Question

1.0 g of ${}_{79}A{u}^{198}$ (half life = 65 hr) decays by $\beta$-emission to produce stable Hg ${(}_{80}H{g}^{198})$. How much Hg (in g) will be present after 260 hr?

• A. $\frac{16}{15}$g
• B. $\frac{15}{16}$g
• C. $\frac{21}{20}$g
• D. $\frac{20}{21}$g

Answer 1

Answer: (B)

$\frac{15}{16}$g

${}_{79}A{u}^{198}\to {}_{80}H{g}^{198}+{}_{-1}{e}^0\left(\beta \right)$

$t_{1/2}=65hr;t=260hr$

$\Rightarrow$ Number of half lives $=\frac{260}{65}=4$

Mass of Au left $=\frac{w}{2^n}=\frac{1}{2^4}=\frac{1}{16}$

Mass of Hg produced $=1-\frac{1}{16}=\frac{15}{16}g$