A radioactive nucleus $$n_{2}$$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $$n_{1}$$ . If initial number of both nuclei are the same, what is the ratio of number of nuclei of $$n_{2}$$ to the number of nuclei of $$n_{1}$$ , after one half-life of $$n_{1}$$ ?

Since the decay constant of $$n_2$$ is three times that of $$n_1$$, $$\lambda_2 = 3\lambda_1$$, we use the radioactive decay law:

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

After one half-life of $$n_1$$ (that is, $$t = T_1 = \frac{\ln 2}{\lambda_1}$$), the remaining amount of $$n_1$$ is

$$ N_1 = N_0 e^{-\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-\ln 2} = \frac{N_0}{2} $$

At the same time, the amount of $$n_2$$ is

$$ N_2 = N_0 e^{-\lambda_2 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\ln 2} = N_0 \cdot 2^{-3} = \frac{N_0}{8} $$

Hence, the ratio of the two amounts is

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

The correct answer is Option 4: 1/4.