Question

The half-life of ⁶⁰Co is 5.27 years. The activity of a ⁶⁰Co

sample is 3.50 x 10⁹ Bq. What is the mass of the sample?

Short answer

Hence the mass of the sample is :$8.361\times {10}^{-8}\mathrm{kg}$

Step-by-step solution

Given information

$$\begin{gathered} t_{1/2}=5.27\text{years} \\ R=3.50\times {10}^9\mathrm{Bq} \end{gathered}$$

Explanation

We will use the formula:

$R=r·N\to \text{(Equation 1.)}$

First, we must determine the value of the decay rate r:

$$\begin{gathered} r=\frac{1}{\tau } \\ r=\frac{\ln \left(2\right)}{t_{1/2}} \\ r=\frac{0.693}{\left(5.27\times 3.15\times {10}^7\right)\mathrm{s}} \\ r=\frac{0.693}{1.663\times {10}^8\mathrm{s}} \\ r=4.17\times {10}^{-9}\mathrm{s} \end{gathered}$$

Next, we must determine the value of N using equation 1:

$$\begin{gathered} R=r·N \\ N=\frac{R}{r} \\ N=\frac{3.50\times {10}^9\text{decays}/\mathrm{s}}{4.17\times {10}^{-9}\mathrm{s}} \\ N=8.39\times {10}^{17}\text{atoms} \end{gathered}$$

Explanation

Finally, we will find the mass:

$$\begin{gathered} M=m·N \\ M=\left(60\times \left(1.661\times {10}^{-27}\mathrm{kg}\right)\right)\times 8.39\times {10}^{17} \\ M=8.361\times {10}^{-8}\mathrm{kg} \end{gathered}$$