Chapter 30, Problem 58P

(a)

To determine

Energy produced by the fission of $1.00\text{kg}$ of ${}^{239}\text{Pu}$.

Answer to Problem 58P

Energy produced by the fission of $1.00\text{kg}$ of ${}^{239}\text{Pu}$ is $2.24\times {10}^7\text{kWh}$.

Explanation of Solution

Write the equation to find the number of nuclie in the given mass.

    $n=\frac{N_{\text{ava}}}{M_{\text{Pu}}}m$

Here, $n$ is the number of nuclei, $N_{\text{ava}}$ is the Avogadro number, $M_{\text{Pu}}$ is the atomic mass of plutonium, and $m$ is the given mass in gram.

Write the equation to find the total energy produced.

    $E_{total}=nE$

Here, $E_{total}$ is the total energy produced and $E$ is the energy produced per fission.

Conclusion:

Substitute $6.02\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_{\text{ava}}$, $239.05\text{g}/\text{mol}$ for $M_{\text{Pu}}$, and $1.00\text{kg}$ for $m$ in the equation for $n$.

    $\begin{array}{c}n=\frac{6.02\times {10}^{23}\text{nuclei}/\text{mol}}{239.05\text{g}/\text{mol}}\left(1.00\text{kg}\left(\frac{{10}^3\text{g}}{1.00\text{kg}}\right)\right)\\ =2.52\times {10}^{24}\text{nuclei}\end{array}$

Substitute $2.52\times {10}^{24}\text{nuclei}$ for $n$ and $200\text{MeV}$ for $E$ in the equation for $E_{total}$.

  $\begin{array}{c}{E}_{total}=\left(2.52\times {10}^{24}\text{nuclei}\left(\frac{\text{1 fission}}{\text{nucleus}}\right)\right)\left(\frac{200\text{ MeV}}{\text{ fission}}\right)\\ =5.04\times {10}^{26}\text{MeV}\left(\frac{4.44\times {10}^{-20}\text{kWh}}{\text{1}\text{MeV}}\right)\\ =2.24\times {10}^7\text{kWh}\end{array}$

Therefore, the energy produced by the fission of $1.00\text{kg}$ of ${}^{239}\text{Pu}$ is $2.24\times {10}^7\text{kWh}$.

(b)

To determine

Energy produced by the deuterium –tritium fusion reaction.

Answer to Problem 58P

Energy produced by the deuterium –tritium fusion reaction is $17.6\text{MeV}/\text{fission}$.

Explanation of Solution

Write the equation to find the Energy produced by the deuterium –tritium fusion reaction.

    $E=\Delta m{c}^2$

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

Write the equation to find $\Delta m$.

    $\Delta m=m_{{}^{2}H}+m_{{}^{3}H}-m_{{}^{4}He}-m_{neutron}$

Here, $m_{{}^{2}H}$ is the mass of deuterium, $m_{{}^{3}H}$ is the mass of tritium, $m_{{}^{4}He}$ is the mass of helium nuclei, and $m_{neutron}$ is the mass of neutron.

Rewrite the expression for $E$ by substituting the equation for $\Delta m$.

    $E=\left(m_{{}^{3}H}+m_{{}^{2}H}-m_{{}^{4}He}-m_{neutron}\right)c^2$

Conclusion:

Substitute $3.016049\text{u}$ for $m_{{}^{3}H}$, $2.014102\text{ u}$ for $m_{{}^{2}H}$, $4.002603\text{u}$ for $m_{{}^{4}He}$, $1.008665\text{u}$ for $m_{neutron}$, and $931.5\text{MeV}/\text{u}$ for $c^2$ in the above equation to find $E$.

    $\begin{array}{c}E=\left(3.016049\text{u}+2.014102\text{u}-4.002603\text{u}-1.008665\text{u}\right){\left(931.5\text{MeV}/\text{u}\right)}^2\\ =17.6\text{MeV}/\text{fission}\end{array}$

Therefore, the energy produced by the deuterium –tritium fusion reaction is $17.6\text{MeV}/\text{fission}$.

(c)

To determine

Energy produced by the deuterium –tritium fusion reaction for $1.00\text{kg}$ of deuterium.

Answer to Problem 58P

Energy produced by the deuterium –tritium fusion reaction for $1.00\text{kg}$ of deuterium is $2.34\times {10}^8\text{kWh}$.

