A sample contains a mixture of ^ 110 Ag and ^ 108 Ag isotopes, each...

The total activity of the mixture is given by $A = A_{10} e^{-\lambda_1 t} + A_{20} e^{-\lambda_2 t}$, where $\lambda_1$ and $\lambda_2$ are the decay constants of ^{110}\text{Ag}$and ^{108}\text{Ag}$ respectively.

Since ^{108}\text{Ag}$has a larger half-life, its decay constant$\lambda_2$is smaller than$\lambda_1$. For large values of time$t$, the term$e^{-\lambda_1 t}$decays much faster and becomes negligible compared to$e^{-\lambda_2 t}$.<br><br>Thus, for large$t$, the total activity is approximately$A \approx A_{20} e^{-\lambda_2 t}$.<br><br>Using the data for large times, at$t_1 = 400 \text{ s}$,$A_1 = 1.1671 \times 10^8 \text{ s}^{-1}$and at$t_2 = 500 \text{ s}$,$A_2 = 0.7212 \times 10^8 \text{ s}^{-1}$.<br><br>Taking the ratio of the activities:<br><br>$\dfrac{A_1}{A_2} = \dfrac{e^{-\lambda_2 t_1}}{e^{-\lambda_2 t_2}} = e^{\lambda_2 (t_2 - t_1)}$<br><br>$\dfrac{1.1671}{0.7212} = e^{\lambda_2 (500 - 400)}$<br><br>$1.618 = e^{100 \lambda_2}$<br><br>Taking the natural logarithm on both sides:<br><br>$100 \lambda_2 = \ln(1.618) \approx 0.481$<br><br>$\lambda_2 = 0.00481 \text{ s}^{-1}$<br><br>The half-life of ^{108}\text{Ag}$ is:

$T_{1/2} = \dfrac{\ln 2}{\lambda_2} = \dfrac{0.693}{0.00481} \approx 144 \text{ s}$