Answer to Question #126049 in Atomic and Nuclear Physics for Muhammad qasim

Question

 Two radioactive nuclides, S and M, both have the same activity X at time t = 0. S and M have a half-life of 26 years and 16 years respectively. Now both samples S and M are mixed. Find out the total activity of the mixture at t = 50 years?

Answer 1

Let initial number of atom of both S and M is $N_0$ .

Since,

$A=\bigg|\frac{dN}{dt}\bigg|=\lambda N$

Thus, initially activity is

$X=\lambda_SN_0+\lambda_MN_0$

Also,

$\lambda=\frac{\ln2}{t_{1/2}}\\ \implies \lambda_S=\frac{\ln2}{26}\\ \lambda_M=\frac{\ln2}{16}$

Hence,

$N_0=\dfrac{208X}{21\ln2}$

Since,

$N=N_0\bigg(\frac{1}{2}\bigg)^{t/t_{\frac{1}{2}}}$

Thus,

$N_S=N_0(1/2)^{50/26}=N_0(1/2)^{25/13}\\N_M=N_0(1/2)^{50/16}=N_0(1/2)^{25/8}$

Thus, activity at t=50 is

$A=\lambda_SN_S+\lambda _MN_M\\ \implies A=\frac{\ln2}{26}N_0(1/2)^{25/13}+\frac{\ln2}{16}N_0(1/2)^{25/8}\\ \implies A=\bigg(\frac{8}{21}(1/2)^{25/13}+\frac{13}{21}(1/2)^{25/8}\bigg)X\approx0.17X$