Chapter 30, Problem 77GP

To determine

To find:The activity of boneafter 60,000 year.

Answer to Problem 77GP

  $0.2\text{decays/s}$

Explanation of Solution

Given data:

The mass of carbon 60,000 years ago was $1.0\text{kg}$ .

Half-life of carbon is $5730\text{yr}$

  $1\text{yr}\text{=}3.16\times {10}^7\text{s}$

Calculation:

Number of atom present in 1.0 kg of carbon

  $N=\frac{\left(1.0\times {10}^3\text{g}\right)\left(6.02\times {10}^{23}\text{ atoms }/\text{mol}\right)}{12\text{g}/\text{mol}}=5.017\times {10}^{25}\text{C}$ atoms

Number of ${}_6^{14}\text{C}$ atom in the bone at that time

  $N_0=N_{{}_6^{14}\text{C}}=\left(5.017\times {10}^{25}\right)\left(1.3\times {10}^{-12}\right)=6.522\times {10}^{13}$ nuclei

Activity of bone after 60000 yr

  $\begin{array}{c}R=\lambda N\\ =\frac{\ln 2}{T_{1/2}}N\\ =\frac{\ln 2}{T_{1/2}}{N}_0{e}^{-\frac{\ln 2}{T_{1/2}}t}\\ =\frac{\ln 2}{(5730\text{yr})\left(3.16\times {10}^7\text{s}/\text{yr}\right)}\left(6.522\times {10}^{13}\right)e^{-\frac{\ln 2(60,000\text{yr})}{(5730\text{yr})}}\\ =0.1759\text{decays/s}\approx 0.2\text{decays/s}\end{array}$

Conclusion:

The activity of bone is $0.2\text{decays/s}$