Question

The decay constant of ${}_{80}{}^{197}\mathrm{Hg}$ (electron capture to ${}_{79}{}^{197}\mathrm{Au}$) is 1.8 × 10⁻⁴ S⁻¹. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

Answer 1

Given:
Decay constant of ${}_{80}{}^{197}\mathrm{Hg}$, $\lambda$ = 1.8 × 10⁻⁴ s^$-1$

(a)
Half-life, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }$

$$\begin{gathered} \Rightarrow T_{1/2}=\frac{0.693}{1.8\times {10}^{-4}} \\ =3850\mathrm{s}=64\mathrm{minutes} \end{gathered}$$

(b)
$$\begin{gathered} \mathrm{Average}\mathrm{life},T_{av}=\frac{T_{1/2}}{0.693} \\ =\frac{64}{0.693} \\ =92\mathrm{minutes} \end{gathered}$$

(c)
Number of active nuclei of mercury at t = 0
= N₀
= 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N = 75

Now, $\frac{N}{N_0}=e^{-\lambda t}$
Here, N = Number of inactive nuclei
N₀ = Number of nuclei at t = 0
$\lambda$ = Disintegration constant

On substituting the values, we get
$\frac{75}{100}=e^{-\lambda t}$
$$\begin{gathered} \Rightarrow 0.75={\mathrm{e}}^{-\lambda x} \\ \Rightarrow \mathrm{In}0.75=-\lambda t \\ \Rightarrow t=\frac{\mathrm{In}0.75}{-0.00018} \\ =1600\mathrm{s} \end{gathered}$$