Question

The half-life of ${}^{40}$K is 1.30 $\times {10}^9$ y. A sample of 1.00 g of pure KCI gives 160 counts $s^{-1}$. Calculate the relative abundance of ${}^{40}$K (fraction of ${}^{40}$K present) in natural potassium.

Answer 1

Given : Half life period

$t_{\frac{1}{2}}=1.30\times {10}^9$ years,

A = 160 counts/s

= $1.30\times {10}^9\times 365\times 86400$

$\therefore A=\lambda$ N

$\Rightarrow 160=\frac{0.693}{t_{\frac{1}{2}}}$N

$\Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^9\times 365\times 86400}\right)$N

$\Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^9}{0.693}$

=$9.5\times {10}^{18}$

$\because 6.023\times {10}^{23}$ No. of present in 40 gm.

$\therefore 9.5\times 10$ present in

= $\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}}$

= $\frac{4\times 9.5\times 10-4}{6.023}$ gm

= $6.309\times {10}^{-4}$ = 0.00063 gm

$\therefore$ The relative abudance at 40 K in natural potassium.

=$(2\times 0.00063\times 100)$ % = 0.12 %