Question:
A radioactive sample is undergoing $\alpha$ decay. At any time $t_{1}$, its activity is $A$ and another time $t_{2}$ the activity is $\frac{A}{5} .$ What is the average life time for the sample?
Correct Option: 1
Solution:
(1)
For activity of radioactivesample
$A=A_{0} e^{-\alpha t_{1}} \ldots(1)$
$\frac{A}{5} A_{0} e^{-\alpha t_{2}} \ldots(2)$
From $(1) /(2)$
$5=e^{-\lambda\left(t_{1}-t_{2}\right)}$
$\ln (5)=\left(t_{2}-t_{1}\right) \lambda \Rightarrow \lambda=\frac{\ln (5)}{t_{2}-t_{1}}$
avg. life $=\frac{1}{\lambda} \Rightarrow \frac{t_{2}-t_{1}}{\ln (5)}$
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