Q154 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
Two radioactive materials A and B having decay constant '$7\lambda$' and '$\lambda$' respectively, initially have same number of nuclei. The time taken to have the ratio of number of nuclei of material B to that of A as '$e$', is
- A$1/\lambda$
- Correct answerB$1/6\lambda$
- C$1/7\lambda$
- D$1/8\lambda$
Solution
Let the initial number of nuclei of both materials be $N_0$.
The number of nuclei of material A at time $t$ is given by $N_A = N_0 e^{-7\lambda t}$.
The number of nuclei of material B at time $t$ is given by $N_B = N_0 e^{-\lambda t}$.
Given that the ratio of the number of nuclei of material B to that of A is $e$:
$\dfrac{N_B}{N_A} = e$
$\dfrac{N_0 e^{-\lambda t}}{N_0 e^{-7\lambda t}} = e$
$e^{6\lambda t} = e^1$
Equating the powers of $e$, we get:
$6\lambda t = 1$
$t = \dfrac{1}{6\lambda}$
Answer: $1/6\lambda$
The number of nuclei of material A at time $t$ is given by $N_A = N_0 e^{-7\lambda t}$.
The number of nuclei of material B at time $t$ is given by $N_B = N_0 e^{-\lambda t}$.
Given that the ratio of the number of nuclei of material B to that of A is $e$:
$\dfrac{N_B}{N_A} = e$
$\dfrac{N_0 e^{-\lambda t}}{N_0 e^{-7\lambda t}} = e$
$e^{6\lambda t} = e^1$
Equating the powers of $e$, we get:
$6\lambda t = 1$
$t = \dfrac{1}{6\lambda}$
Answer: $1/6\lambda$
Hence, the correct answer is option: 'B'.