Q147 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
The activity of a radioactive sample is measured as $N_0$ counts per minute at time $t = 0$ and $\dfrac{N_0}{e}$ counts per minute at time $t = 3$ minute. The time (in minute) in which the activity reduces to half the value, is
- A$\dfrac{1}{3}\log_e 2$
- Correct answerB$3\log_e 2$
- C$3\log_{10} 2$
- D$\dfrac{3}{\log_e 2}$
Solution
The activity of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$.
Given that at $t = 0$, $A = N_0$.
At $t = 3$ minutes, $A = \dfrac{N_0}{e}$.
Substituting these values into the decay equation:
$\dfrac{N_0}{e} = N_0 e^{-3\lambda}$
$e^{-1} = e^{-3\lambda}$
Comparing the powers of $e$, we get:
$3\lambda = 1 \Rightarrow \lambda = \dfrac{1}{3} \text{ min}^{-1}$
The time in which the activity reduces to half its initial value is the half-life $T_{1/2}$, which is given by:
$T_{1/2} = \dfrac{\log_e 2}{\lambda}$
Substituting the value of $\lambda$:
$T_{1/2} = \dfrac{\log_e 2}{\dfrac{1}{3}} = 3 \log_e 2 \text{ min}$
Answer: $3\log_e 2$
Given that at $t = 0$, $A = N_0$.
At $t = 3$ minutes, $A = \dfrac{N_0}{e}$.
Substituting these values into the decay equation:
$\dfrac{N_0}{e} = N_0 e^{-3\lambda}$
$e^{-1} = e^{-3\lambda}$
Comparing the powers of $e$, we get:
$3\lambda = 1 \Rightarrow \lambda = \dfrac{1}{3} \text{ min}^{-1}$
The time in which the activity reduces to half its initial value is the half-life $T_{1/2}$, which is given by:
$T_{1/2} = \dfrac{\log_e 2}{\lambda}$
Substituting the value of $\lambda$:
$T_{1/2} = \dfrac{\log_e 2}{\dfrac{1}{3}} = 3 \log_e 2 \text{ min}$
Answer: $3\log_e 2$
Hence, the correct answer is option: 'B'.