Question

1 g of ${}_{79}^{198}Au$ $(T_{50}=65h.)$ decays by $\beta$-emission to produce stable Hg.
How much Hg will be present after 260 h?

• A. 0.9375 g
• B. 0.7375 g
• C. 0.6375 g
• D. None of these

Answer 1

The correct option is A 0.9375 g

${}_{79}^{198}Au⟶{}_{80}^{198}Hg+{}_{-1}^{0}e\left(\beta \right)$
Au and Hg are isobars.
Half-life = 65 days, t= 260 days
$y=\frac{totaltime}{half-life}=4$
Au left $C=C_0{\left(\frac{1}{2}\right)}^y=1{\left(\frac{1}{2}\right)}^4=\frac{1}{16}g$
Hg formed $C_0-C=1-\frac{1}{16}=\frac{15}{16}=0.9375g$