Chapter 30, Problem 48P

To determine

Part a:

The decay constant for a sample of ${}_{92}{}^{233}U$ with $T_{1/2}=1.59\times {10}^5$ yr?

Answer to Problem 48P

Solution:

Decay constant $=1.38\times {10}^{-13}{s}^{-1}$

Explanation of Solution

Formula used:

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

Calculation:

We have decay constant, $\lambda =\frac{\ln 2}{T_{1/2}}$

$=\frac{\ln 2}{1.59\times {10}^5\times 365\times 24\times 60\times 60}=1.38\times {10}^{-13}{s}^{-1}$

To determine

Part b:

Approximately how much disintegrations will occur per minute for a sample of ${}_{92}{}^{233}U$ with $T_{1/2}=1.59\times {10}^5$ yr and containing $4.50\times {10}^{18}$ nuclei?

Answer to Problem 48P

Solution:

Number of disintegrations $=3.73\times {10}^{7}decays/\min$

Explanation of Solution

Formula used:

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 $=\lambda N$

$=1.38\times {10}^{-13}\times 4.50\times {10}^{18}=6.215\times {10}^{5}decays/s$

$=3.73\times {10}^{7}decays/\min$