Chapter 14, Problem 34P

(a)

To determine

Calculate the energy in kWh released by the fission of  ${}^{239}\text{Pu}$.

Answer to Problem 34P

Energy released is $22.4\times {10}^6\text{kWh}$.

Explanation of Solution

Write the equation to find the number of Pu nuclei in one kilogram.

   $n=N_A\left(\frac{1\text{kg}}{M_{\text{Pu}}}\right)$

Here, $n$ the number of nuclei, $N_A$ the Avogadro number and the atomic mass of $M_{\text{Pu}}$.

Write the equation to find the energy released.

    $E=n{E}_{\text{per}\text{fission}}$

Here, $E$ is the energy released and $E_{\text{per}\text{fission}}$ the energy released per fission.

Conclusion:

Substitute $6.023\times {10}^{23}\text{nuclie}/\text{mole}$ $N_A$ $239.05\text{g}/\text{mole}$ $M_{\text{Pu}}$ in the equation for $n$.

    $\begin{array}{c}n=\left(6.023\times {10}^{23}\text{nuclie}/\text{mole}\right)\left(\frac{1\text{kg}}{239.05\text{g}/\text{mole}}\right)\\ =25.2\times {10}^{23}\text{nuclei}\end{array}$

Substitute $25.2\times {10}^{23}\text{nuclei}$ for $n$ and 200 MeV in the equation for $E$.

    $\begin{array}{c}E=\left(25.2\times {10}^{23}\text{nuclei}\right)\left(200\text{MeV}\right)\\ =\left(5.04\times {10}^{26}\text{MeV}\right)\left(4.45\times {10}^{-20}\text{kWh}/\text{MeV}\right)\\ =22.4\times {10}^6\text{ kWh}\end{array}$

Thus, the energy released is $22.4\times {10}^6\text{kWh}$.

(b)

To determine

Energy in MeV in D-T fusion.

Answer to Problem 34P

Energy released is $17.6\text{MeV}$.

Explanation of Solution

Write the equation for mass-energy equivalence.

  $E=\Delta m{c}^2$

Here, $E$ the energy, $\Delta m$ the mass defect and is the $c$ the speed of light in vacuum.

Write the equation for $\Delta m$ the reaction ${}_1^2\text{H}\text{+}{}_1^3\text{H}{\to }_2^4\text{He}\text{+}{}_0^1\text{n}$.

    $\Delta m=m\left({}_1^2\text{H}\right)\text{+}m\left({}_1^3\text{H}\right)-m\left({}_2^4\text{He}\right)-\text{m}\left({}_0^1\text{n}\right)$

Rewrite the equation for $E$ substituting the above equation.

    $E=\left(m\left({}_1^2\text{H}\right)\text{+}m\left({}_1^3\text{H}\right)-m\left({}_2^4\text{He}\right)-\text{m}\left({}_0^1\text{n}\right)\right)c^2$

Conclusion:

Substitute $3.016050\text{u}$ $m\left({}_1^2\text{H}\right)$, $2.014102\text{u}$ $m\left({}_1^3\text{H}\right)$, $4.002603\text{u}$ $m\left({}_2^4\text{He}\right)$, $1.008665\text{u}$ $\text{m}\left({}_0^1\text{n}\right)$   $931\text{MeV}$ for $c^2$ in the above equation.

    $\begin{array}{c}E=\left(3.016050\text{u}\text{+}2.014102\text{u}-4.002603\text{u}-1.008665\text{u}\right)\left(931\text{MeV}\right)\\ =17.6\text{MeV}\end{array}$

Thus, the energy released is $17.6\text{MeV}$.

(c)

To determine

Energy in kWh released during in deuterium fusion.

Answer to Problem 34P

Energy released is $2.18\times {10}^7\text{ kWh}$.

Explanation of Solution

Write the equation to find the energy released in deuterium fusion.

  $E=n_{\text{d}}{E}_{\text{per}\text{fusion}}$

Here, $n_{\text{d}}$ the number of deuterium nuclei and $E_{\text{per}\text{fusion}}$ the energy generated per fusion.

Write the equation to find $n_{\text{d}}$.

