Chapter 30, Problem 50P

To determine

The mass of the sample of ${}_{92}{}^{238}U$ is decaying at a rate of $4.2\times {10}^2\text{decays/s}$.

Answer to Problem 50P

Solution:

The mass of the sample of ${}_{92}{}^{238}U$ = 33.8 mg

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:

We have Activity $=\lambda N$

$=\frac{\ln 2}{T_{1/2}}\times N=4.2\times {10}^{2}decays/s$

We know half -life for ${}_{92}{}^{238}U$$=4.468\times {10}^{9}yr$

⇨ $\frac{\ln 2}{4.468\times {10}^9\times 365\times 24\times 60\times 60}\times N=4.2\times {10}^{2}decays/s$

We will get $N=8.54\times {10}^{19}nuclei$

Mass of the sample $=\frac{8.54\times {10}^{19}}{6.023\times {10}^{23}}\times 238=3.38\times {10}^{-2}g=33.8mg$