Q24.35 P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the specific activity (in $\mathbf{\text{Bq}}\mathbf{/}\mathbf{\text{g}}$ ) if  $\mathbf{8}\mathbf{.58}\mathit{\mu }\mathbf{\text{g}}$ of an isotope emits  $\mathbf{7}\mathbf{.4}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{\alpha }$particles per minute?

Step-by-Step Solution

Answer

The specific activity is $1.43\times {10}^{-4} \text{Bq}/\text{g.}$

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: Calculate the specific activity

An isotope's mass is calculated as follows:

 $\begin{array}{c}\text{m}=8.58\mu \text{g}\\ =8.58\times {10}^{-6} \text{g.}\end{array}$

The radioactive isotope's activity is :

 $\text{A}=7.4\times {10}^4$decays$/\text{min}$ .

 $1 \text{Bq}=1\text{decay}/\text{s}$

We need to find the specific activity (in Bq/g)

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

$\begin{array}{c}\text{A}=7.4\times {10}^4\frac{\text{ decays }}{\min }\times \frac{1 \text{min}}{60 \text{s}}\\ =1.23\times {10}^3\frac{\text{ decays }}{\text{s}}\times \frac{1 \text{Bq}}{1\text{ decay }/\text{s}}\\ =1.23\times {10}^3 \text{Bq.}\end{array}$ 

We can now compute the specific activity.

Therefore, the specific activity is:

$\begin{array}{c}specificactivity=\frac{\text{A}}{\text{m}}\\ =\frac{1.23\times {10}^3 \text{Bq}}{8.58\times {10}^6 \text{g}}\\ =1.43\times {10}^{-4} \text{Bq}/\text{g.}\end{array}$