Question

What is the half-life in days of a radioactive sample with 5.0 x 10¹⁵

atoms and an activity of 5.0 x 10⁸ Bq?

Short answer

Therefore the half life in days is$80.2\text{days}$

Step-by-step solution

Given information

$$\begin{gathered} N=5.0\times {10}^{15}\text{atoms} \\ R=5.0\times {10}^8\mathrm{Bq} \end{gathered}$$

Explanation

The activity of a radioactive sample is calculated using

$R=r·N\to \left(\text{Equation 1.}\right)$

The decay rate is:

$$\begin{gathered} r=\frac{1}{\tau } \\ r=\frac{\ln \left(2\right)}{t_{1/2}} \\ r=\frac{0.693}{t_{1/2}}(Equation2.) \end{gathered}$$

Calculation

Substituting the equation (2) in equation (1), we get

$$\begin{gathered} R=r·N \\ R=\frac{0.693}{t_{1/2}}·N \\ {t}_{1/2}·R=(0.693)·N \\ {t}_{1/2}=\frac{(0.693)·N}{R} \\ {t}_{1/2}=\frac{(0.693)·\left(5.0\times {10}^{15}\text{atoms}\right)}{\left(5.0\times {10}^8\mathrm{Bq}\right)} \\ {t}_{1/2}=6.93\times {10}^6\mathrm{s} \\ {\mathrm{t}}_{1/2}=6.93\times {10}^6\mathrm{s}\times \frac{1\text{day}}{86400\mathrm{s}} \\ {\mathrm{t}}_{1/2}=80.2\text{days} \end{gathered}$$