Chapter 30, Problem 51P

To determine

Age of the fossils if the rubidium isotope ${}_{37}{}^{87}Rb$ a $\beta$ emitter with a half-life of $4.75\times {10}^{10}yr$ is used to determine the age of rocks and fossils.

Answer to Problem 51P

Solution:

Age of the fossils $=1.78\times {10}^{9}yr$

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

Calculation:

Assume that the amount of ${}_{37}{}^{87}Rb$ decayed is equal to amount of ${}_{38}{}^{87}Sr$ formed.

So that $N_{Sr}=-\Delta N_{Rb}=\lambda N_{Rb}\Delta t$

⇨ $\Delta t=\frac{N_{Sr}}{\lambda N_{Rb}}=\frac{0.0260\times T_{1/2}}{\ln 2}=\frac{0.0260\times 4.75\times {10}^{10}yr}{\ln 2}$

Age of the fossils $=1.78\times {10}^{9}yr$