Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where $$\lambda$$ is the decay constant)

We begin with the radioactive-decay law, which states that the number of undecayed nuclei present at any time $$t$$ is

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

where $$N_0$$ is the initial number of nuclei and $$\lambda$$ is the decay constant.

First we locate the instant when exactly one-quarter of the original nuclei have decayed. If one-quarter (that is, $$\dfrac14$$) have disappeared, then three-quarters (that is, $$\dfrac34$$) are still present. Hence, at this instant (let us call the time $$t_1$$) we have

$$N(t_1)=\frac34\,N_0.$$

Substituting this in the decay law gives

$$\frac34\,N_0 \;=\; N_0\,e^{-\lambda t_1}.$$

We can cancel the common factor $$N_0$$ on both sides:

$$\frac34 \;=\; e^{-\lambda t_1}.$$

Now we take the natural logarithm (ln) of both sides. Using the algebraic fact $$\ln e^x = x,$$ we get

$$\ln\!\left(\frac34\right) \;=\; -\lambda t_1.$$

Solving for $$t_1$$ gives

$$t_1 \;=\; -\frac1\lambda\,\ln\!\left(\frac34\right) \;=\; \frac1\lambda\,\ln\!\left(\frac43\right).$$

Next we locate the instant when exactly one-half of the nuclei have decayed. If one-half have disappeared, the remaining fraction is $$\dfrac12$$. Let the corresponding time be $$t_2$$. Thus

$$N(t_2)=\frac12\,N_0.$$

Again substituting in the decay law, we write

$$\frac12\,N_0 \;=\; N_0\,e^{-\lambda t_2}.$$

Cancelling $$N_0$$ yields

$$\frac12 \;=\; e^{-\lambda t_2}.$$

Taking natural logarithms once more:

$$\ln\!\left(\frac12\right) \;=\; -\lambda t_2.$$

Therefore,

$$t_2 \;=\; -\frac1\lambda\,\ln\!\left(\frac12\right) \;=\; \frac1\lambda\,\ln 2.$$

The required quantity is the time gap between these two instants, i.e.,

$$\Delta t=t_2-t_1.$$

Substituting the expressions for $$t_2$$ and $$t_1$$, we obtain

$$\Delta t \;=\; \frac1\lambda\,\ln 2 \;-\; \frac1\lambda\,\ln\!\left(\frac43\right).$$

Factoring out the common factor $$\dfrac1\lambda$$:

$$\Delta t \;=\; \frac1\lambda\left[\ln 2 - \ln\!\left(\frac43\right)\right].$$

By the logarithm subtraction rule $$\ln a - \ln b = \ln\!\left(\dfrac{a}{b}\right),$$ this simplifies to

$$\Delta t \;=\; \frac1\lambda\,\ln\!\left(\frac{2}{\frac43}\right).$$

Simplifying the fraction inside the logarithm:

$$\frac{2}{\frac43} \;=\; 2 \times \frac34 \;=\; \frac32.$$

Thus,

$$\Delta t \;=\; \frac1\lambda\,\ln\!\left(\frac32\right).$$

So the time interval between the moment when a quarter of the nuclei have decayed and the moment when half of them have decayed is

$$\boxed{\displaystyle \frac{\ln\!\left(\frac32\right)}{\lambda}}.$$

Hence, the correct answer is Option D.