Chapter 30, Problem 45P

To determine

Part a:

How many Cs nuclei are present if we have $8.7\mu g$ initially for ${}_{55}{}^{124}Cs$ which has a half-life of 30.8 s?

Answer to Problem 45P

Solution:

Initial number of nuclei = $=4.2\times {10}^{16}$

Explanation of Solution

We have initial number of nuclei, $N_0=\frac{8.7\times {10}^{-6}}{124}\times 6.02\times {10}^{23}=4.2\times {10}^{16}nucei$

To determine

Part b:

How many are present 2.6 min later?

Answer to Problem 45P

Solution:

The number of nuclei remaining after a time 2.6 minute = $1.3\times {10}^{15}$

Explanation of Solution

Formula used:

The number of nuclei remaining after a time t, $N=N_0{e}^{-\lambda t}$, $N_0$ is the number of nuclei present after time t = 0 and $\lambda$ is the decay constant.

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

Calculation:

We have Decay constant, $\lambda =\frac{\ln 2}{T_{1/2}}=\frac{\ln 2}{30.8}=\text{0}\text{.022505 }{s}^{-1}$

The number of nuclei remaining after a time 2.6 minute, $N=N_0{e}^{-\lambda t}$

$=4.2\times {10}^{16}\times e^{-\text{0}\text{.022505}\times \text{2}\text{.6}\times \text{60}}=1.3\times {10}^{15}nuclei$

To determine

Part c:

What is the activity at time 2.6 min?

Answer to Problem 45P

Solution:

Activity $=2.8\times {10}^{13}$ decays/s

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:

Activity $=\lambda N=\text{0}\text{.022505}\times 1.3\times {10}^{15}=2.8\times {10}^{13}$ decays/s

To determine

Part d:

After how much time will the activity drop to less than about 1 per second?

Answer to Problem 45P

Solution:

Time required to drop activity to less than about 1 per second = 26 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

The number of nuclei remaining after a time t, $N=N_0{e}^{-\lambda t}$, $N_0$ is the number of nuclei present after time t = 0 and $\lambda$ is the decay constant.

Calculation:

Activity $=\lambda N=1$ decays/s

⇨ $N=\frac{1}{\text{0}\text{.022505}}=44.435$ nuclei

Now we have $N=N_0{e}^{-\lambda t}$

⇨ $44.435=4.2\times {10}^{16}{e}^{-\text{0}\text{.022505}t}$

So, time t = 1532 s $\approx 26$ min