Chapter 29, Problem 85P

(a)

To determine

The percentage of the activity of ${}^{238}\text{Pu}$ source decreases in $10\text{ yr}$.

Answer to Problem 85P

The percentage of the activity of ${}^{238}\text{Pu}$ source decreases in $10\text{ yr}$ is $8\%$.

Explanation of Solution

Radioactive ${}^{238}\text{Pu}$ is used as a power supply for a pacemaker and the replacement period of the pacemaker is $10.0\text{ yr}$ and the radioactive ${}^{238}\text{Pu}$ nuclide decay via $\alpha$ decay with half-life $T_{1/2}=87.7\text{ yr}$.

Write the formula for the activity

    $R=R_0\left(2^{-t/T_{1/2}}\right)$                                                                                            (I)

Here, $R$ is the initial activity, $R_0$ is the activity after a time $t$, $t$ is the number of days for the activity and $T_{1/2}$ is the half-life.

Conclusion:

Substitute $87.7\text{ yr}$ for $T_{1/2}$ and $10.0\text{ yr}$ for $t$ in equation (I) to find the ratio of $R/R_0$

$\begin{array}{l}\frac{R}{R_0}=\left(2^{-10.0\text{ yr}/87.7\text{ yr}}\right)\\ \frac{R}{R_0}=\left(\frac{1}{2^{10.0\text{ yr}/87.7\text{ yr}}}\right)\\ \frac{R}{R_0}=0.92\end{array}$

The percentage of decrease in the acitivity is $100\%-92\%=8\%$.

Thus, the percentage of the activity of ${}^{238}\text{Pu}$ source decreases in $10\text{ yr}$ is $8\%$.

(b)

To determine

The power output initially and after $10.0\text{ yr}$

Answer to Problem 85P

The power output initially is $0.57\text{ mW}$ and after $10.0\text{ yr}$ is $0.52\text{ mW}$.

Explanation of Solution

The energy of $\alpha$ particle emitted is $5.6\text{ MeV}$ and assume all the energy is used to run the pacemaker. And the pacemaker starts with $m=1.0\text{ mg}$ of ${}^{238}\text{Pu}$. The molar pass of ${}^{238}\text{Pu}$ is $238\text{g}/\text{mol}$

Write the formula for half-life

    $T_{1/2}=\tau \ln 2$                                                                                           (I)

Here, $T_{1/2}$ is the half-life and $\tau$ is the time constant.

Write the formula for the initial number of nuclei

    $N_0=\frac{m{N}_A}{M}$                                                                                           (II)

Here, $N_0$ is the initial number of nuclei, $m$ is the mass, $N_A$ is the Avogadro’s number and $M$ is the molar mass.

Write the formula for the initial activity

    $R_0=\frac{N_0}{\tau }$                                                                                                 (III)

Here, $R_0$ is the initial activity.

Write the formula for power output

    $P_0=R_{0}E\times \text{efficiency}$                                                                                 (IV)

Here, $E$ is the energy per decay and $P_0$ is the initial power output

Conclusion:

The half-life of ${}^{238}\text{Pu}$ is $87.7\text{ yr}$. One year has $\text{3}.\text{1557}\times {10}^7\text{ s}$.

Substitute $238\text{g}/\text{mol}$ for $M$, $1.0\text{ mg}$ for $m$ and $6.022\times {10}^{23}\text{atoms}/\text{mol}$ for $N_A$ in equation (II) to find the value of $N_0$

$\begin{array}{l}{N}_0=\frac{1.0\times {10}^{-3}\text{ g}\times 6.022\times {10}^{23}\text{atoms}/\text{mol}}{238\text{g}/\text{mol}}\\ N_0=2.53\times {10}^{18}\text{ atoms}\end{array}$

The initial number of atoms is $2.53\times {10}^{18}\text{ atoms}$.

Substitute equation (I) in (III) and solve for $R_0$

$R_0=\frac{N_0}{\tau }=\frac{N_0\ln 2}{T_{1/2}}$

Substitute $2.53\times {10}^{18}\text{ atoms}$ for $N_0$ and $87.7\text{ yr}$ for $T_{1/2}$ in the above equation to find the value of $R_0$

$\begin{array}{l}{R}_0=\frac{2.53\times {10}^{18}\text{ atoms}\times \ln 2}{87.7\text{ yr}\times \text{3}.\text{1557}\times {10}^7\text{ s}}\\ R_0=6.34\times {10}^8\text{ Bq}\end{array}$

The initial activity is $6.34\times {10}^8\text{ Bq}$.

Substitute $6.34\times {10}^8\text{ Bq}$ for $R_0$, $5.6\text{ MeV}$ for $E$ and $1$ for the efficiency in equation (IV) to find the value of $P_0$

$\begin{array}{l}{P}_0=6.34\times {10}^8\text{ Bq}\times 5.6\text{ MeV}\times 1\\ P_0=5.7\times {10}^{-4}\text{W}\\ P_0=0.57\text{ mW}\end{array}$

The initial power output is $0.57\text{ mW}$.

The power output after $10.0\text{ yr}$ is $92\%$ of $P_0$ : $\begin{array}{l}{P}_{t=10\text{ yr}}=0.57\text{ mW}\times 0.92\\ P_{t=10\text{ yr}}=0.52\text{ mW}\end{array}$

The power output after $10.0\text{ yr}$ is $0.52\text{ mW}$.

Thus, the power output initially is $0.57\text{ mW}$ and after $10.0\text{ yr}$ is $0.52\text{ mW}$.