MHT CET · Physics · Nuclear Physics
If ' \(T\) ' is the half life of a radioactive substancethen its instantaneous rate of change of activity is proportional to
- A T
- B \(\mathrm{T}^{-2}\)
- C \(\mathrm{T}^{+2}\)
- D \(\mathrm{T}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{T}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\text {Activity, } R=\frac{-d N}{d t}=\lambda N \)
\( \frac{d R}{d t}=\frac{d}{d t}(\lambda N)=\lambda \frac{d N}{d t}=\lambda(-\lambda N)=-\lambda^2 N \)
\( \lambda=\frac{\log _{\mathrm{e}} 2}{T_{1 / 2}} \)
\( \therefore \lambda^2 N=\frac{\left(\log _{\mathrm{e}}\right)^2 N}{\mathrm{~T}_{1 / 2}^2} \)
\( \therefore \frac{\mathrm{dR}}{\mathrm{dt}} \propto \frac{1}{\mathrm{~T}_{1 / 2}^2}\)
\( \frac{d R}{d t}=\frac{d}{d t}(\lambda N)=\lambda \frac{d N}{d t}=\lambda(-\lambda N)=-\lambda^2 N \)
\( \lambda=\frac{\log _{\mathrm{e}} 2}{T_{1 / 2}} \)
\( \therefore \lambda^2 N=\frac{\left(\log _{\mathrm{e}}\right)^2 N}{\mathrm{~T}_{1 / 2}^2} \)
\( \therefore \frac{\mathrm{dR}}{\mathrm{dt}} \propto \frac{1}{\mathrm{~T}_{1 / 2}^2}\)
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