Question

The decay constant of ${}_{80}^{197}Hg$ (electron capture to ${}_{79}^{197}Au$) is $1.8\times {10}^{-4}{s}^{-1}$. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert $25$% of this isotope of mercury into gold?

Answer 1

Decay constant is denoted by $\lambda$

$\lambda =1.8\times {10}^{-4}$

$\left(a\right)$Half life period means the time in which half of the sample will decay,

Half Life Period $=T=\frac{0.693}{\lambda }=\frac{0.693\times {10}^4}{1.8}=3850sec$

$\left(b\right)$Average life $T_A$is the reciprocal of the decay constant

$\therefore T_A=\frac{1}{\lambda }=\frac{1}{1.8\times {10}^{-4}}=5555.55sec$

$\left(c\right)$25% of isotpe will convert means, 75% will remain

means $N=0.75N_0$

${\log }_e\left(\frac{N}{N_0}\right)=-\lambda t$

$\therefore {\log }_{e}0.75=-\lambda t$

$\therefore t=\frac{{\log }_e\left(0.75\right)}{-\lambda }=\frac{-0.2877}{-1.8\times {10}^{-4}}=1598.33sec$