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

The first step of the proton–proton chain isHere $d$ is the deuteron. A weak charged current converts one proton into a neutron, while the resulting proton–neutron pair becomes bound. The Q value follows from the difference of initial and final rest energies:This includes the emitted Positron's rest energy in the final-state mass and excludes its subsequent electron-positron annihilation. The nuclear binding energy of the deuteron makes this reaction possible although the isolated conversion $p\to n+e^{+}+{\nu }_e$ is energetically forbidden. The released energy is shared by final-state kinetic energies, including the escaping neutrino.

The radiation registered in the stated solar luminosity includes two Positron annihilations but excludes the two escaping neutrinos. Thus the visible energy per completed helium-producing chain isThe signs in this energy ledger matter: adding the neutrino energy would overestimate the energy available to sustain the measured radiation. Using $1\mathrm{MeV}=1.602176634\times 10^{-13}$ J, the chain rate and hydrogen-consumption rate areWith a year taken as $365.25$ days, the stated elapsed time is $1.4516496\times 10^{17}$ s. Consequently,At the same fixed luminosity, the remaining time would beOnly about $5.93\%$ of the stipulated initial reservoir has been consumed under these assumptions. This is a calculation for an entirely accessible hydrogen reservoir with constant luminosity. It is not a prediction of the Sun's remaining main-sequence lifetime: only part of the hydrogen participates in core burning, and stellar structure and luminosity evolve.

Interpret the displayed semi-empirical mass formula as an equation for mass energy, using $c=1$ for the mass terms. Its seven contributions are as follows.

The term $Z(m_p+m_e)$ is the rest energy of the $Z$ protons and $Z$ atomic electrons; $(A-Z)m_n$ is the rest energy of the neutrons. Including $Z{m}_e$ makes this an atomic-mass approximation. Small atomic electronic binding energies are neglected.

The volume contribution $-a_{v}A$ represents the short-range attractive strong interaction. Saturation gives each nucleon roughly a fixed number of neighbors, so the nuclear binding energy is approximately proportional to $A$; binding lowers the mass. The surface contribution $+a_s{A}^{2/3}$ restores the binding missing from nucleons near the surface. Approximately constant nuclear density gives radius proportional to $A^{1/3}$ and area proportional to $A^{2/3}$, explaining that power and the positive mass correction.

The Coulomb potential energy contribution $+a_c{Z}^2/A^{1/3}$ is positive because proton charges repel. A sphere's electrostatic self-energy scales as $Z^2/R$, and $R\propto A^{1/3}$. The more refined proton-pair count $Z(Z-1)$ differs from the $Z^2$ approximation explicitly used here.

The nuclear symmetry energy contribution $+a_a(Z-A/2{)}^2/A$ penalizes unequal neutron and proton populations. By the Pauli exclusion principle, filling one species' higher-momentum states is energetically more expensive than balancing the two Fermi seas; nuclear interactions also contribute to the empirical coefficient. The stated convention is equivalent to $23.285(N-Z{)}^2/A$ MeV, since $Z-A/2=(Z-N)/2$. It must not be combined with the usual coefficient near $23$ MeV without this factor of four.

Finally, nuclear pairing favors paired protons and paired neutrons. In the mass formula,The signs reverse when this correction is written as a contribution to positive nuclear binding energy. This smooth liquid-drop model omits detailed nuclear shell corrections.

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.

The supplied average retained-energy fraction in one elastic collision isThe mean-energy neutron moderation count follows by iterating the stated mean-energy model for neutron moderation, giving $E_n=f^n{E}_0$. Equivalently, take logarithms to solve the geometric decrease:Hence the estimated integer count isAfter $117$ collisions the predicted mean energy is about $0.0285$ eV; after $118$ it is about $0.0244$ eV, bracketing the target. This is a mean-energy estimate using the given average retention factor. Individual scattering angles and energy losses fluctuate, so a specified single neutron is not guaranteed to reach the target on exactly its $118$th collision. An estimate based on the average logarithmic energy decrement is a different statistical convention.