Solution
= Solution
\b[Smooth-model fission energy]. Use <natural units> $c=1$, so the quoted nuclear <masses> are expressed as rest <energies>. Conservation of <mass number> and <atomic number> fixes the second fragment to $(A_2,Z_2)=((1-y)A,(1-y)Z)$. First omit <nuclear pairing> to compare a continuous range of $y$. Pairing can subsequently add $\delta_1+\delta_2-\delta_{\rm parent}$ for integer fragments; it has no smooth parity assignment for arbitrary real $y$.
In the parent-minus-daughters mass difference, the <nucleon> <rest energy> and volume terms cancel. The <nuclear symmetry energy> also cancels because both fragments have the same $Z/A$:
$$
\frac{(A_1-2Z_1)^2}{A_1}+\frac{(A_2-2Z_2)^2}{A_2}
=[y+(1-y)]\frac{(A-2Z)^2}{A}.
$$
The remaining <liquid-drop fission energy> is
$$
\boxed{\begin{aligned}
Q(y)={}&a_SA^{2/3}\left[1-y^{2/3}-(1-y)^{2/3}\right]\\
&+\frac{a_CZ^2}{A^{1/3}}\left[1-y^{5/3}-(1-y)^{5/3}\right].
\end{aligned}}
$$
Splitting increases the <surface energy> cost, but reduces <nuclear Coulomb energy>. Both contributions are symmetric under $y\leftrightarrow1-y$, so the symmetric split is stationary. Symmetry alone would not establish a maximum, since the two contributions compete.
For a full check, put $S=a_SA^{2/3}$, $x=a_CZ^2/(a_SA)$ and $t=y^{1/3}+(1-y)^{1/3}$. On $0\leq y\leq1/2$, $t$ increases from $1$ to $2^{2/3}$. Expressing the two fractional-power sums through $t$ gives
$$
\frac QS=1+x-\frac{t^2+2/t}{3}+\frac{x}{9}(t^5-5t^2-5/t),
\qquad
\frac{d(Q/S)}{dt}=\frac{t^3-1}{9t^2}\left[5x(t^3-1)-6\right].
$$
The derivative can change sign only from negative to positive, so any interior stationary point in $t$ is a minimum. The largest value is therefore either the no-split endpoint $Q=0$ or the symmetric split. Thus <symmetric fission maximizes the positive liquid-drop energy release>: whenever this smooth model permits an exothermic split, its largest release occurs at $y=1/2$.
There is a qualification to the paper's unqualified wording. For arbitrary low $Z^2/A$, the symmetric split need not maximize $Q$; indeed it is a local minimum when $Z^2/A<(2/5)a_S/a_C=10$. The symmetric-maximum conclusion is valid in the heavy, fission-favorable regime relevant here, rather than a universal property of the formula.
At $y=1/2$,
$$
Q_{\rm sym}=-a_SA^{2/3}(2^{1/3}-1)
+\frac{a_CZ^2}{A^{1/3}}(1-2^{-2/3}).
$$
Requiring $Q_{\rm sym}>0$ gives
$$
\boxed{\frac{Z^2}{A}>\frac{a_S}{a_C}
\frac{2^{1/3}-1}{1-2^{-2/3}}
=\frac{18.0}{0.72}\frac{0.259921}{0.370039}
\simeq17.6.}
$$
\b[Barrier and fragment yields]. Positive final-state <Q value> does not remove the <fission barrier> along a deformation path. <Surface energy> initially rises as the nucleus deforms; <spontaneous fission> then requires collective <quantum tunnelling> unless enough excitation is present. Its rate is exponentially sensitive to barrier height, width and collective inertia. Thus nuclei close to the exothermicity threshold can have immeasurably long spontaneous-fission <mean lifetimes>, while sufficiently heavier nuclei have smaller effective <fission barriers> or larger <quantum tunnelling> probabilities.
