NBPhO 2025

Overview

The problem is a four-part tour of thermal-reactor physics, glued together by non-relativistic mechanics and the ideal gas law. Two small parameters control the kinematics:

$$\frac{E_f}{m_n c^{2}} \;=\; \frac{0.025\,\text{eV}}{940\,\text{MeV}} \;\approx\; 2.7\times 10^{-11}, \qquad \frac{E_0}{m_n c^{2}} \;=\; \frac{2\,\text{MeV}}{940\,\text{MeV}} \;\approx\; 2.1\times 10^{-3}.$$

The first tells us the slow (“thermalised”) neutron is fantastically non-relativistic; the second is small enough that even the fast 2 MeV neutron is non-relativistic to better than $1\,\%$. Throughout we may use $E_k = \tfrac{1}{2}m_n v^{2}$ without apology.

The physical content of each part is:

Part	Physics
(i)	Non-relativistic kinematics; equipartition $\langle E\rangle = \tfrac{3}{2}k_B T$
(ii)	Same, with a check that $v/c \ll 1$ at the higher energy
(iii)	1D elastic central collisions; optimal mass-matching for energy transfer
(iv)	Boyle’s law for the helium across a volume change, then Dalton’s law for the mixture

A subtle conceptual point pervades part (i): the mode and mean of a Maxwell–Boltzmann distribution differ by a factor $3/2$, and the famous ”$0.025\,\text{eV}$ thermal energy” is the kinetic energy at the most-probable speed at room temperature, not the average kinetic energy. Reading the problem literally — “average kinetic energy is $E_f$” — yields $T_f \approx 193\,\text{K}$, not the more commonly quoted $290\,\text{K}$.

A pleasing structural feature of the problem is that the $M = 135\,m_n$ moderator in part (iii) corresponds in mass number to the xenon-135 isotope that builds up in part (iv): the same nucleus that is hard to use as a moderator (because $M\gg m_n$) is, in real reactors, a fission product that poisons the chain reaction by absorbing neutrons. The problem and the periodic table conspire.

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Part (i) — Slow-neutron speed and effective temperature

Justifying the non-relativistic limit

The kinetic energy of a relativistic particle is $E_k = (\gamma - 1) m c^{2}$ with $\gamma = 1/\sqrt{1-(v/c)^{2}}$. For $E_k \ll m c^{2}$ one expands

$$\gamma - 1 \;=\; \frac{1}{2}\left(\frac{v}{c}\right)^{2} + \frac{3}{8}\left(\frac{v}{c}\right)^{4} + \dots,$$

and truncating at the leading term gives $E_k = \tfrac{1}{2}m v^{2}$. The relative error of this truncation is $\sim (v/c)^{2}/4 = E_k/(2 m c^{2})$. For the slow neutron, $E_f/(m_n c^{2}) \approx 3\times 10^{-11}$, so non-relativistic mechanics is exact to better than one part in $10^{10}$.

Speed at energy $E_f$

Solving $E_f = \tfrac{1}{2} m_n v_f^{2}$ for $v_f$ and writing $m_n = (m_n c^{2})/c^{2}$,

$$v_f \;=\; c\sqrt{\frac{2 E_f}{m_n c^{2}}}.$$

This form is convenient because $m_n c^{2}$ is given directly. With $E_f = 0.025\,\text{eV}$ and $m_n c^{2} = 940\,\text{MeV} = 9.4\times 10^{8}\,\text{eV}$,

$$v_f \;=\; (3.00\times 10^{8}\,\text{m s}^{-1})\sqrt{\frac{2 \times 0.025}{9.4\times 10^{8}}} \;=\; (3.00\times 10^{8})(7.29\times 10^{-6})\,\text{m s}^{-1}.$$ $$\boxed{\;v_f \;\approx\; 2.19\times 10^{3}\,\text{m s}^{-1} \;\approx\; 2.2\,\text{km s}^{-1}\;}$$

Note $v_f/c \approx 7\times 10^{-6}$ — well within the non-relativistic regime.

