Chapter 30, Problem 49P

To determine

How many disintegrations occur per second for a 345-g sample of pure carbon contains 1.3 parts in ${10}^{12}$ (atoms)of ${}_6{}^{14}C$?

Answer to Problem 49P

Solution:

Disintegrations occur per second = 86

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:

$N_0$ of pure carbon $=\frac{345}{12}\times 6.023\times {10}^{23}=1.73\times {10}^{25}nuclei$

$N_0$ of ${}_6{}^{14}C$= $\frac{1.3}{{10}^{12}}\times 1.73\times {10}^{25}=2.25\times {10}^{13}nuclei$

We have half-life for ${}_6{}^{14}C$=5730 yr

⇨ Decay constant, $\lambda =\frac{\ln 2}{T_{1/2}}=\frac{\ln 2}{5730\times 365\times 24\times 60\times 60}$

Activity $=\lambda N_0=\frac{\ln 2}{5730\times 365\times 24\times 60\times 60}\times 2.25\times {10}^{13}$

$=86$ decays/s