Question

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

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

Answer 1

The correct option is B

$9.2\times {10}^9\mathrm{s}$

1. This reaction belongs to the category of parallel reactions in which a single reactant simultaneously forms two product.
2. The reaction is a radioactive decay and all radioactive decays follow first order kinetics.

Step 1: Formula for parallel first order kinetics- $\mathrm{N}={\mathrm{N}}_{\circ }{\mathrm{e}}^{-\left({\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2\right)\mathrm{t}}$ where, $\mathrm{N}$ is the final concentration of the sample, ${\mathrm{N}}_{\circ }$ is the initial concentration of the sample, $\mathrm{t}$ is the time taken, $\mathrm{\lambda }$ is the disintegration constant or the rate constant of the reaction.

We can write- $\frac{\mathrm{N}}{{\mathrm{N}}_{\circ }}=e^{-\left({\lambda }_1+{\lambda }_2\right)t}$

Step 2 : For $99\%$ completion we can write- $\frac{\mathrm{N}}{{\mathrm{N}}_{\circ }}=\frac{1}{100}$

${\mathrm{\lambda }}_{\mathrm{Ca}}=4.5\times {10}^{-10}{\mathrm{s}}^{-1}$

${\mathrm{\lambda }}_{\mathrm{Ar}}=0.5\times {10}^{-10}{\mathrm{s}}^{-1}$

Step 3: Calculation of the time after which $99\%$ of ${}_{19}\mathrm{K}^{40}$gets decayed -

$\frac{1}{100}=e^{-\left(\left(4.5+0.5\right){10}^{-10}\right)t}$

$-\ln 100=-\left(5\times {10}^{-10}\right)\mathrm{t}$

$\mathrm{t}=\frac{\ln 100}{5\times {10}^{-10}}$ s

$\mathrm{t}=9.2\times {10}^9$ s

So, the time after which $99\%$ of ${}_{19}\mathrm{K}^{40}$gets decayed is - $9.2\times {10}^9$ s.

Hence, option B is the correct answer.