Two radioactive substances A and B have decay constants $$5\lambda$$ and $$\lambda$$ respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $$\frac{1}{e^2}$$ will be

The behaviour of any radioactive substance is governed by the exponential decay law, which is written as $$N = N_0 e^{-\lambda t},$$ where $$N_0$$ is the initial number of undecayed nuclei, $$N$$ is the number remaining after time $$t,$$ and $$\lambda$$ is the decay constant of that substance.

In the present question we have two different radioactive substances:

 • Substance A has decay constant $$5\lambda.$$
 • Substance B has decay constant $$\lambda.$$

At the initial moment $$t = 0$$ we are told that the two samples contain the same number of nuclei, so we can denote that common number as $$N_0.$$ Thus,

$$N_{A0} = N_0,\qquad N_{B0} = N_0.$$

After a time $$t,$$ we apply the decay law to each substance separately.

For substance A (decay constant $$5\lambda$$):

$$N_A = N_0 e^{-5\lambda t}.$$

For substance B (decay constant $$\lambda$$):

$$N_B = N_0 e^{-\lambda t}.$$

We are asked to find the time when the ratio $$\dfrac{N_A}{N_B}$$ becomes $$\dfrac{1}{e^2}.$$ Let us therefore form that ratio using the expressions derived above:

$$\frac{N_A}{N_B} \;=\; \frac{N_0 e^{-5\lambda t}} {N_0 e^{-\lambda t}} \;=\; e^{-5\lambda t}\,e^{+\lambda t}.$$

Simplifying the exponentials we obtain

$$\frac{N_A}{N_B} = e^{-4\lambda t}.$$

By the condition given in the question, this ratio must equal $$\dfrac{1}{e^2}=e^{-2}.$$ Hence we set

$$e^{-4\lambda t} = e^{-2}.$$

Because the bases of the two exponentials are equal, we equate their exponents directly:

$$-4\lambda t = -2.$$

Now divide both sides by $$-4\lambda$$ to isolate $$t$$:

$$t = \frac{-2}{-4\lambda} = \frac{2}{4\lambda} = \frac{1}{2\lambda}.$$

Thus the required time is $$\displaystyle t = \frac{1}{2\lambda}.$$

Hence, the correct answer is Option B.