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?

Short answer

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

Step-by-step solution

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.

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

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