A radioactive sample is undergoing α decay. At any time t1, its activity is A and another time t2, the activity is A/5. What is the average life time for the sample? (a) (ln 5)/(t₂-t₁) (b) (ln (t₂+t₁))/(2) (c) (t₁-t₂)/(ln 5) (d) (t₂-t₁)/(ln 5)

Answer: (d). A = A₀e-λ t, so A/5/A = e-λ(t₂-t₁). λ(t₂ - t₁) = ln 5 ⇒ λ = (ln 5)/(t₂-t₁). Average life τ = (1)/(λ) = (t₂-t₁)/(ln 5), so the answer is (d).

A radioactive sample is undergoing α decay. At any time t1, its… | JEE Main 2021 (26 Feb, Shift 2)

Q63JEE Main 2021 · 26 FEB · SHIFT 2Medium

Law of Radioactive Decay

A radioactive sample is undergoing $\alpha$ decay. At any time t1, its activity is A and another time t2, the activity is $\frac{A}{5}$ . What is the average life time for the sample?

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Answer: (d)

$A = A_0e^{-\lambda t}$ , so $\frac{A/5}{A} = e^{-\lambda(t_2-t_1)}$ .
$\lambda(t_2 - t_1) = \ln5 \Rightarrow \lambda = \frac{\ln5}{t_2-t_1}$ .
Average life $\tau = \frac{1}{\lambda} = \frac{t_2-t_1}{\ln5}$ , so the answer is (d).

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