Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 3 / d / Solution

The bookkeeping reaction isBoth mass number and electric charge are conserved: $236=92+140+4$ and $92=37+55$. Express each atomic mass as the sum of its constituent rest masses minus its nuclear binding energy. The total constituent masses, including the atomic-electron masses, cancel between the two sides. Neglecting incident-neutron kinetic energy, the Q value is thereforeThe relevant initial binding is that of uranium-235. Introducing uranium-236 as an intermediate state is possible only if the neutron-capture energy is also accounted for.

In the stated coefficient convention,Uranium-235 has odd $A$, hence ${\delta }_B=0$. Both products are odd–odd: rubidium has $(Z,N)=(37,55)$ and caesium has $(55,85)$. Each therefore has ${\delta }_B=-12/\sqrt{A}$ MeV. Direct substitution givesThis is the estimate from the supplied smooth mass formula, rather than a value imported from measured isotope masses.

For $100$ MW recovered as the stated fission energy, the annual energy output is $10^8(365.25)(86400)$ J. ThusEach event consumes one uranium-235 atom. With Avogadro constant $N_A=6.02214076\times 10^{23}$ mol${}^{-1}$ and molar mass approximated as $235$ g mol${}^{-1}$,This uses the stated power directly as recovered fission power; an unspecified conversion efficiency or additional heat-loss correction cannot be inferred from it.