\right)
\]
If \( t_{1} \) is time for \( 50 \% \) decay and \( t_{2} \) is time for \( 87.5 \% \) decay, we have
\[
\begin{array}{l}
t_{2}-t_{1}=-\frac{T_{1 / 2}}{\ln 2} \ln \left(\frac{N_{2}}{N_{0}}\right)+\frac{T_{1 / 2}}{\ln 2} \ln \left(\frac{N_{1}}{N_{0}}\right) \\
=\frac{T_{1 / 2}}{\ln 2}\left(-\ln N_{2}+\ln N_{0}+\ln N_{1}-\ln N_{0}\right)=\frac{T_{1 / 2}}{\ln 2}\left(\ln N_{1}-\ln N_{2}\right) \\
=\frac{T_{1 / 2}}{\ln 2} \ln \left(\frac{N_{1}}{N_{2}}\right)
\end{array}
\]
Substituting the values, we get
\( t_{2}-t_{1}=\frac{20 \mathrm{~min}}{0.693} \ln \left(\frac{50}{12.5}\right)=\frac{20}{0.693} \times \ln 4=40 \) minutes
Therefore, time taken between \( 50 \% \) decay and \( 87.5 \% \) decay will be \( 40 \) minutes.