Question

A sample of ${}_{19}{K}^{40}$ disintegrates into two nuclei Ca and Ar with decay constant ${\lambda }_{ca}=4.5\times {10}^{-10}{s}^{-1}$ and ${\lambda }_{Ar}=0.5\times {10}^{-10}{s}^{-1}$ respectively. The time after which $99\%$ of ${}_{19}{K}^{40}$ get decayed is

• A. $6.2\times {10}^{9}sec$
• B. $9.2\times {10}^{9}sec$
• C. $7.2\times {10}^{9}sec$
• D. $4.2\times {10}^{9}sec$

Answer 1

The correct option is B $9.2\times {10}^{9}sec$

As one nuclie is disintegrates into two nuclei so the disintegration constant for both of them will add up
$\frac{dN}{dt}=-({\lambda }_{ca}+{\lambda }_{ar})N$
${\int }_{N_0}^N\frac{dN}{N}=-({\lambda }_{ca}+{\lambda }_{ar}){\int }_0^{t}dt$
$lnN{|}_{N_0}^N=-({\lambda }_{ca}+{\lambda }_{ar})t{|}_0^t$
$ln\frac{N}{N_0}=-({\lambda }_{ca}+{\lambda }_{ar})t$
$ln\frac{N}{N_0}=(4.5\times {10}^{-10}+0.5\times {10}^{-10})t$
$2.303{\log }_{10}\left(\frac{N_0}{N}\right)=5\times {10}^{-10}t$
When the potassium disintegrtate about $99\%$
$2.303{\log }_{10}\left(\frac{N_0}{N_0/100}\right)=5\times {10}^{-10}t$
$2.303{\log }_{10}\left({10}^2\right)=5\times {10}^{-10}t$
$2.303\times 2{\log }_{10}\left(10\right)=5\times {10}^{-10}t$
$2.303\times 2\times 1=5\times {10}^{-10}t$
$t=\frac{2.303\times 2}{5\times {10}^{-10}}$
$t=9.2\times {10}^{9}s$