Q173 / 180 · KCET 2015 · Physics
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
\( 87.5 \% \) decay of the substance will be
- A\( 30 \) minutes
- Correct answerB\( 40 \) minutes
- C\( 25 \) minutes
- D\( 10 \) minutes
Solution
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.
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.
Hence, the correct answer is option: 'B'.