Question

The half-life of ¹⁹⁹Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of ¹⁹⁸Au. (b) What will be the activity after 7 days? Take the atomic weight of ¹⁹⁸Au to be 198 g mol⁻¹.

Answer 1

Given:
Half-life of ¹⁹⁹Au, T_1/2= 2.7 days

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle 1/2}}}$ = $\frac{0.693}{2.7\times 24\times 60\times 60}=2.97\times {10}^{-6}{\mathrm{s}}^{-1}$

Number of atoms left undecayed, N = $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}$

Now, activity, $A_0=\lambda N$

= $\frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}\times 2.97\times {10}^{-6}$
= 0.244 Ci

(b) After 7 days,
Activity, $A=A_0{e}^{-\lambda t}$
Here, A₀ = 0.244 Ci
$$\begin{gathered} \therefore A=0.244\times e^{-2.97\times {10}^{-6}\times 7\times 3600\times 24} \\ =0.244\times e^{17962.56\times {10}^{-4}} \\ =0.040\mathrm{Ci} \end{gathered}$$