Question

There are two radioactive substances $A$ and $B$. Decay constant of $B$ is two times that of $A$. Initially, both have an equal number of nuclei. After n half-lives of $A$, the rate of disintegration of both is equal. The value of $n$ is

• A. $4$
• B. $2$
• C. $1$
• D. $5$

Answer 1

The correct option is C

$1$

Step 1: Given data:

Let decay constant of A and B are ${\lambda }_{A}and{\lambda }_B$ respectively.

$$\begin{gathered} Let{\lambda }_A=\lambda \\ Therefore{\lambda }_B=2\lambda \end{gathered}$$

If N_o is the number of atoms in A and B at t=0, then

Initial rate of disintegration of $A=\lambda N_o$

Initial rate of disintegration of $B=2\lambda N_o$

Step 2: Calculation

Since ${\lambda }_B=2{\lambda }_A$

So ${\left(T_{\frac{1}{2}}\right)}_B=\frac{1}{2}{\left(T_{\frac{1}{2}}\right)}_A$

After one-half life of A

${\left(-\frac{dN}{dt}\right)}_A=\frac{\lambda N_o}{2}.............\left(1\right)$

Similarly after two-half lives of B

${\left(-\frac{dN}{dt}\right)}_B=\frac{2\lambda N_o}{4}=\frac{\lambda N_o}{2}............\left(2\right)$

From equations (1) and (2)

${\left(-\frac{dN}{dt}\right)}_A={\left(-\frac{dN}{dt}\right)}_B$

Hence the value of n=1

This implies that after one-half life of A, the rate of disintegration of both will be equal

So the correct option is (C).