Temperature

For an ideal monatomic gas in three dimensions, equipartition gives

$$\langle E_k\rangle \;=\; \frac{3}{2}k_B T,$$

with each of the three translational degrees of freedom contributing $\tfrac{1}{2}k_B T$. The problem states that the average kinetic energy is $E_f$, so

$$T_f \;=\; \frac{2 E_f}{3 k_B} \;=\; \frac{2 \times (0.025\times 1.602\times 10^{-19})\,\text{J}}{3 \times 1.38\times 10^{-23}\,\text{J K}^{-1}} \;=\; \frac{8.01\times 10^{-21}}{4.14\times 10^{-23}}\,\text{K}.$$ $$\boxed{\;T_f \;\approx\; 193\,\text{K}\;}$$

Educational remark — why “thermal” is sometimes quoted as 290 K

The standard slogan in reactor physics is that “thermal neutrons have $E \approx 0.025\,\text{eV}$ at room temperature ($T \approx 290\,\text{K}$)”. This corresponds to $E = k_B T$, not $\tfrac{3}{2}k_B T$:

$$k_B T \;=\; (1.38\times 10^{-23}\,\text{J K}^{-1})(290\,\text{K}) \;\approx\; 4.0\times 10^{-21}\,\text{J} \;\approx\; 0.025\,\text{eV}.$$

Why the missing factor of $3/2$? Because "$k_B T$" identifies the kinetic energy of a particle moving at the most-probable speed of a 3D Maxwell–Boltzmann distribution. From $f(v) \propto v^{2}\exp[-mv^{2}/(2k_B T)]$, $\partial_v f = 0$ gives $v_{\text{mode}}^{2} = 2 k_B T/m$, hence $\tfrac{1}{2}m v_{\text{mode}}^{2} = k_B T$. The average kinetic energy is $\tfrac{3}{2}k_B T$, larger by exactly $3/2$.

The problem says “average”, so $\tfrac{3}{2}k_B T_f = E_f$ and $T_f \approx 193\,\text{K}$ — colder than room temperature. Physically: if the average neutron carries only $0.025\,\text{eV}$, the gas as a whole is colder than 290 K, because in a 290 K gas the average energy would be the larger value $\tfrac{3}{2}k_B(290) \approx 0.0375\,\text{eV}$. Reactor textbooks finesse this by tracking only the most-probable speed; the present problem, more carefully, tracks the average.

Consistency checks

• Dimensions. $[E_k/m_n] = \text{J}/\text{kg} = \text{m}^{2}/\text{s}^{2}$; square root has units of m/s. ✓
• Limits. As $E_f \to 0$, both $v_f$ and $T_f$ vanish — a frozen neutron gas. As $E_f$ approaches $m_n c^{2}$, the non-relativistic formula breaks down, and one would need the full relativistic dispersion.

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Part (ii) — Fast-neutron speed

Re-checking the non-relativistic regime

At $E_0 = 2\,\text{MeV}$,

$$\frac{E_0}{m_n c^{2}} \;=\; \frac{2}{940} \;\approx\; 2.13\times 10^{-3},$$

so $(v/c)^{2} \approx 2 E_0/(m_n c^{2}) \approx 4.26\times 10^{-3}$ and $v/c \approx 0.065$. The non-relativistic correction in $E_k = \tfrac{1}{2}m v^{2}$ is at the few-tenths-of-a-percent level, which we shall accept and verify a posteriori.

Calculation

$$v_0 \;=\; c\sqrt{\frac{2 E_0}{m_n c^{2}}} \;=\; c\sqrt{\frac{2 \times 2}{940}} \;=\; c\sqrt{4.26\times 10^{-3}} \;\approx\; 0.0652\,c.$$ $$\boxed{\;v_0 \;\approx\; 1.96\times 10^{7}\,\text{m s}^{-1} \;\approx\; 6.5\,\%\;\text{of}\;c\;}$$

Cross-check against the exact relativistic result

Were one nervous about the non-relativistic approximation, the exact relativistic kinetic energy gives

$$\gamma \;=\; 1 + \frac{E_0}{m_n c^{2}} \;=\; 1.00213, \qquad \frac{v_0}{c} \;=\; \sqrt{1 - \gamma^{-2}} \;\approx\; 0.0651.$$

The relativistic and non-relativistic results agree to within $0.2\,\%$, confirming that $E_k = \tfrac{1}{2}m v^{2}$ is fine even at $2\,\text{MeV}$.

