Question

If the mean life of a radioactive sample is 10 minutes, then the time (in minutes) taken for the activity of the radioactive sample to become $\mathrm{e}^{21}$ times its initial activity is

• A. 10
• B. 20
• C. 30
• D. 40

Hint:

Activity follows $A = A_0 \mathrm{e}^{-\lambda t}$, with $\tau = \frac{1}{\lambda}$. For reduction by $\mathrm{e}^{-n}$, time is $t = n \tau$. Half-life is $t_{1/2} = \ln 2 \cdot \tau \approx 0.693 \tau$. Always check if the problem implies activity decrease or increase.

Answer 1

The Correct Option is B

1. The activity $A$ of a radioactive sample decays exponentially: $A = A_0 \mathrm{e}^{-\lambda t}$, where $\lambda$ is the decay constant.
2. The mean life $\tau = 1/\lambda = 10$ minutes, so $\lambda = 0.1$ min$^{-1}$.
3. The problem likely intends ``to become $\frac{1}{\mathrm{e}^{2}}$ times its initial activity'' (typographical error in ``$\mathrm{e}^{21}$'' for $\mathrm{e}^{2}$), as activity decreases in radioactive decay.
4. For $A = \frac{A_0}{\mathrm{e}^{2}} = A_0 \mathrm{e}^{-2}$, we have $-\lambda t = -2$, so $\lambda t = 2$.
5. Thus, $t = \frac{2}{\lambda} = 2 \tau = 2 \times 10 = 20$ minutes.
6. If interpreted literally as $\mathrm{e}^{21}$ times (increase), it would not make sense for decay, confirming the error in the problem statement. Therefore, the correct option is (2) 20.