The radioactive sources A and B have half-liv | vedclass

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Question

(C) The rate of disintegration $R$ is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms remain

Vedclass · Accepted Answer

$\sqrt{2} : 1$

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(C) The rate of disintegration $R$ is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms remaining at time $t$.Given: $T_{1/2, A} = 2 \ hr$,$T_{1/2, B} = 4 \ hr$.Decay constants: $\lambda_A = \frac{\ln 2}{2}$ and $\lambda_B = \frac{\ln 2}{4}$.Initial number of atoms: $N_0(A) = N_0(B) = N_0$.At $t = 2 \ hr$,the number of atoms remaining is $N = N_0 e^{-\lambda t}$.For source $A$: $N_A = N_0 e^{-(\frac{\ln 2}{2}) 	imes 2} = N_0 e^{-\ln 2} = \frac{N_0}{2}$.For source $B$: $N_B = N_0 e^{-(\frac{\ln 2}{4}) 	imes 2} = N_0 e^{-\frac{\ln 2}{2}} = \frac{N_0}{\sqrt{2}}$.The rates of disintegration are $R_A = \lambda_A N_A = (\frac{\ln 2}{2}) 	imes (\frac{N_0}{2}) = \frac{N_0 \ln 2}{4}$ and $R_B = \lambda_B N_B = (\frac{\ln 2}{4}) 	imes (\frac{N_0}{\sqrt{2}}) = \frac{N_0 \ln 2}{4\sqrt{2}}$.The ratio $R_A / R_B = \frac{N_0 \ln 2 / 4}{N_0 \ln 2 / 4\sqrt{2}} = \sqrt{2} : 1$.

The radioactive sources $A$ and $B$ have half-lives of $2 \ hr$ and $4 \ hr$ respectively and initially contain the same number of radioactive atoms. At the end of $2 \ hr$,the ratio of their rates of disintegration is:

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