Educational remark — speed ratio

Comparing parts (i) and (ii):

$$\frac{v_0}{v_f} \;=\; \sqrt{\frac{E_0}{E_f}} \;=\; \sqrt{\frac{2\times 10^{6}}{0.025}} \;\approx\; 8.9\times 10^{3}.$$

The fast neutron is nearly $10^{4}$ times faster than the thermalised one — and equivalently, $E_0/E_f = 8\times 10^{7}$ is the energy ratio the moderator must dissipate. That dissipation factor is exactly what part (iii) asks us to bookkeep.

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Part (iii) — Optimal moderator mass and number of collisions

Setting up the 1D elastic central collision

A neutron of mass $m_n$ moving at speed $v$ strikes a stationary moderator nucleus of mass $M$, head-on. Both velocities lie along one axis. Conservation of momentum and kinetic energy give

$$m_n v \;=\; m_n v' + M V', \qquad \tfrac{1}{2}m_n v^{2} \;=\; \tfrac{1}{2}m_n v'^{\,2} + \tfrac{1}{2}M V'^{\,2}.$$

Eliminating $V'$ — equivalently, using the 1D-elastic identity $v - 0 = -(v' - V')$ that the relative velocity reverses — yields the classic result

$$v' \;=\; \frac{m_n - M}{m_n + M}\, v \;\equiv\; \alpha\, v, \qquad V' \;=\; \frac{2 m_n}{m_n + M}\, v.$$

The neutron’s energy after the collision is therefore

$$E' \;=\; \alpha^{2}\, E, \qquad \alpha \;=\; \frac{m_n - M}{m_n + M},$$

and the energy-transfer fraction is

$$\frac{\Delta E}{E} \;=\; 1 - \alpha^{2} \;=\; \frac{4 m_n M}{(m_n + M)^{2}}.$$

Optimal moderator mass

Treat $\Delta E/E$ as a function of $M$. Differentiating,

$$\frac{d}{dM}\!\left[\frac{4 m_n M}{(m_n+M)^{2}}\right] \;=\; \frac{4 m_n\bigl[(m_n+M)^{2} - 2M(m_n+M)\bigr]}{(m_n+M)^{4}} \;=\; \frac{4 m_n (m_n - M)}{(m_n+M)^{3}},$$

which vanishes (and switches sign from positive to negative) at $M = m_n$. There the energy transfer is $100\,\%$: after one head-on collision the neutron is at rest and the moderator nucleus carries away the entire kinetic energy.

$$\boxed{\;M_{\text{opt}} \;=\; m_n\;}$$

Educational remark — why hydrogen wins

This is the most familiar billiard-ball result: equal masses in a head-on elastic collision exchange velocities. In reactor practice it explains why water ($\text{H}_2\text{O}$) is such an efficient moderator — each collision with a proton ($M \approx m_n$) can drain a large fraction of the neutron’s energy. The competing real-world considerations are:

• Neutron absorption. Hydrogen captures neutrons via $n + p \to d + \gamma$, removing them from the chain reaction. Heavy water ($\text{D}_2\text{O}$, $M \approx 2\,m_n$) needs slightly more collisions per thermalisation but absorbs far less, allowing reactors to run on natural uranium (CANDU design).
• Practicality. Graphite ($M \approx 12\,m_n$) is solid, requires no pressurisation, but takes many more collisions per thermalisation. It was the moderator of the first chain reaction (Chicago Pile-1, 1942).
• Phase. The moderator must remain dense and stable across the operating temperature range; this rules out pure hydrogen gas in practice.

