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Initial decay rate = $\text{2.00}脳{\text{10}}^{\text{11}}{\text{鈥塻}}^{鈭�\text{1}}$

Decay rate after half hour =$\text{6.42}脳{\text{10}}^{\text{10}}{\text{鈥塻}}^{鈭�\text{1}}$.

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

role="math" localid="1658465944953" ${\mathbf{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.693}}{\mathbf{位}}$ 鈥︹�︹�︹��.. (1)

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

Expression for the activity of radioactivity sample is given by:

$\mathbf{R}\mathbf{=}\mathbf{位狈}$ 鈥︹�︹�︹�︹�︹��. (2)

And radioactive decay equation is given by:

$\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{鈭�}\mathbf{位迟}}$

The decay rate is related to the decay constant by:

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

The ratio of the number of particles is:

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

Substitute $\text{6.42}脳{\text{10}}^{\text{10}}{\text{鈥塻}}^{鈭�\text{1}}$for ${\mathrm{R}}_1$and $\text{2.00}脳{\text{10}}^{\text{11}}{\text{鈥塻}}^{鈭�\text{1}}$ for $R_0$ in the above equation, we get:

$\begin{array}{c}\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}=\frac{6.42脳{10}^{10}{\text{鈥塻}}^{鈭�\text{1}}}{2.00脳{10}^{11}{\text{鈥塻}}^{鈭�\text{1}}}\\ =0.321\end{array}$

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

$\begin{array}{c}{\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{鈭�\mathrm{位迟}}\\ \frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}={\mathrm{e}}^{鈭�\mathrm{位迟}}\end{array}$

Now applying logarithms on both sides, we will get:

$\begin{array}{c}\mathrm{In}\left(\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}\right)=鈭�\mathrm{位迟}\\ \mathrm{位}=鈭�\frac{\mathrm{In}\left(\frac{{\mathrm{N}}_1}{{\mathrm{N}}_0}\right)}{\mathrm{t}}\end{array}$

Substitute$\text{2.00}脳{\text{10}}^{\text{11}}{\text{鈥塻}}^{鈭�\text{1}}$ for${\mathrm{N}}_1$ and$\text{6.42}脳{\text{10}}^{\text{10}}{\text{鈥塻}}^{鈭�\text{1}}$for${\mathrm{N}}_0$in the above equation, we get,

$\begin{array}{c}\mathrm{位}=鈭�\frac{\ln (0.321)}{30}\left(\frac{1.0}{60}\right)\\ =\frac{鈭�(鈭�1.136)}{0.5脳60}\\ =0.0378\text{鈥�}{\min }^{鈭�1}\end{array}$

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

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

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

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

Now, substitute $\text{0.0378鈥�}{\min }^{鈭�1}$for disintegration constant in the above equation :

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

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