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

The total activity $A(t)$ of the mixture is the sum of the activities of the two isotopes:

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

Given that the initial activities are equal, $A_{01} = A_{02} = \dfrac{A_0}{2}$, where $A_0$ is the total initial activity.

$A(t) = \dfrac{A_0}{2} (e^{-\lambda_1 t} + e^{-\lambda_2 t})$

$\dfrac{A(t)}{A_0} = \dfrac{1}{2} (e^{-\lambda_1 t} + e^{-\lambda_2 t})$

Taking the natural logarithm:

$\ln\left(\dfrac{A}{A_0}\right) = \ln\left(e^{-\lambda_1 t} + e^{-\lambda_2 t}\right) - \ln 2$

The slope of the curve is given by the derivative with respect to time:

$\text{Slope} = \dfrac{d}{dt} \ln\left(\dfrac{A}{A_0}\right) = \dfrac{-\lambda_1 e^{-\lambda_1 t} - \lambda_2 e^{-\lambda_2 t}}{e^{-\lambda_1 t} + e^{-\lambda_2 t}}$

At $t = 0$, the initial slope is:

$m_{\text{initial}} = -\dfrac{\lambda_1 + \lambda_2}{2}$

Since ^{108}\text{Ag}$has a larger half-life than ^{110}\text{Ag}$, its decay constant is smaller, so $\lambda_2 < \lambda_1$.

As $t \to \infty$, the term with the larger decay constant ($e^{-\lambda_1 t}$) approaches zero much faster than $e^{-\lambda_2 t}$. The slope asymptotically approaches:

$m_{\text{final}} = \lim_{t \to \infty} \dfrac{-\lambda_1 e^{-(\lambda_1 - \lambda_2) t} - \lambda_2}{e^{-(\lambda_1 - \lambda_2) t} + 1} = -\lambda_2$

Comparing the magnitudes of the slopes:

$|m_{\text{initial}}| = \dfrac{\lambda_1 + \lambda_2}{2} > \lambda_2 = |m_{\text{final}}|$

The initial slope is more negative (steeper) than the final slope. Therefore, the curve starts with a steeper negative slope and asymptotically approaches a straight line with a less steep negative slope.