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

The total activity of the sample at any time $t$ is given by the sum of the activities of the two isotopes:

$A(t) = A_1(t) + A_2(t) = A_{01} e^{-\lambda_1 t} + A_{02} e^{-\lambda_2 t}$

Given the initial activities are equal: $A_{01} = A_{02} = 8.0 \times 10^8 \text{ s}^{-1}$.

Since ^{108}\text{Ag}$has a larger half-life than ^{110}\text{Ag}$, its decay constant is smaller ($\lambda_2 < \lambda_1$). At large times ($t \ge 300 \text{ s}$), the faster-decaying isotope ^{110}\text{Ag}$will have almost completely decayed, and the total activity will be dominated by ^{108}\text{Ag}$.

Using the activity values at $t = 400 \text{ s}$ and $t = 500 \text{ s}$ to find $\lambda_2$:

$\dfrac{A(500)}{A(400)} = \dfrac{0.7212}{1.1671} \approx 0.6179$

$\dfrac{A_{02} e^{-500 \lambda_2}}{A_{02} e^{-400 \lambda_2}} = e^{-100 \lambda_2} = 0.6179$

$-100 \lambda_2 = \ln(0.6179) \approx -0.4814 \Rightarrow \lambda_2 \approx 0.004814 \text{ s}^{-1}$

Now, we calculate the activity of ^{108}\text{Ag}$at an earlier time, say$t = 20 \text{ s}$:<br><br>$A_2(20) = 8.0 \times e^{-20 \times 0.004814} = 8.0 \times 0.9082 = 7.266 \times 10^8 \text{ s}^{-1}$<br><br>The activity of ^{110}\text{Ag}$ at $t = 20 \text{ s}$ is the total activity minus the activity of ^{108}\text{Ag}$:<br><br>$A_1(20) = A(20) - A_2(20) = 11.799 - 7.266 = 4.533 \times 10^8 \text{ s}^{-1}$<br><br>Using the decay equation for ^{110}\text{Ag}$ at $t = 20 \text{ s}$:

$A_1(20) = A_{01} e^{-20 \lambda_1}$

$4.533 = 8.0 e^{-20 \lambda_1}$

$e^{-20 \lambda_1} = \dfrac{4.533}{8.0} \approx 0.5666$

$-20 \lambda_1 = \ln(0.5666) \approx -0.568 \Rightarrow \lambda_1 \approx 0.0284 \text{ s}^{-1}$

Finally, the half-life of ^{110}\text{Ag}$is:<br><br>$T_{1/2} = \dfrac{\ln 2}{\lambda_1} = \dfrac{0.693}{0.0284} \approx 24.4 \text{ s}$<br><br>Answer:$24.4 \text{ s}