The <fission-fragment mass distribution> of low-excitation actinide fission is usually double peaked: complementary fragments have mass fractions roughly $0.4$ and $0.6$ rather than both exactly $1/2$. <Nuclear shell model> effects and the fission dynamics help favor asymmetric fragments. Maximizing the final smooth-model energy release is not a calculation of the fragment yield. The following yield curve is schematic, with no measured counts implied:
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2018-fission-energy-and-yield.png]
{title=Smooth-model fission energy versus a schematic asymmetric fragment-yield distribution}
\b[Capture excitation energies]. Capturing a negligible-energy neutron forms a compound nucleus with the rest energy of the target plus the free <neutron>. Relative to the compound <ground state>, the <neutron-capture excitation energy> is therefore
$$
E^*=M(A,92)+m_n-M(A+1,92)
=B(A+1,92)-B(A,92)=S_n(A+1,92),
$$
where $B$ is positive <nuclear binding energy> and $S_n$ the compound nucleus's <neutron separation energy>. The free <neutron> <rest energy> cancels in the binding-energy difference. With the supplied <semi-empirical mass formula>,
$$
\begin{aligned}
S_n={}&a_V-a_S[(A+1)^{2/3}-A^{2/3}]\\
&-a_CZ^2[(A+1)^{-1/3}-A^{-1/3}]\\
&-a_A\left[\frac{(A+1-2Z)^2}{A+1}-\frac{(A-2Z)^2}{A}\right]
+\delta(A+1,Z)-\delta(A,Z).
\end{aligned}
$$
At $Z=92$, <uranium-235> has odd $A$, while <uranium-236> is even-even. The pairing contribution is $+33.5(236)^{-3/4}=+0.55637\,\mathrm{MeV}$. By contrast <uranium-238> is even-even and <uranium-239> has odd $A$, giving $-33.5(238)^{-3/4}=-0.55286\,\mathrm{MeV}$. The individual contributions are
$$
\begin{array}{c|rr}
\text{Contribution (MeV)}&{}^{235}\mathrm U+n&{}^{238}\mathrm U+n\\\hline
\text{Volume}&15.80000&15.80000\\
\text{Surface}&-1.94320&-1.93502\\
\text{Coulomb}&1.39680&1.37342\\
\text{Asymmetry}&-9.15424&-9.51287\\
\text{Pairing}&0.55637&-0.55286
\end{array}
$$
Thus the model predicts
$$
\boxed{E^*({}^{236}\mathrm U)=6.65573\,\mathrm{MeV}\simeq6.66\,\mathrm{MeV},\qquad
E^*({}^{239}\mathrm U)=5.17268\,\mathrm{MeV}\simeq5.17\,\mathrm{MeV}.}
$$
Of the $1.48305\,\mathrm{MeV}$ predicted difference, $1.10922\,\mathrm{MeV}$ comes from the change in <nuclear pairing>. It is the primary reason the even-even capture product is more strongly excited; the smooth terms account for the remaining $0.37383\,\mathrm{MeV}$.
\b[Thermal versus fast-neutron fission]. The last comparison uses the supplied observed capture excitations, $6.5$ and $4.8\,\mathrm{MeV}$, rather than replacing them by the preceding smooth-model estimates. Interpreting the quoted $6.2$ and $6.6\,\mathrm{MeV}$ activation energies as the barriers in the respective capture channels gives
$$
\boxed{6.5>6.2:\quad\text{thermal capture on }{}^{235}\mathrm U\text{ can lead to prompt fission},}
$$
$$
\boxed{4.8<6.6:\quad\text{thermal capture on }{}^{238}\mathrm U\text{ is below the fission barrier}.}
$$
A simple <neutron-induced fission threshold> estimate for the second channel requires about $6.6-4.8=1.8\,\mathrm{MeV}$ additional centre-of-mass kinetic energy. Sub-barrier tunnelling is possible, but it does not give the large thermal-neutron fission probability of the first channel.
The barrier labels in the original PDF need care: it assigns them to $^{235}\mathrm U$ and $^{238}\mathrm U$, whereas the states formed after capture are $^{236}\mathrm U^*$ and $^{239}\mathrm U^*$. The comparison above uses the intended capture-channel interpretation. If the quoted numbers are literally barriers of the uncaptured target nuclei, the compound-nucleus barriers are missing and this numerical comparison does not follow from those data alone.
In a thermal reactor, <neutron moderation> slows fission neutrons so the large low-energy fission cross section of <uranium-235> can support a <nuclear chain reaction>. Fuel composition, moderator choice and geometry must balance production against capture and leakage; capture resonances of <uranium-238> during slowing are important losses from that neutron population. The reaction rate per target nucleus depends on the spectrum through $\int\phi(E)\sigma_f(E)\,dE$, where $\phi(E)$ is differential <neutron flux> per unit <energy>. A fast spectrum can access the <uranium-238> fission channel and changes this balance. Also, <uranium-238> is a <fertile material>: capture followed by two <beta-minus decays> can produce <plutonium-239>, a <fissile material>, so capture can contribute to breeding even when it causes no immediate fission.