Question

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

• A. 4
• B. 2
• C. 1
• D. 5

Answer 1

The Correct Option is C

Let $\lambda_{A} = \lambda \therefore \lambda_{B} = 2\lambda$ If $N_{0}$ is total number of atoms in A and B at $t = 0$, then initial rate of disintegration of $A = \lambda N_{0}$, and initial rate of disintegration of $B = 2\lambda N_{0}$ As $\lambda_{B} = 2\lambda _{A} \quad\left(\because \lambda = \frac{ln 2}{T_{1/2}} \right)$ $\therefore\quad\left(T_{1/2}\right)_{B} = \frac{1}{2}\left(T_{1/2}\right)_{A}$ i.e., half-life of B is half the half-life of A. After one half-life of A $\left(-\frac{dN}{dt}\right)_{A} = \frac{\lambda N_{0}}{2}\quad \dots\left(i\right)$ Equivalently, after two half lives of B $\left(-\frac{dN}{dt}\right)_{B} =\frac{2\lambda N_{0}}{4} = \frac{\lambda N_{0}}{2}\quad \dots \left(ii\right)$ From $\left(i\right)$ and $\left(ii\right)$, we get $\left(-\frac{dN}{dt}\right)_{A} = \left(-\frac{dN}{dt}\right)_{B},$ After $n = 1$, i.e., one half-life of A, the rate of disintegration of both will be equal.