    $n_{\text{d}}=\frac{1}{2}{N}_A\frac{m}{M}$

Here, $m$ the given mass of deuterium and $M$ the atomic mass of deuterium.

Rewrite the equation for $E$ substituting the above equation.

    $E=\left(\frac{1}{2}{N}_A\frac{m}{M}\right)E_{\text{per}\text{fusion}}$

Conclusion:

Substitute $6.023\times {10}^{23}\text{atoms}/\text{mole}$ $N_A$, $1\text{kg}$ $m$, $\text{2}\text{.014 }\text{g}/\text{mole}$ $M$ and $3.27\text{ MeV}$ $E_{\text{per}\text{fusion}}$ the above equation.

    $\begin{array}{c}{n}_{\text{d}}=\frac{1}{2}\left(6.023\times {10}^{23}\text{atoms}/\text{mole}\right)\left(\frac{1\text{kg}\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\right)}{\text{2}\text{.014 }\text{g}/\text{mole}}\right)\left(3.27\text{MeV}\left(\frac{4.45\times {10}^{-20}\text{kWh}/\text{MeV}}{1\text{MeV}}\right)\right)\\ =2.18\times {10}^7\text{ kWh}\end{array}$

Thus, the energy released is $2.18\times {10}^7\text{ kWh}$.

(d)

To determine

Energy released in the combustion process.

Answer to Problem 34P

Energy released is $9.4\text{ kWh}$.

Explanation of Solution

Write the equation to find the energy released during the combustion.

  $E=n_c{E}_{\text{per}}$

Here, $n_c$ the number of carbon atoms and $E_{\text{per}}$ the energy generated by the combustion of one carbon atom.

Write the equation to find $n_c$.

    $n_c=\frac{N_A}{M}$

Here, $M$ the atomic mass of carbon.

Rewrite the equation for $E$ substituting the above equation.

    $E=\frac{N_A}{M}{E}_{\text{per}}$

Conclusion:

Substitute $6.023\times {10}^{23}\text{atoms}/\text{mole}$ $N_A$, $12.01\text{kg}/\text{mole}$ $M$ $4.2\text{ eV}$ $E_{\text{per}}$ the above equation.

  $\begin{array}{c}{E}_n=\left(\frac{6.02\times {10}^{26}\text{atoms}/\text{kg mole}}{12.01\text{kg}/\text{mole}}\right)\left(4.2\text{ eV}\left(\frac{{10}^{-6}\text{MeV}}{1\text{eV}}\right)\left(\frac{4.45\times {10}^{-20}\text{kWh }}{1\text{MeV}}\right)\right)\\ =9.4\text{ kWh}\end{array}$

Thus, the energy released is $9.4\text{ kWh}$.

(e)

To determine

Compare and contrast the advantages and disadvantages of nuclear fission, fusion and the combustion.

Answer to Problem 34P

Nuclear fission uses only less fuel, but produce radioactive pollutants, fusion gives clean and safe energy, but technology is insufficient and in case of combustion, method is cheap and no series safety issues, but air pollution is severe.

Explanation of Solution

The advantage of nuclear fission process is that energy can be constructed without less environmental damage compared to that of working of a hydrothermal power generation. Another point is that it uses only very less amount of fuel. But it accompanies certain disadvantages too. First one is that it produces radioactive pollutants. It is a very risky challenge to properly dispose it. Another point is that the fuel must be highly purified.

Nuclear fusion is process which requires temperatures of order of million kelvins. Current technology is insufficient to produce energy using fusion. Moreover, plant production cost must be very high than nuclear fission power plants. Also, it requires lithium and helium s fuel for superconducting magnets. Both are very rare resources. It accompanies also huge thermal pollution. But safety wise and pollution, fusion is far better than other processes.

Combustion is a cheap way of energy production and safety is not a serious issue. The main disadvantage is the heavy air pollution.

Thus, nuclear fission uses only less fuel, but produce radioactive pollutants, fusion gives clean and safe energy, but technology is insufficient and in case of combustion, method is cheap and no series safety issues, but air pollution is severe.

Conclusion: