QJEEM25Ph08b1: JEE Main 2025 Physics - Class 12 JEEM25Ph08b1 Solution: A radioactive nucleus has 3 times the decay constant as…

Let's work through the problem step by step.

The decay of a radioactive nucleus is given by the formula:

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

where:

$N(t)$ is the number of nuclei remaining at time $t$ ,

$N_0$ is the initial number of nuclei,

$\lambda$ is the decay constant.

For nucleus $n_1$ with decay constant $\lambda_1$ , its half-life $t_{1/2}$ is:

$$t_{1/2}=\frac{\ln 2}{{\lambda }_1}.$$

After one half-life of $n_1$ , the number of $n_1$ nuclei remaining is:

$$N_1=N_0{e}^{-{\lambda }_1{t}_{1/2}}=N_0{e}^{-{\lambda }_1\frac{\ln 2}{{\lambda }_1}}=N_0{e}^{-\ln 2}=\frac{N_0}{2}.$$

Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$ :

$${\lambda }_2=3{\lambda }_1.$$

For nucleus $n_2$ , the number of nuclei remaining after one half-life of $n_1$ (which is $t = \frac{\ln2}{\lambda_1}$ ) is:

$N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}.$

Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time:

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

Therefore, the correct ratio is $\frac{1}{4}$ , which corresponds to Option A.