The decay constant of ^ 197 _ 80 Hg (which decays by electron...

Given that $25\%$ of the mercury isotope is converted into gold, the amount of mercury remaining is $75\%$ of the initial amount.

Let the initial amount be $N_0$. The remaining amount at time $t$ is $N = 0.75 N_0 = \dfrac{3}{4} N_0$.

Using the radioactive decay law:

$N = N_0 e^{-\lambda t}$

$\dfrac{3}{4} N_0 = N_0 e^{-\lambda t}$

$e^{\lambda t} = \dfrac{4}{3}$

Taking the natural logarithm on both sides:

$\lambda t = \ln\left(\dfrac{4}{3}\right) = \ln 4 - \ln 3$

Substituting the standard values $\ln 4 \approx 1.386$ and $\ln 3 \approx 1.098$:

$\lambda t \approx 1.386 - 1.098 = 0.288$

Given the decay constant $\lambda = 1.8 \times 10^{-4} \text{ s}^{-1}$, the time $t$ is:

$t = \dfrac{0.288}{1.8 \times 10^{-4}}$

$t = 0.16 \times 10^4 \text{ s} = 1600 \text{ s}$