Chapter 45, Problem 60AP

(a)

To determine

The edge dimension of the cube.

Answer to Problem 60AP

The edge dimension of the cube is $15.4\text{cm}$.

Explanation of Solution

Write the equation for volume of a cube.

    $V=l^3$                                                                                                                  (I)

Here, $V$ is the volume of a cube and $l$ is the edge dimension of the cube.

Write the equation connecting mass, density and volume.

    $V=\frac{m}{\rho }$                                                                                                                (II)

Here, $m$ is the mass and $\rho$ is the density.

Write the equation for the edge dimension of the cube from equations (I) and (II).

    $\begin{array}{c}{l}^3=\frac{m}{\rho }\\ l={\left(\frac{m}{\rho }\right)}^{1/3}\end{array}$                                                                                                         (III)

Conclusion:

Substitute $70.0\text{kg}$ for $m$  and $19.1\times {10}^3{\text{kg/m}}^3$ for $\rho$ in equation (III) to find $l$.

    $\begin{array}{c}l={\left(\frac{70.0\text{kg}}{19.1\times {10}^3{\text{kg/m}}^3}\right)}^{1/3}\\ ={\left(0.00366\right)}^{1/3}\\ =\left(0.154\text{m}\right)\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =15.4\text{cm}\end{array}$

Thus, the edge dimension of the cube is $15.4\text{cm}$.

(b)

To determine

The net decay energy.

Answer to Problem 60AP

The net decay energy is $51.7\text{MeV}$.

Explanation of Solution

Write the given reaction.

    $\text{U}\to \text{8}\left(\text{H}\text{e}\right)+6\left(\text{e}\right)+\text{P}\text{b}+6\overline{\nu }+Q_{net}$

Add 92 electrons on both sides of the reaction to obtain the new nuclear reaction.

    $\text{U}\text{atom}\to \text{8}\text{H}\text{e}\text{atom}+\text{P}\text{b}\text{atom}$                                                          (IV)

Write the equation for Q value.

    $Q=\left[M_{\text{U}}-8M_{\text{H}\text{e}}-M_{\text{P}\text{b}}\right]c^2$                                                                        (V)

Conclusion:

Substitute $238.050783\text{u}$ for $M_{\text{U}}$, $4.002603\text{u}$ for $M_{\text{H}\text{e}}$ and $205.974449\text{u}$ for $M_{\text{P}\text{b}}$ in equation (V) to find $Q$.

    $\begin{array}{c}Q=\left[238.050783\text{u}-8\left(4.002603\text{u}\right)-205.974449\text{u}\right]c^2\\ =\left(0.05551\text{u}\right)c^2\left(\frac{931.5\text{MeV/}{c}^2}{\text{u}}\right)\\ =51.7\text{MeV}\end{array}$

Thus, the net decay of energy is $51.7\text{MeV}$.

(c)

To determine

Prove that the power output $P=QR$.

Answer to Problem 60AP

The power output is $P=QR$.

Explanation of Solution

The number of decays per second is equal to the decay rate given by $R$.

The energy released in each decay is $Q$.

Conclusion:

The energy released per unit time interval is the power output.

Write the equation for the power output.

    $P=QR$

Thus, the power output is $p=QR$.

(d)

To determine

The power output due to the radioactivity of the uranium and its daughters.

Answer to Problem 60AP

The power output due to the radioactivity of the uranium and its daughters is $2.27\times {10}^5\text{J/yr}$.

Explanation of Solution

Find the amount of the nuclei.

    $\begin{array}{c}N=\left(\frac{70.0\text{kg}}{238\text{g/mol}}\right)\\ =\left(\frac{\left(70.0\text{kg}\right)\left(\frac{{10}^{-3}\text{g}}{1\text{kg}}\right)}{238\text{g/mol}}\right)\left(6.02\times {10}^{23}\text{nuclei/mol}\right)\\ =1.77\times {10}^{26}\text{nuclei}\end{array}$

Write the equation for the decay constant.

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

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

Write the equation for the rate of decay.

    $R=\lambda N$                                                                                                          (VII)

Here, $R$ is the rate of decay.

Write the equation for the power output.

    $P=QR$                                                                                                         (VIII)

Here, $P$ is the power output and $Q$ is the Q factor.

Conclusion:

Substitute $4.47\times {10}^9\text{yr}$ for $T_{1/2}$ in equation (VI) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{\ln 2}{4.47\times {10}^9\text{yr}}\\ =\frac{0.693}{4.47\times {10}^9\text{yr}}\\ =1.55\times {10}^{-10}\frac{1}{\text{yr}}\end{array}$

Substitute $1.55\times {10}^{-10}\frac{1}{\text{yr}}$ for $\lambda$ and $1.77\times {10}^{26}\text{nuclei}$ for $N$ in equation (VII) to find $R$.

    $\begin{array}{c}R=\left(1.55\times {10}^{-10}\frac{1}{\text{yr}}\right)\left(1.77\times {10}^{26}\text{nuclei}\right)\\ =2.75\times {10}^{16}\text{decays/yr}\end{array}$

Substitute $2.75\times {10}^{16}\text{decays/yr}$ for $R$ and $51.7\text{MeV}$ for $Q$ in equation (VIII) to find $P$.

    $\begin{array}{c}P=\left(51.7\text{MeV}\right)\left(2.75\times {10}^{16}\text{decays/yr}\right)\\ =\left(51.7\text{MeV}\right)\left(2.75\times {10}^{16}\text{decays/yr}\right)\left(\frac{1.60\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\\ =2.27\times {10}^5\text{J/yr}\end{array}$

Thus, the power output due to the radioactivity of the uranium and its daughters is $2.27\times {10}^5\text{J/yr}$.

(e)

To determine

The rate per year at which the person absorb the energy of radiation.

Answer to Problem 60AP

The rate per year at which the person absorb the energy of radiation is $3.18\text{J/yr}$.

Explanation of Solution

Find the dose in rad/yr.

    $\text{dose in rad/yr}=\frac{\text{dose in rem/yr}}{\text{RBE}}$                                                                     (IX)

Here, $\text{RBE}$ is the relative biological effectiveness.

Write the equation for the allowed whole-body dose.

    $\text{dose}=m\left(\text{dose in rad/yr}\right)$                                                                                 (X)

Conclusion:

Substitute $5.00\text{rem/yr}$ for $\text{dose in rem/yr}$ and $1.10$ for $\text{RBE}$ in equation (IX) to find $\text{dose in rad/yr}$$\lambda$.

    $\begin{array}{c}\text{dose in rad/yr}=\frac{5.00\text{rem/yr}}{\text{1}\text{.10}}\\ =4.55\text{rad/yr}\end{array}$

Substitute $4.55\text{rad/yr}$ for $\text{dose in rad/yr}$ and $70.0\text{kg}$ for $m$ in equation (X) to find the dose.

    $\begin{array}{c}\text{dose}=\left(70.0\text{kg}\right)\left(4.55\text{rad/yr}\right)\\ =\left(70.0\text{kg}\right)\left(4.55\text{rad/yr}\right)\left(\frac{{10}^{-2}\text{J/yr}}{1\text{rad}}\right)\\ =3.18\text{J/yr}\end{array}$

Thus, the rate per year at which the person absorb the energy of radiation is $3.18\text{J/yr}$.