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?

Short answer

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

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:

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

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:

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