Number of head-on collisions for $M = 135\,m_n$

For the hypothetical “heavy moderator” of part (iii) — coincidentally the mass of the xenon nucleus that will appear in part (iv):

$$\alpha \;=\; \frac{1 - 135}{1 + 135} \;=\; -\frac{134}{136} \;=\; -\frac{67}{68} \;\approx\; -0.9853.$$

The negative sign tells us the neutron bounces back (whenever $M > m_n$ the neutron recoils). After $N$ head-on collisions its energy is

$$E_N \;=\; \alpha^{2N}\, E_0 \;=\; \left(\frac{67}{68}\right)^{2N} E_0.$$

The thermalisation criterion is $E_N \le E_f$, i.e. $\alpha^{2N} = E_f/E_0$. Taking logarithms (both sides negative),

$$N \;=\; \frac{\ln(E_f/E_0)}{\ln \alpha^{2}} \;=\; \frac{\ln(E_0/E_f)}{2\,\ln(1/|\alpha|)}.$$

Plugging in numbers:

$$\frac{E_0}{E_f} \;=\; \frac{2\times 10^{6}}{0.025} \;=\; 8\times 10^{7}, \qquad \ln(8\times 10^{7}) \;\approx\; 18.20,$$ $$\ln\!\frac{1}{|\alpha|} \;=\; \ln\!\frac{68}{67} \;\approx\; 0.01481,$$ $$N \;=\; \frac{18.20}{2 \times 0.01481} \;\approx\; 614.$$ $$\boxed{\;N \;\approx\; 614\;}$$

A useful approximation

When $M \gg m_n$, $|\alpha| = (M - m_n)/(M + m_n) \approx 1 - 2 m_n/M$, so

$$\ln\!\frac{1}{|\alpha|} \;\approx\; \frac{2 m_n}{M}, \qquad N \;\approx\; \frac{M}{4\,m_n}\,\ln\!\frac{E_0}{E_f}.$$

For $M = 135\,m_n$: $N \approx 135 \times 18.20/4 \approx 614$, agreeing with the direct calculation.

The closed-form scaling $N \propto M$ for $M \gg m_n$ is illuminating. A xenon-moderated chain reaction would need on the order of 600 head-on collisions, where a hydrogen-moderated one would need one. (The exactly-zero answer for $M = m_n$ is a singularity of the head-on assumption — in practice, isotropic CoM-frame scattering means even hydrogen needs $\sim 18$ collisions to thermalise on average.)

Pitfall — head-on vs realistic angular distributions

The problem specifies head-on collisions, which is the best-case energy transfer per collision. In reality scattering is roughly isotropic in the centre-of-mass frame, and the relevant average is the logarithmic energy decrement

$$\xi \;\equiv\; \langle\ln(E/E')\rangle \;=\; 1 + \frac{\alpha^{2}}{1-\alpha^{2}}\ln\alpha^{2} \;\xrightarrow[M \gg m_n]{}\; \frac{2}{M/m_n + 2/3}.$$

The realistic number of collisions to thermalise is $\ln(E_0/E_f)/\xi$, roughly twice the head-on count. The simpler head-on assumption is what the problem asks for.

Sign / range checks

• For $M < m_n$ (light moderator): $\alpha > 0$, the neutron continues forward but slower. Always loses energy.
• For $M = m_n$: $\alpha = 0$, the neutron stops dead in one collision (entire energy transferred).
• For $M > m_n$ (heavy moderator): $\alpha < 0$, the neutron bounces back. Magnitude of $|\alpha|$ approaches 1 as $M \to \infty$ — a brick wall reflects the neutron with no energy loss, which is consistent with the diverging $N$ in the limit.

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Part (iv) — Helium and xenon inside the fuel rod

Setup and what is given

Quantity	Value
Initial pressure (helium only)	$p_0 = 2.5\,\text{MPa}$
Final total pressure (He + Xe)	$p = 6.5\,\text{MPa}$
Initial gas volume	$V_0 = 18\,\text{cm}^{3}$
Final gas volume (after pellet swelling)	$V = 9\,\text{cm}^{3}$
Temperature (constant)	$T_0 = 20\,°\text{C} = 293.15\,\text{K}$

The amount of helium is fixed (it is a noble gas, and the rod is sealed). The amount of xenon is what we wish to compute.

