Breeding plutonium produces energy even before any plutonium is fissioned. (The primary purpose of the four nuclear reactors at Chernobyl was breeding plutonium for weapons. Electrical power was a by-product used by the civilian population.) Calculate the energy produced in each of the reactions listed for plutonium breeding just following Example 32.4. The pertinent masses are \(m\left( {{\rm{ }}239{\rm{ U}}} \right){\rm{ }} = {\rm{ }}239.054289{\rm{ u }},{\rm{ }}m\left( {{\rm{ }}239{\rm{ Np}}} \right){\rm{ }} = {\rm{ }}239.052932{\rm{ u }},{\rm{ and }}m\left( {{\rm{ }}239{\rm{ Pu}}} \right){\rm{ }} = {\rm{ }}239.052157{\rm{ u}}\)

Amount of plutonium produced to the amount of uranium-235 needed this process is called plutonium breeding.

In this problem, we will calculate the energy released in the plutonium breeding cycle:

\(^{238}{\rm{U}} + n{ \to ^{239}}{\rm{U}} + \gamma \)

We are to find the energy released in this neutron-induced reaction. The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energy difference can be calculated by the free mass-energy formula:

\(E = \Delta m{c^2}\)

To get these mass differences we take parent nuclei and subtract from them daughter nuclei and masses of all other byproducts of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit we should remember how to convert it to\({\rm{eV}}\):

\(lu = \frac{{931.5\,{\rm{MeV}}}}{{{c^2}}}\)

To find the mass difference we subtract the atomic masses of parent and daughter nuclei. Trough Appendix A we can find the needed atomic masses:

\(\begin{array}{c}{\rm{Neutron }} = 1.008665\,{\rm{u}}\\^{238}{\rm{U}} = 238.05078\,{\rm{u}}\\^{239}{\rm{U}} = 239.054289\,{\rm{u}}\end{array}\)

So our expression for the mass difference will be:

\(\Delta m = m(n) + m\left( {^{238}{\rm{U}}} \right) - m\left( {^{238}{\rm{U}}} \right)\)

Putting in all of the numbers we got:

\(\begin{array}{l}\Delta m = 1.008645\,{\rm{u}} + 238.05078\,{\rm{u}} - 239.054289\,{\rm{u}}\\\Delta m = 0.005136{\rm{u}}\end{array}\)

From the mass-energy equivalence relation we have:

\(E = \Delta m{c^2} = \left( {0.005136 \times \frac{{931.5\,{\rm{MeV}}}}{{{c^2}}}} \right)\left( {{c^2}} \right)\)

And we get the result of:

\(E = \Delta m{c^2} = 4.78\,{\rm{MeV}}\)

For the second reaction we have\(\beta \)decay of uranium:

\(^{239}{\rm{U}}{ \to ^{239}}\;{\rm{Np}} + {\beta ^ - } + {v_e}\)

We just subtract the mass of neptunium from the mass of the uranium:

\(\Delta m = m(n) + m\left( {^{238}{\rm{U}}} \right) - m\left( {^{239}\;{\rm{Np}}} \right)\)

From the Appendix we can find the masses, so when we put in the numbers we got:

\(\Delta m = m(n) + m\left( {^{238}{\rm{U}}} \right) - m\left( {^{239}\;{\rm{Np}}} \right)\)

Which gives the result of:

\(\Delta m = 0.001357\,{\rm{u}}\)

To get the energy we follow the same procedure as for the first reaction:

\(E = \Delta m{c^2} = \left( {0.001357 \times \frac{{931.5\,{\rm{MeV}}}}{{{c^2}}}} \right) \times {c^2}\)

And we get the result of:

\(E = \Delta m{c^2} = 1.26\,{\rm{MeV}}\)

For the last reaction were plutonium is made we have:

\(^{239}\;{\rm{Np}}{ \to ^{239}}{\rm{Pu}} + {\beta ^ - } + {v_e}\)

For the mass difference we got:

\(\Delta m = m\left( {^{239}{\rm{Np}}} \right) - m\left( {^{239}{\rm{Pu}}} \right)\)

Putting in the numbers we have:

\(\Delta m = 239.052932\,{\rm{u}} - 239.052157\,{\rm{u}}\)

Which gives the result of:

\(\Delta m = 0.000775\,{\rm{u}}\)

To get the energy we follow the same procedure as for the first reaction:

\(E = \Delta m{c^2} = \left( {0.000775 \times \frac{{931.5\,{\rm{MeV}}}}{{{c^2}}}} \right)\left( {{c^2}} \right)\)

And we get the result of:

\(E = \Delta m{c^2} = 0.722\,{\rm{MeV}}\)

(a)\(E = 4.78\,{\rm{MeV}}\)

(b)\(E = 1.26\,{\rm{MeV}}\)

(c) \(E = 0.722\,{\rm{MeV}}\)