At a given instant, say $$t = 0$$, two radioactive substance A and B have equal activities. The ratio $$\frac{R_B}{R_A}$$ of their activities after time $$t$$ itself decays with time $$t$$ as $$e^{-3t}$$. If the half-life of A is $$\ln 2$$, the half-life of B is:

We begin with the basic law of radioactive decay. For any radioactive substance the activity $$R(t)$$ at time $$t$$ is given by the exponential law

$$R(t)=R_0\,e^{-\lambda t},$$

where $$R_0$$ is the activity at $$t=0$$ and $$\lambda$$ is the decay constant of the nuclide. The half-life $$T_{1/2}$$ is connected to the decay constant by the well-known formula

$$T_{1/2}=\frac{\ln 2}{\lambda}\;. \quad -(1)$$

Let the decay constants of substances A and B be $$\lambda_A$$ and $$\lambda_B$$ respectively. Their activities at time $$t$$ are therefore

$$R_A(t)=R_A(0)\,e^{-\lambda_A t},\qquad R_B(t)=R_B(0)\,e^{-\lambda_B t}.$$

We are told that at $$t=0$$ the activities are equal, so

$$R_B(0)=R_A(0).$$

Hence the ratio of the activities at a general time $$t$$ becomes

$$\frac{R_B(t)}{R_A(t)}=\frac{R_B(0)\,e^{-\lambda_B t}}{R_A(0)\,e^{-\lambda_A t}} =e^{-(\lambda_B-\lambda_A)t}.$$

The problem states that this ratio itself decays with time as $$e^{-3t}$$. Equating the two expressions gives

$$e^{-(\lambda_B-\lambda_A)t}=e^{-3t}\quad\Longrightarrow\quad \lambda_B-\lambda_A=3. \quad -(2)$$

Next we use the given half-life of substance A. The statement “the half-life of A is $$\ln 2$$” means

$$T_{1/2,A}=\ln 2.$$

Substituting this value into formula (1) for substance A yields

$$\ln 2=\frac{\ln 2}{\lambda_A}\quad\Longrightarrow\quad\lambda_A=1.$$

Now we substitute $$\lambda_A=1$$ into equation (2) to find $$\lambda_B$$:

$$\lambda_B-1=3\quad\Longrightarrow\quad\lambda_B=4.$$

Finally, the half-life of substance B is obtained from formula (1) applied to B:

$$T_{1/2,B}=\frac{\ln 2}{\lambda_B}=\frac{\ln 2}{4}.$$

Hence, the correct answer is Option C.