Chapter 30, Problem 56P

To determine

Part a:

How long do we have to wait for the decay rate to drop to 25 per secondif ${}_4{}^{7}Be$ decays with a half-life of about 53 d.

Answer to Problem 56P

Solution:

Time taken = 200 days

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

Activity at a time t, $R=R_0{e}^{-\lambda t}$, $R_0$ is the activity at time t=0 and $\lambda$ is the decay constant

Calculation:

We have $R=R_0{e}^{-\lambda t}$

⇨ $t=-\frac{1}{\lambda }\ln \left(\frac{R}{R_0}\right)=-\frac{T_{1/2}}{\ln 2}\ln \left(\frac{25}{350}\right)$

$t=-\frac{58days}{\ln 2}\ln \left(\frac{25}{350}\right)=201.79days\approx 200days$

To determine

Part b:

The initial mass of ${}_4{}^{7}Be$ on the leaf if ${}_4{}^{7}Be$ decays with a half-life of about 53 d.

Answer to Problem 56P

Solution:

Mass of ${}_4{}^{7}Be$$=2.7\times {10}^{-14}g$

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

Activity at a time t, $R=R_0{e}^{-\lambda t}$, $R_0$ is the activity at time t=0 and $\lambda$ is the decay constant

Calculation:

We have Activity $=\lambda N$

$350=\frac{\ln 2}{T_{1/2}}\times \frac{m\times 6.02\times {10}^{23}}{7}$

⇨ Mass of ${}_4{}^{7}Be$, $m=2.7\times {10}^{-14}g$