Chapter 44, Problem 63AP

(a)

To determine

Find the number of ${}_{91}^{239}\text{Pu}$ nuclei at $t=0$.

Answer to Problem 63AP

The number of ${}_{91}^{239}\text{Pu}$ nuclei at $t=0$ is $2.52\times {10}^{24}$.

Explanation of Solution

Write the equation for number of nuclei,

    $N_0=\frac{\text{mass}}{\text{mass per atom}}$                                                     (I)

Conclusion:

Substitute $1.00\text{ kg}$ for mass and $(\text{239}\text{.05 u})\left(1.66\times {10}^{-27}\text{kg}/\text{u}\right)$ for mass per atom in equation I.

    $\begin{array}{c}{N}_0=\frac{1.00\text{ kg}}{(\text{239}\text{.05 u})\left(1.66\times {10}^{-27}\text{kg}/\text{u}\right)}\\ =2.52\times {10}^{24}\end{array}$

Therefore, the number of ${}_{91}^{239}\text{Pu}$ nuclei at $t=0$ is $2.52\times {10}^{24}$.

(b)

To determine

Find the initial activity in the sample.

Answer to Problem 63AP

The initial activity in the sample is $2.29\times {10}^{12}\text{ Bq}$.

Explanation of Solution

Write the equation for decay constant.

    $\lambda =\frac{\ln 2}{T_{1/2}}$                                                     (II)

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

Write the equation for initial activity.

    $R_0=\lambda N_0$                                                     (III)

Here, $R_0$ is the initial activity and $N_0$ is the number of nuclei.

Conclusion:

Substitute $2.412\times {10}^4\text{ yr}$ for $T_{1/2}$ in equation II.

    $\begin{array}{c}\lambda =\frac{\ln 2}{\left(2.412\times {10}^4\text{ yr}\right)\left(3.156\times {10}^7\text{s}/\text{yr}\right)}\\ =9.106\times {10}^{-13}{\text{ s}}^{-1}\end{array}$

Substitute $9.106\times {10}^{-13}{\text{ s}}^{-1}$ for $\lambda$ and $2.52\times {10}^{24}$ for $N_0$ in equation III.

    $\begin{array}{c}{R}_0=\left(9.106\times {10}^{-13}{\text{ s}}^{-1}\right)\left(2.52\times {10}^{24}\right)\\ =2.29\times {10}^{12}\text{ Bq}\end{array}$

Therefore, the initial activity in the sample is $2.29\times {10}^{12}\text{ Bq}$.

(c)

To determine

Find the time interval does the sample have to be stored.

Answer to Problem 63AP

The time interval does the sample have to be stored is $1.07\times {10}^6\text{ yr}$.

Explanation of Solution

Write the equation for radioactive decay.

    $R=R_0{e}^{-\lambda t}$                                                    (IV)

Here, $R$ is the decay rate, $R_0$ is the initial decay, $\lambda$ is the decay constant and $t$ is the time.

Write the equation for decay constant.

    $\lambda =\frac{\ln 2}{t_{1/2}}$                                                    (V)

Here, $t{1/2}_{}$ is the half life time.

Substitute equation V in IV and rewrite to get time.

    $t=\frac{1}{\lambda }\ln \left(\frac{R_0}{R}\right)$                                                    (VI)

Conclusion:

Substitute $9.106\times {10}^{-13}{\text{ s}}^{-1}$ for $\lambda$, $2.29\times {10}^{12}\text{ Bq}$ for $R_0$ and $0.100\text{ Bq}$ for $R$ in equation VI.

    $\begin{array}{c}t=\frac{1}{9.106\times {10}^{-13}{\text{ s}}^{-1}}\ln \left(\frac{2.29\times {10}^{12}\text{ Bq}}{\text{0}\text{.100 Bq}}\right)\\ =3.38\times {10}^{13}\text{ s}\left(\frac{\text{1 yr}}{\text{3}\text{.156}\times {\text{10}}^{\text{7}}\text{ s}}\right)\\ =1.07\times {10}^6\text{ yr}\end{array}$

Therefore, the time interval does the sample have to be stored is $1.07\times {10}^6\text{ yr}$.