Chapter 30, Problem 53P

To determine

Age assuming that in living trees the ratio of ${}^{14}C{/}^{12}C$ atoms is about $1.3\times {10}^{-12}$ if an ancient wooden club is found that contains 73 g of carbon and has an activity of 7.0 decays per second.

Answer to Problem 53P

Solution:

Age =7900 yr

Explanation of Solution

Formula used:

Decay constant, $\lambda =\frac{\ln 2}{T_{1/2}}$, $T_{1/2}$ is the half life

Activity $=\lambda N$, where $\lambda$ is the decay constant and N is the number of nuclei remaining after a time t

Activity at a time t, $R=R_0{e}^{-\lambda t}$, $R_0$ is the activity at time t=0 and $\lambda$ is the decay constant

Calculation:

$N_{{}_6{}^{12}C}=\frac{73}{12}\times 6.02\times {10}^{23}=\text{3}\text{.662}\times {\text{10}}^{24}atoms$

$N_{{}_6{}^{14}C}=\text{3}\text{.662}\times {\text{10}}^{24}\times 1.3\times {10}^{-12}=\text{4}\text{.761}\times {10}^{12}atoms$

We also have $R=R_0{e}^{-\lambda t}$

⇨ $R=R_0{e}^{-\lambda t}$

$(\lambda N)=(\lambda N_0)e^{-\lambda t}$

i.e. $7=\frac{\ln 2}{5730\times 365\times 24\times 60\times 60}\times \text{4}\text{.761}\times {10}^{12}{e}^{-\frac{\ln 2}{5730}t}$

⇨ $t=7921yr\approx 7900\text{yr}$