24.33 P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: What is the specific activity (in $\mathbf{\text{Ci}}\mathbf{/}\mathbf{\text{g}}$) if 1.65 mg of an isotope emits $\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{\alpha }$ particles per second?

Step-by-Step Solution

Answer

The specific activity is $2.56\times {10}^{2}Ci/g$.

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:

 $\begin{array}{rcl}\text{m}& =& 1.65\text{mg}\\ & =& 1.65\text{mg}\times \frac{1 \text{g}}{1000\text{mg}}\\ & =& 1.65\times {10}^3 \text{g}\end{array}$

The radioactive isotope's activity is :

 $\text{A}=1.56\times {10}^6$ decays .

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

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

Let's begin by converting the activity from decays/s to  

 $\begin{array}{rcl}\text{A}& =& 1.56\times {10}^6\text{decays}/\text{s}\times \frac{1\text{Ci}}{3.70\times {10}^{10}\text{decays}/\text{s}}\\ & =& 4.216\times {10}^5\text{Ci.}\end{array}$

We can now compute the specific activity.

 $\begin{array}{rcl}\text{specific activity}& =& \frac{\text{A}}{\text{m}}\\ & =& \frac{4.216\times {10}^5\text{Ci}}{1.65\times {10}^3 \text{g}}\\ & =& 2.56\times {10}^2\text{Ci}/\text{g}\end{array}$

Therefore, the required value is $2.56\times {10}^2\text{Ci}/\text{g}$.