Chapter 30, Problem 74GP

To determine

To find:The half-life of the decay

Answer to Problem 74GP

  $1.1\times {10}^{11}\text{yr}$

Explanation of Solution

Given data:

Mass of natural samarium is 1.00 g

Rate of emit $\alpha$ particles is $120{\text{s}}^{-1}$

The natural abundance of ${}_{62}^{147}\text{Sm}$ is $15\%.$

Atomic mass of Natural samarium is 150.36 grams per mole.

Calculation:

We first find the number of ${}^{147}\text{Sm}$ nuclei from the mass and proportion information.

  $\begin{array}{c}{N}_{\text{S}\text{m}}=(0.15)N_{\text{natural }}\\ =\frac{(0.15)(1.00\text{g})\left(6.02\times {10}^{23}\text{ nuclei }/\text{mol}\right)}{150.36\text{g}/\text{mol}}\\ =6.006\times {10}^{20}{\text{ nuclei of }}_{62}^{147}\text{Sm}\end{array}$

The activity level is used to calculate the half-life.

  $\text{Activity }=R=\lambda N=\frac{\ln 2}{T_{1/2}}N$

Half-life of the decay

  $\begin{array}{c}{T}_{1/2}=\frac{\ln 2}{R}N\\ =\frac{\ln 2}{120\text{ decays }/\text{s}}\left(6.006\times {10}^{20}\right)\\ =3.469\times {10}^{18}\text{s}\left(\frac{1\text{yr}}{3.156\times {10}^7\text{s}}\right)\\ =1.1\times {10}^{11}\text{yr}\end{array}$

Conclusion:

The half-life of the decay is $1.1\times {10}^{11}\text{yr}$