Step 1 — Boyle’s law for the helium

The temperature is the same in initial and final states, and the helium content does not change. With $n_{\text{He}}$ and $T_0$ fixed, the ideal gas law $pV = nRT$ reduces to Boyle’s law $pV = \text{const}$ for the helium subsystem alone:

$$p_0 V_0 \;=\; p_{\text{He}}^{f}\, V \quad\Longrightarrow\quad p_{\text{He}}^{f} \;=\; p_0 \frac{V_0}{V}.$$

With the given numbers,

$$p_{\text{He}}^{f} \;=\; (2.5\,\text{MPa}) \times \frac{18}{9} \;=\; 5.0\,\text{MPa}.$$

The helium partial pressure has risen even though no helium has been added — simply because the available volume halved.

Step 2 — Dalton’s law for the mixture

In the final state He and Xe coexist in volume $V$ at the same temperature. Dalton’s law says the total pressure is the sum of the partial pressures (each gas exerts the pressure it would exert alone in the volume):

$$p \;=\; p_{\text{He}}^{f} + p_{\text{Xe}}^{f} \quad\Longrightarrow\quad p_{\text{Xe}}^{f} \;=\; p - p_{\text{He}}^{f} \;=\; 6.5 - 5.0 \;=\; 1.5\,\text{MPa}.$$

Step 3 — Moles of xenon

Apply the ideal gas law to the xenon subsystem alone:

$$n_{\text{Xe}} \;=\; \frac{p_{\text{Xe}}^{f}\, V}{R\, T_0} \;=\; \frac{(1.5\times 10^{6}\,\text{Pa})(9\times 10^{-6}\,\text{m}^{3})}{(8.31\,\text{J mol}^{-1}\text{K}^{-1})(293.15\,\text{K})} \;=\; \frac{13.5}{2436.5}\,\text{mol}.$$ $$\boxed{\;n_{\text{Xe}} \;\approx\; 5.54\times 10^{-3}\,\text{mol} \;=\; 5.54\,\text{mmol}\;}$$

A more compact closed form, useful for checking, is

$$n_{\text{Xe}} \;=\; \frac{p\,V - p_0\,V_0}{R\,T_0} \;=\; \frac{(6.5)(9) - (2.5)(18)}{R\,T_0}\,\text{MPa}\!\cdot\!\text{cm}^{3} \;=\; \frac{13.5\,\text{J}}{R\,T_0}.$$

Step 4 — He : Xe ratio

For two gases in the same volume at the same temperature, the ratio of partial pressures equals the ratio of mole numbers:

$$\frac{n_{\text{He}}}{n_{\text{Xe}}} \;=\; \frac{p_{\text{He}}^{f}}{p_{\text{Xe}}^{f}} \;=\; \frac{5.0}{1.5} \;=\; \frac{10}{3}.$$ $$\boxed{\;n_{\text{He}} : n_{\text{Xe}} \;=\; 10 : 3 \;\approx\; 3.33 : 1\;}$$

Numerically, $n_{\text{He}} = p_0 V_0/(R T_0) = 45/2436.5 \approx 1.85\times 10^{-2}\,\text{mol} = 18.5\,\text{mmol}$, consistent with the ratio above.

Educational remark — why the volume change matters

A common slip is to use the initial volume when computing the final helium partial pressure — as though the helium still occupied $V_0 = 18\,\text{cm}^{3}$. That would give $p_{\text{He}}^{f} = p_0 = 2.5\,\text{MPa}$ and $p_{\text{Xe}}^{f} = 4.0\,\text{MPa}$, overestimating the xenon by a factor

$$\frac{4.0\,\text{MPa}}{1.5\,\text{MPa}} \;=\; \frac{8}{3}.$$

The pellet swelling shrinks the void volume available to the gases from $V_0 = 18$ to $V = 9\,\text{cm}^{3}$, compressing the helium even before any xenon is added. Only after that compression does Dalton’s law correctly describe the mixture.

