Question

1 g of ${}_{79}^{200}Au(T_{50}=13days)$ emits a $\beta$-particle forming a stable Hg atom. Hg formed at the end of 52 days is:

• A. 0.0625 g
• B. 0.9375 g
• C. 0.7500 g
• D. 0.2500 g

Answer 1

Answer: (B)

0.9375 g

$\beta$-particle emission gives isobar.
${}_{79}^{200}Au⟶{}_{82}^{200}Hg+{}_{-1}^{0}e$
$C=C_0{\left(\frac{1}{2}\right)}^y$
$y=\frac{52}{13}=4$
$C=1{\left(\frac{1}{2}\right)}^4=\frac{1}{16}$
$\therefore Auleft=\frac{1}{16}g$
Au changed to Hg $=1-\frac{1}{16}=\frac{15}{16}=0.9375g$