Q. $1\ g\ of\ \ _{79}{Au}^{198}\left( t_{1/2} = 65\ h \right)$ give stable mercury by $\beta$- emission. What amount of mercury will left after 260 h?
A) 0.9375 g
B) 0.3758 g
C) 0.7586 g
D) 0.9000 g
Explanation
(a)
$\ _{79}{Au}^{198}\overset{\ \ \ \ - B\ \ \ \ \ }{\rightarrow}\ _{80}{Hg}^{198}$
$\ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{65}$
After 260 hr,
$\ k = \frac{2.303}{260}\log\frac{a}{a - x}$
$\frac{0.693}{65} = \frac{2.303}{260}\log\frac{a}{a - x}$
$\frac{a}{a - x} = 16$
$\ \frac{1}{1 - x} = 16$
$x = \frac{15}{16}\ g = 0.9375\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.