Question

The initial decay rate of a sample of a certain radioactive isotope is $\mathbf{\text{2.00}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{11}}}{\mathbf{\text{\hspace{0.17em}s}}}^{\mathbf{-}\mathbf{\text{1}}}$. After half an hour, the decay rate is$\mathbf{6.42}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$. Determine the half-life of the isotope.

Short answer

The half-life of the isotope is 18.3 minutes.

Step-by-step solution

Given data and the formula used.

Initial decay rate = $\text{2.00}\times {\text{10}}^{\text{11}}{\text{\hspace{0.17em}s}}^{-\text{1}}$

Decay rate after half hour =$\text{6.42}\times {\text{10}}^{\text{10}}{\text{\hspace{0.17em}s}}^{-\text{1}}$.

As we know, the Mathematical expression for the half-life is given by:

role="math" localid="1658465944953" ${\mathbf{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.693}}{\mathbf{\lambda }}$ ………….. (1)

Where$\mathbf{\lambda }$= disintegration constant.

Expression for the activity of radioactivity sample is given by:

$\mathbf{R}\mathbf{=}\mathbf{\lambda N}$ ………………. (2)

And radioactive decay equation is given by:

$\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda t}}$

Half-life of the isotope

The decay rate is related to the decay constant by:

$\begin{array}{l}{\mathrm{R}}_0={\mathrm{\lambda N}}_0\\ {\mathrm{R}}_1={\mathrm{\lambda N}}_1\end{array}$

The ratio of the number of particles is:

$\begin{array}{c}\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}=\frac{{\mathrm{R}}_1/\mathrm{\lambda }}{{\mathrm{R}}_0/\mathrm{\lambda }}\\ =\frac{{\mathrm{R}}_1}{{\mathrm{R}}_0}\end{array}$

Substitute $\text{6.42}\times {\text{10}}^{\text{10}}{\text{\hspace{0.17em}s}}^{-\text{1}}$for ${\mathrm{R}}_1$and $\text{2.00}\times {\text{10}}^{\text{11}}{\text{\hspace{0.17em}s}}^{-\text{1}}$ for $R_0$ in the above equation, we get:

$\begin{array}{c}\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}=\frac{6.42\times {10}^{10}{\text{\hspace{0.17em}s}}^{-\text{1}}}{2.00\times {10}^{11}{\text{\hspace{0.17em}s}}^{-\text{1}}}\\ =0.321\end{array}$

In half an hour, the number of the particle will change as follows:

$\begin{array}{c}{\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}\\ \frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}={\mathrm{e}}^{-\mathrm{\lambda t}}\end{array}$

Now applying logarithms on both sides, we will get:

$\begin{array}{c}\mathrm{In}\left(\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}\right)=-\mathrm{\lambda t}\\ \mathrm{\lambda }=-\frac{\mathrm{In}\left(\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}\right)}{\mathrm{t}}\end{array}$

Substitute$\text{2.00}\times {\text{10}}^{\text{11}}{\text{\hspace{0.17em}s}}^{-\text{1}}$ for${\mathrm{N}}_1$ and$\text{6.42}\times {\text{10}}^{\text{10}}{\text{\hspace{0.17em}s}}^{-\text{1}}$for${\mathrm{N}}_0$in the above equation, we get,

$\begin{array}{c}\mathrm{\lambda }=-\frac{\ln (0.321)}{30}\left(\frac{1.0}{60}\right)\\ =\frac{-(-1.136)}{0.5\times 60}\\ =0.0378{\min }^{-1}\end{array}$

The relation between the half-life period and disintegration constant is:

$\mathrm{\lambda }=\frac{\text{In}2}{{\mathrm{T}}_{1/2}}$

Rearranging the above equation for half-time, we get:

${\mathrm{T}}_{1/2}=\frac{\text{In}2}{\mathrm{\lambda }}$

Now, substitute $\text{0.0378\hspace{0.17em}}{\min }^{-1}$for disintegration constant in the above equation :

$\begin{array}{c}{\mathrm{T}}_{\frac{1}{2}}=\frac{\text{In}2}{0.0378{\min }^{-1}}\\ =18.3\min \end{array}$

Therefore the half-life of the isotope is 18.3 minutes.