Educational remark — what these numbers tell us about the rod

The number of $^{135}\text{Xe}$ atoms released per rod is

$$N_{\text{Xe}} \;=\; n_{\text{Xe}}\, N_A \;=\; (5.54\times 10^{-3})(6.022\times 10^{23}) \;\approx\; 3.3\times 10^{21}.$$

Each $^{135}\text{Xe}$ atom is the end-product of (or, more often, an intermediate decay product on) a fission event. With a fission yield of order $6\,\%$ for the mass-135 chain, a few times $10^{22}$ fissions have occurred in the rod over its lifetime. That is consistent with a few percent of the original $^{235}\text{U}$ inventory being burnt — a sensible number for a commercial fuel rod at end-of-life.

In the operating reactor itself, $^{135}\text{Xe}$ is also a famous neutron poison, with a thermal-neutron capture cross-section of $\sigma \sim 3\times 10^{6}\,\text{barn}$ — orders of magnitude larger than $^{235}\text{U}$ fission. The build-up and decay of xenon-135 produces the well-known “xenon transient” that complicates reactor restart after shutdown, and was a contributing factor in the Chernobyl accident. None of this dynamics is visible in the static measurement of part (iv), but the static measurement would be the diagnostic an inspector uses to estimate burn-up after the rod is pulled.

Sanity checks

• Pressure budget. $p_{\text{He}}^{f} + p_{\text{Xe}}^{f} = 5.0 + 1.5 = 6.5\,\text{MPa} = p$. ✓
• Limit $V \to V_0$ (no swelling). Boyle gives $p_{\text{He}}^{f} = p_0$, so $p_{\text{Xe}}^{f} = p - p_0 = 4.0\,\text{MPa}$ and $n_{\text{Xe}} = 4 p_{\text{Xe}}^{f}V_0/(R T_0)$ would be $\sim 5\times$ larger. The volume change matters a lot.
• Limit $V \to V_0/(p/p_0) = 18/2.6 \approx 6.9\,\text{cm}^{3}$. At this volume, Boyle alone would push the helium pressure up to the measured $p = 6.5\,\text{MPa}$, leaving no room for xenon — i.e. $n_{\text{Xe}} \to 0$. Below this critical volume the formula would give a negative $p_{\text{Xe}}^{f}$, meaning the assumption that all extra pressure comes from xenon would have failed. Our $V = 9\,\text{cm}^{3}$ is comfortably above the threshold.
• Order of magnitude of $n_{\text{Xe}}$. Roughly, the extra pressure $\Delta p \approx p - p_0 = 4\,\text{MPa}$ in $V \approx 10\,\text{cm}^{3}$ at $T_0 \approx 300\,\text{K}$ gives $n \sim p V/(R T) \sim 4\times 10^{6} \times 10^{-5}/(8.3 \times 300) \sim 1.6\times 10^{-2}\,\text{mol}$. This naive estimate misses the volume-change correction by a factor $\sim 3$, which is exactly the discrepancy we resolved with Boyle’s law.

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Summary of results

Part	Quantity	Result	Numerical value
(i)	Slow-neutron speed	$v_f = c\sqrt{2 E_f/(m_n c^{2})}$	$\approx 2.19\times 10^{3}\,\text{m s}^{-1}$
(i)	Effective temperature	$T_f = 2 E_f/(3 k_B)$	$\approx 193\,\text{K}$
(ii)	Fast-neutron speed	$v_0 = c\sqrt{2 E_0/(m_n c^{2})}$	$\approx 1.96\times 10^{7}\,\text{m s}^{-1}$
(iii)	Optimal moderator mass	$M_{\text{opt}} = m_n$	(hydrogen)
(iii)	Collisions for $M = 135\,m_n$	$N = \dfrac{\ln(E_0/E_f)}{2,\ln(1/	\alpha
(iv)	Xenon released	$n_{\text{Xe}} = (pV - p_0 V_0)/(R T_0)$	$\approx 5.54\,\text{mmol}$
(iv)	He : Xe ratio	$p_{\text{He}}^{f}/p_{\text{Xe}}^{f}$	$10 : 3 \approx 3.33$

One sentence to take away. The problem stitches together three independent pieces of physics — non-relativistic kinematics, 1D elastic-collision energy transfer (peaked at $M = m_n$), and the ideal-gas pair “Boyle then Dalton” — and the only place where one must think twice is the difference between the mode and mean of a Maxwell–Boltzmann distribution in part (i).