Chapter 30, Problem 76GP

To determine

To find:The activity of an average banana due to ${}_{19}^{40}\text{K}$

Answer to Problem 76GP

  $13\text{ decays }/\text{s}$

Explanation of Solution

Given data:

A typical banana contains $400\text{mg}$ of potassium

From the Appendix B, $N_{\text{K}}=(0.000117)N_{\text{naturally }}$

Molar mass of potassium is $39.0983\text{g}/\text{mol}$

Calculation:

Number of the radioactive isotope ${}_{19}^{40}\text{K}$

  $\begin{array}{c}{N}_{\text{K}}=(0.000117)N_{\text{naturally }}\\ =(0.000117)\frac{\left(400\times {10}^{-3}\text{g}\right)\left(6.02\times {10}^{23}\text{ atoms }/\text{mol}\right)}{39.0983\text{g}/\text{mol}}\\ =7.2\times {10}^{17}\end{array}$

Activity of banana

  $\begin{array}{c}R=\lambda N\\ =\frac{\ln 2}{T_{1/2}}N\\ =\frac{\ln 2}{1.265\times {10}^9\text{yr}}\left(7.2\times {10}^{17}\right)\left(\frac{1\text{yr}}{3.156\times {10}^7\text{s}}\right)\\ =12.5\text{ decays }/\text{s}\\ \approx 13\text{ decays }/\text{s}\end{array}$

Conclusion:

The activity of an average banana due to ${}_{19}^{40}\text{K}$ is $13\text{ decays }/\text{s}$