Solveeit Logo

Question

Chemistry Question on Decay Rate

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

Explanation

Solution

Let λA=λλB=2λ\lambda_{A} = \lambda \therefore \lambda_{B} = 2\lambda If N0N_{0} is total number of atoms in A and B at t=0t = 0, then initial rate of disintegration of A=λN0A = \lambda N_{0}, and initial rate of disintegration of B=2λN0B = 2\lambda N_{0} As λB=2λA(λ=ln2T1/2)\lambda_{B} = 2\lambda _{A} \quad\left(\because \lambda = \frac{ln 2}{T_{1/2}} \right) (T1/2)B=12(T1/2)A\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 (dNdt)A=λN02(i)\left(-\frac{dN}{dt}\right)_{A} = \frac{\lambda N_{0}}{2}\quad \dots\left(i\right) Equivalently, after two half lives of B (dNdt)B=2λN04=λN02(ii)\left(-\frac{dN}{dt}\right)_{B} =\frac{2\lambda N_{0}}{4} = \frac{\lambda N_{0}}{2}\quad \dots \left(ii\right) From (i)\left(i\right) and (ii)\left(ii\right), we get (dNdt)A=(dNdt)B,\left(-\frac{dN}{dt}\right)_{A} = \left(-\frac{dN}{dt}\right)_{B}, After n=1n = 1, i.e., one half-life of A, the rate of disintegration of both will be equal.