JEE Physics 2025 Question with Solution

Given: The decay constant of $n_2$ is three times that of $n_1$, so $\lambda_2 = 3\lambda_1$. The initial number of nuclei is the same for both, say $N_0$.

Find: The ratio $\frac{N_2}{N_1}$ after one half-life of $n_1$.

For radioactive decay,

$$N = N_0 e^{-\lambda t}$$

The half-life is related to decay constant by

$$T_{1/2} = \frac{\ln 2}{\lambda}$$

For nucleus $n_1$,

$$T_{1/2,1} = \frac{\ln 2}{\lambda_1}$$

After one half-life of $n_1$,

$$N_1\left(T_{1/2,1}\right) = \frac{N_0}{2}$$

For nucleus $n_2$ over the same time interval,

$$N_2\left(T_{1/2,1}\right) = N_0 e^{-\lambda_2 T_{1/2,1}}$$

Substituting $\lambda_2 = 3\lambda_1$ and $T_{1/2,1} = \frac{\ln 2}{\lambda_1}$,

$$N_2\left(T_{1/2,1}\right) = N_0 e^{-3\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\ln 2} = N_0\left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$$

Therefore,

$$\frac{N_2\left(T_{1/2,1}\right)}{N_1\left(T_{1/2,1}\right)} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1}{4}$$

Therefore, the correct option is A, and the required ratio is $\frac{1}{4}$.

Given: $\lambda_2 = 3\lambda_1$ and both nuclei start with the same initial number $N_0$.

Find: The ratio $\frac{N_2}{N_1}$ after one half-life of $n_1$.

After one half-life of $n_1$, nucleus $n_1$ becomes

$$N_1 = \frac{N_0}{2}$$

Since $n_2$ has three times the decay constant, in the same time it effectively undergoes three half-life exponents:

$$N_2 = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$$

Hence,

$$\frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1}{4}$$

This works because the decay factor over time $t$ is determined by $e^{-\lambda t}$, so tripling $\lambda$ triples the exponent for the same time interval. Therefore, the correct option is A.