Chapter 30, Problem 71GP

(a)

To determine

To find: Time taken before there is only one $\text{C}$ nucleus left.

Answer to Problem 71GP

Time taken before there is only one $\text{C}$ nucleus left is $t=2.4\times {10}^5\text{yr}$

Explanation of Solution

Given:

Mass of sample, $m=92\text{ g}$

Calculation:

The fraction of 14-carbon to the 12-carbon is given as: $1.3\times {10}^{-12}$

So, the number of carbon atoms can be calculated as follows:

  $\begin{array}{l}N=\left(\frac{m}{M}\right)N_A\\ N=\left(\frac{92\text{ g}}{12\text{ g/mol}}\right)\left(6.022\times {10}^{23}\text{atoms/mol}\right)\\ N=4.62\times {10}^{24}\text{atoms}\end{array}$

Now, the number of nuclei of the sample can be given as:

  $\begin{array}{l}{N}_s=\left(1.3\times {10}^{-12}\right)\left(4.62\times {10}^{24}\right)\\ N_s=6\times {10}^{12}\text{nuclei}\end{array}$

Also, the half-life can be given as:

  $\begin{array}{l}\frac{N}{N_s}={\left(\frac{1}{2}\right)}^n\\ n\log 2=\log N_s\\ n=\log \left(\frac{N_s}{2}\right)\\ n=\log \left(\frac{6\times {10}^{12}\text{nuclei}}{2}\right)\\ n=42.4\end{array}$

So, the time taken will be:

  $\begin{array}{l}t=n{T}_{\frac{1}{2}}\\ t=\left(42.4\right)\left(5730\text{ yr}\right)\\ t=2.4\times {10}^5\text{yr}\end{array}$

Conclusion:

Time taken before there is only one $\text{C}$ nucleus left is $t=2.4\times {10}^5\text{yr}$

(b)

To determine

To find: Time taken before there is only one $\text{C}$ nucleus left if the mass of the sample is changed.

Answer to Problem 71GP

Time taken before there is only one $\text{C}$ nucleus left if the mass of the sample is changed is $t=2.5\times {10}^5\text{yr}$ and the limit of the carbon dating is about ${10}^5\text{yr}$

Explanation of Solution

Given:

Mass of the sample is $m=280\text{ g}$

Calculation:

The fraction of 14-carbon to the 12-carbon is given as: $1.3\times {10}^{-12}$

So, the number of carbon atoms can be calculated as follows:

  $\begin{array}{l}N=\left(\frac{m}{M}\right)N_A\\ N=\left(\frac{280\text{ g}}{12\text{ g/mol}}\right)\left(6.022\times {10}^{23}\text{atoms/mol}\right)\\ N=1.41\times {10}^{25}\text{atoms}\end{array}$

Now, the number of nuclei of the sample can be given as:

  $\begin{array}{l}{N}_s=\left(1.3\times {10}^{-12}\right)\left(1.41\times {10}^{25}\right)\\ N_s=1.83\times {10}^{13}\text{nuclei}\end{array}$

Also, the half-life can be given as:

  $\begin{array}{l}\frac{N}{N_s}={\left(\frac{1}{2}\right)}^n\\ n\log 2=\log N_s\\ n=\log \left(\frac{N_s}{2}\right)\\ n=\log \left(\frac{1.83\times {10}^{13}}{2}\right)\\ n=44.1\end{array}$

So, the time taken will be:

  $\begin{array}{l}t=n{T}_{\frac{1}{2}}\\ t=\left(44.1\right)\left(5730\text{ yr}\right)\\ t=2.5\times {10}^5\text{yr}\end{array}$

This tells us that the limit of carbon dating is about ${10}^5\text{yr}$

Conclusion:

Time taken before there is only one $\text{C}$ nucleus left if the mass of the sample is changed is $t=2.5\times {10}^5\text{yr}$ and the limit of the carbon dating is about ${10}^5\text{yr}$