Q24.34 P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the specific activity (in  $\mathbf{\text{Ci}}\mathbf{/}\mathbf{\text{g}}$ ) if  $\mathbf{2}\mathbf{.6}\mathbf{ }\mathbf{\text{g}}$of an isotope emits$\mathbf{4}\mathbf{.13}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}{\mathit{\beta }}^{\mathbf{-}}$  particles per hour?

Step-by-Step Solution

Answer

The specific activity is $1.19\times {10}^6\text{Ci}/\text{g.}$   

Step 1: Definition of Kinetics of Radioactive Decay

The decrease in the number of radioactive nuclei per unit time is the rate of decay, or activity, of a sample of a radioactive substance.

 

Step 2: Calculation of the specific activity

An isotope's mass is calculated as follows: $\text{m}=2.6 \text{g}$ .

The radioactive isotope's activity is :

$\text{A}=4.13\times {10}^8$  decays/hour.

$1\text{Ci}=3.70\times {10}^{10}$ decays .

We need to find the specific activity (in $\text{Ci}/\text{g}$ )

Let's begin by converting the activity from decays/s to  $\text{Ci}$

 $\begin{array}{c}\text{A}=4.13\times {10}^8\frac{\text{ decays }}{\text{ hour }}\times \frac{1\text{ hour }}{3600\text{ seconds }}\\ =1.147\times {10}^5\frac{1\text{Ci}}{\text{ decays }}\times \frac{1}{3.70\times {10}^{10}\text{ decays/second }}\\ =3.1\times {10}^6\text{Ci.}\end{array}$

Now we can calculate the specific activity

 $\begin{array}{c}\text{specific activity}=\frac{\text{A}}{\text{m}}\\ =\frac{3.1\times {10}^6\text{Ci}}{2.6 \text{g}}\\ =1.19\times {10}^6\text{Ci}/\text{g.}\end{array}$

Therefore, the required value is  $1.19\times {10}^6\text{Ci}/\text{g.}$