Chemistry - Chemical Kinetics Question with Solution | TestHub

Initial amount of this isotope$N_0=100$

Final amount of the isotope$N=25$

We know that$N={\left(\frac{1}{2}\right)}^n{N}_0$
So,
$25={\left(\frac{1}{2}\right)}^n\times 100$or$\frac{25}{100}={\left(\frac{1}{2}\right)}^n$

or$\frac{1}{4}={\left(\frac{1}{2}\right)}^{n}or{\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^n$

or$n=2$

Time taken$T=n\times t_{\frac{1}{2}}$

$=2\times 2.5\times {10}^5=5\times {10}^5$years