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

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

$A(t) = A_{110}(t) + A_{108}(t)$

For ^{108}\text{Ag}$, the initial activity is$A_{108}(0) = 8.0 \times 10^8 \text{ disintegrations s}^{-1}$and its half-life is$T_{1/2} = 144 \text{ s}$. The activity of ^{108}\text{Ag}$ at $t = 100 \text{ s}$ is:

$A_{108}(100) = 8.0 \times 10^8 \times e^{-\dfrac{100 \ln 2}{144}} = 8.0 \times 10^8 \times e^{-0.4813} \approx 4.943 \times 10^8 \text{ disintegrations s}^{-1}$

From the given table, the total activity at $t = 100 \text{ s}$ is $A(100) = 5.4115 \times 10^8 \text{ disintegrations s}^{-1}$. Thus, the activity of ^{110}\text{Ag}$at$t = 100 \text{ s}$is:<br><br>$A_{110}(100) = A(100) - A_{108}(100) = (5.4115 - 4.943) \times 10^8 = 0.4685 \times 10^8 \text{ disintegrations s}^{-1}$<br><br>Since radioactive decay follows an exponential law ($A = A_0 e^{-\lambda t}$), the activity at the midpoint$t = 50 \text{ s}$is the geometric mean of the activities at$t = 0 \text{ s}$and$t = 100 \text{ s}$:<br><br>$A_{110}(50) = \sqrt{A_{110}(0) \times A_{110}(100)}$<br><br>$A_{110}(50) = \sqrt{8.0 \times 10^8 \times 0.4685 \times 10^8} = \sqrt{3.748} \times 10^8 \approx 1.93 \times 10^8 \text{ disintegrations s}^{-1}$<br><br>This is closest to$1.9 \times 10^8 \text{ disintegrations per second}$.<br><br>Answer:$1.9 \times 10^8 \text{ disintegrations per second}