Explanation of Solution

Write the equation to find the energy produced by the deuterium –tritium fusion reaction for given mass of deuterium.

    $E_{total}=n_{de}E$                                                                                                  (I)

Here, $E_{total}$ is the energy produced in fusion and $n_{de}$ is the total number of deuterium nuclie.

Write the equation to find the number of nuclie in the given mass.

    $n=\frac{N_{\text{ava}}}{m_{{}^{2}H}}m$

Here, $n$ is the number of nuclie in the given mass.

Rewrite equation (I) by substituting the above relation for $n$.

  $E_{total}=\left(\frac{N_{\text{ava}}}{m_{{}^{2}H}}m\right)E$

Conclusion:

Substitute $6.02\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_{\text{ava}}$, $239.05\text{g}/\text{mol}$ for $M_{\text{Pu}}$, $1.00\text{kg}$ for $m$, and $17.6\text{MeV}/\text{fission}$ for $E$ in the above equation.

  $\begin{array}{c}{E}_{total}=\left(\frac{6.02\times {10}^{23}}{2.014\text{ g/mol}}\left(1.00\text{kg}\left(\frac{{10}^3\text{g}}{1.00\text{kg}}\right)\right)\right)\left(17.6\text{MeV}/\text{fission}\left(\frac{1\text{kWh}}{4.44\times {10}^{-20}\text{MeV}}\right)\right)\\ =2.34\times {10}^8\text{kWh}\end{array}$

Therefore, the energy produced by the deuterium –tritium fusion reaction for $1.00\text{kg}$ of deuterium is $2.34\times {10}^8\text{kWh}$.

(d)

To determine

Energy produced by the combustion of $1.00\text{kg}$ of carbon.

Answer to Problem 58P

Energy produced by the combustion of $1.00\text{kg}$ carbon is $9.36\text{ kWh}$.

Explanation of Solution

Write the equation to find the energy produced by the deuterium –tritium fusion reaction for given mass of deuterium.

    $E_{total}=n_C{E}_{com}$                                                                                                   (I)

Here, $E_{total}$ is the energy produced in combustion, $n_C$ is the number of carbon atoms, and $E_{com}$ is the energy produced by the formation of one $C{O}_2$ molecule.

Write the equation to find the number of nuclie in the given mass.

    $n=\frac{N_{\text{ava}}}{M_C}{m}_c$

Here, $M_C$ is the atomic mass of carbon and $m_c$ is the given mass in gram.

Rewrite the equation for $E_{total}$ by substituting the above relation for $n$.

  $E_{total}=\left(\frac{N_{\text{ava}}}{M_C}{m}_c\right)E_{com}$

Conclusion:

Substitute $6.02\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_{\text{ava}}$, $12\text{g}/\text{mol}$ for $M_{\text{C}}$, $1.00\text{kg}$ for $m$, and $4.20\text{MeV}$ for $E_{com}$ in the above equation.

    $\begin{array}{c}{E}_{total}=\left(\frac{6.02\times {10}^{23}\text{nuclei}/\text{mol}}{12\text{g}/\text{mol}}\left(1.00\text{kg}\right)\right)\left(4.20\text{MeV}\left(\frac{1\text{kWh}}{4.44\times {10}^{-20}\text{MeV}}\right)\right)\\ =9.36\text{ kWh}\end{array}$

Therefore, the energy produced by the combustion of $1.00\text{kg}$ carbon is $9.36\text{ kWh}$.

(e)

To determine

The pros and corns of energy production by fission, fusion, and combustion.

Explanation of Solution

To produce energy by combustion, coal is used. Coal is abundant form of fossil fuel and it is very cheap. The disadvantage is the high carbon emission and thereby acts as a major contributor in global warming. Energy production by nuclear fission cannot produce carbon and the working of reactor cannot cause any pollution. But the disposal of radioactive waste materials from the reactor is a very big challenge.

Nuclear fusion is better than fission since it does not produce radioactive by-products. But researches on nuclear fusion reactor are in developing stage only and it requires extreme high temperature for its working that cannot be achieved by man at present. Plutonium is a very risky material to handle in fission process.

Therefore, combustion is cheaper, but carbon emission is high, fission cannot cause global warming but produce radioactive bye products, and nuclear fusion cannot produce radioactive pollution but requires very large temperature that cannot be achieved in a laboratory at present.