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?
Show Hint
- \(\frac{\ln 5}{t_2 - t_1}\)
- \(\frac{\ln(t_2 + t_1)}{2}\)
- \(\frac{t_1 - t_2}{\ln 5}\)
- \(\frac{t_2 - t_1}{\ln 5}\)
The Correct Option is D
Solution and Explanation
At \(t_1\): \(A = A_0 e^{-\lambda t_1}\). At \(t_2\): \(\frac{A}{5} = A_0 e^{-\lambda t_2}\).
Dividing: \(\frac{1}{5} = e^{-\lambda (t_2 - t_1)}\), so \(\lambda (t_2 - t_1) = \ln 5\).
\(\lambda = \frac{\ln 5}{t_2 - t_1}\).
Average life \(\tau = \frac{1}{\lambda} = \frac{t_2 - t_1}{\ln 5}\).
Answer: (4) \(\frac{t_2 - t_1}{\ln 5}\)
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