Q24.36 P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the specific activity (in  $\mathbf{\text{Bq}}\mathbf{/}\mathbf{\text{g}}$) if  of an isotope emits  $\mathbf{3}\mathbf{.77}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}{\mathit{\beta }}^{\mathbf{-}}$particles per minute?

Step-by-Step Solution

Answer

The specific activity is $5.87\cdot {10}^2 \text{Bq}/\text{g.}$

 

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:

 $\begin{array}{c}\text{m}=1.07 \text{kg}\\ =1.07 \text{kg}\times \frac{{10}^3 \text{g}}{1 \text{kg}}\\ =1.07\times {10}^3 \text{g.}\end{array}$

The radioactive isotope's activity is :

 $\text{A}=3.77\times {10}^7$decays  $/\text{min}$.

 $1 \text{Bq}=1$decay .

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

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

 $\begin{array}{c}\text{A}=3.77\times {10}^7\frac{\text{ decays }}{\text{min}}\times \frac{1 \text{min}}{60 \text{s}}\\ =6.28\times {10}^5\frac{\text{ decays }}{\text{s}}\times \frac{1 \text{Bq}}{1\text{ decay }/\text{s}}\\ =6.28\times {10}^5 \text{Bq.}\end{array}$

We can now compute the specific activity$=\frac{\text{A}}{\text{m}}$.

Therefore, the specific activity 

 $\begin{array}{c}=\frac{6.28\times {10}^5 \text{Bq}}{1.07\times {10}^3 \text{g}}\\ =5.87\times {10}^2 \text{Bq}/\text{g.}\end{array}$