Question

What is the specific activity (in $\mathbf{\text{Ci}}\mathbf{/}\mathbf{\text{g}}$) if $\mathbf{2}\mathbf{.6}\mathbf{}\mathbf{\text{g}}$of an isotope emits$\mathbf{4}\mathbf{.13}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}{\mathit{\beta }}^{\mathbf{-}}$ particles per hour?

Short answer

The specific activity is $1.19\times {10}^6\text{Ci}/\text{g.}$

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.

Calculation of the specific activity

An isotope's mass is calculated as follows: $\text{m}=2.6\text{g}$.

The radioactive isotope's activity is :

$\text{A}=4.13\times {10}^8$decays/hour.

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

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

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

$\begin{array}{c}\text{A}=4.13\times {10}^8\frac{\text{decays}}{\text{hour}}\times \frac{1\text{hour}}{3600\text{seconds}}\\ =1.147\times {10}^5\frac{1\text{Ci}}{\text{decays}}\times \frac{1}{3.70\times {10}^{10}\text{decays/second}}\\ =3.1\times {10}^6\text{Ci.}\end{array}$

Now we can calculate the specific activity

$\begin{array}{c}\text{specificactivity}=\frac{\text{A}}{\text{m}}\\ =\frac{3.1\times {10}^6\text{Ci}}{2.6\text{g}}\\ =1.19\times {10}^6\text{Ci}/\text{g.}\end{array}$

Therefore, the required value is $1.19\times {10}^6\text{Ci}/\text{g.}$