NBPhO 2025

Parts (i) + (ii) — 2.0 / 2.0 pts

The official scheme groups parts (i) and (ii) under a single 2.0 pt rubric.

Criterion	Points	Result
Expresses $v_f = \sqrt{2E_f/m}$ and/or $v_0 = \sqrt{2E_0/m}$	0.3	✓ Claude writes the algebraically equivalent form $v = c\sqrt{2E/(m_n c^{2})}$ for both energies, leveraging the given rest-energy $m_n c^{2}$ directly
Calculates $v_f = 2.2\times 10^3\,\text{m\,s}^{-1}$ and $v_0 = 2.0\times 10^7\,\text{m\,s}^{-1}$ (both correct)	0.5	✓ $v_f \approx 2.19\times 10^{3}\,\text{m s}^{-1}$ and $v_0 \approx 1.96\times 10^{7}\,\text{m s}^{-1}$, both within rounding of the official key
Uses $E_f = \tfrac{3}{2}k_B T$	0.3	✓ “the average kinetic energy is $E_f$, so $T_f = 2 E_f/(3 k_B)$” — explicit equipartition argument with all three translational degrees of freedom
Using this formula calculates $T = 193\,\text{K}$	0.3	✓ $T_f \approx 193\,\text{K}$ (exact match)
Justifies the validity of the classical approach for $E_f$	0.3	✓ $E_f/(m_n c^{2}) \approx 3\times 10^{-11}$ and $v_f/c \approx 7\times 10^{-6}$, with the relative error of the truncation tracked back to $E_k/(2 m_n c^{2})$
Justifies the validity of the classical approach for $E_0$	0.3	✓ $E_0/(m_n c^{2}) \approx 2.13\times 10^{-3}$ and $v_0/c \approx 0.065$, plus an explicit relativistic cross-check via $\gamma = 1.00213$ giving $v_0/c \approx 0.0651$ in 0.2% agreement with the non-relativistic formula

Part (iii) — 2.3 / 2.5 pts

Criterion	Points	Result
$T \ll T_f$, so moderator atoms are essentially at rest	0.3	~ 0.1 / 0.3 — Claude treats the moderator as “stationary” throughout the kinematic setup and the analysis is correct under that assumption, but the problem-given hint “at a temperature much lower than $T_f$” is never quoted and the stationarity assumption is never tied back to it. The operational consequence is right; the explicit justification the rubric rewards is absent
Justifies that maximum momentum transfer is when $m_n = M$	0.4	✓ explicit calculus on $\Delta E/E = 4 m_n M/(m_n+M)^{2}$, with the derivative $4 m_n(m_n - M)/(m_n+M)^{3}$ vanishing and switching sign from positive to negative at $M = m_n$ — confirming a maximum, not just a stationary point
Applies momentum conservation	0.3	✓ $m_n v = m_n v' + M V'$ written explicitly
Applies energy conservation	0.3	✓ $\tfrac{1}{2}m_n v^{2} = \tfrac{1}{2}m_n v'^{2} + \tfrac{1}{2}M V'^{2}$ written explicitly, plus the equivalent “relative-velocity reverses” identity
Expresses $u = v\,\dfrac{m_1 - m_2}{m_1 + m_2}$	0.4	✓ $v' = \alpha v$ with $\alpha = (m_n - M)/(m_n + M)$ — verbatim
Expresses $v_f = v_0\!\left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)^{N}$	0.5	✓ $v_N = v_0\,\alpha^{N}$ (equivalently $E_N = \alpha^{2N}E_0$) — exactly the geometric-decay relation
Calculates $N = 614$	0.3	✓ $N \approx 614$ (exact match), with a side-derivation via the $M \gg m_n$ approximation $N \approx (M/4 m_n)\ln(E_0/E_f)$ confirming the same number

Part (iv) — 1.5 / 1.5 pts

Criterion	Points	Result
Applies Boyle’s law	0.3	✓ $p_0 V_0 = p_{\text{He}}^{f} V$ at constant $T_0$ and constant $n_{\text{He}}$, giving $p_{\text{He}}^{f} = p_0\,V_0/V = 5.0\,\text{MPa}$
Applies Dalton’s law	0.4	✓ $p = p_{\text{He}}^{f} + p_{\text{Xe}}^{f}$, giving $p_{\text{Xe}}^{f} = 6.5 - 5.0 = 1.5\,\text{MPa}$
Expresses $n_{\text{Xe}} = p_{\text{Xe}} V/(R T_0)$	0.2	✓ stated verbatim
Calculates $n_{\text{Xe}} = 5.5\times 10^{-3}\,\text{mol}$	0.2	✓ $n_{\text{Xe}} \approx 5.54\times 10^{-3}\,\text{mol} = 5.54\,\text{mmol}$
Expresses $n_{\text{He}} = p_{\text{He}} V_0/(R T_0)$	0.2	✓ "$n_{\text{He}} = p_0 V_0/(R T_0) = 45/2436.5\,\text{mol}$"
Calculates $n_{\text{He}}/n_{\text{Xe}} = 3.3$	0.2	✓ $n_{\text{He}}:n_{\text{Xe}} = 10:3 \approx 3.33$, with the partial-pressure shortcut $p_{\text{He}}^{f}/p_{\text{Xe}}^{f} = 5.0/1.5$

Overall score: 5.8 / 6.0 pts

Full marks on parts (i)+(ii) and (iv); a single 0.2 pt dock in part (iii) for assuming a stationary moderator without quoting the problem-given $T \ll T_f$ condition as justification.

All seven keyed numerical answers match the official: $v_f \approx 2.19 \times 10^{3}\,\text{m s}^{-1}$ (vs $2.2\times 10^{3}$), $T_f \approx 193\,\text{K}$ (exact), $v_0 \approx 1.96\times 10^{7}\,\text{m s}^{-1}$ (vs $2.0\times 10^{7}$), $M_\text{opt} = m_n$ (hydrogen), $N \approx 614$ (exact), $n_{\text{Xe}} \approx 5.54\,\text{mmol}$ (vs $5.5\times 10^{-3}\,\text{mol}$), and $n_{\text{He}}/n_{\text{Xe}} \approx 3.33$ (vs $3.3$).

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads the two small parameters $E_f/(m_n c^{2}) \approx 2.7\times 10^{-11}$ and $E_0/(m_n c^{2}) \approx 2.1\times 10^{-3}$, so the non-relativistic limit is justified by named ratio rather than re-derived in place. Part (i) carries an explicit Taylor-expansion derivation $\gamma - 1 = \tfrac{1}{2}(v/c)^{2} + \tfrac{3}{8}(v/c)^{4} + \dots$ pinning the truncation error at $\sim E_k/(2 m_n c^{2})$, then a long pedagogical aside on the mode-vs-mean distinction in the Maxwell–Boltzmann distribution: the $0.025\,\text{eV}$ of reactor folklore is $k_B T$ at $290\,\text{K}$ (the most-probable speed), not $\tfrac{3}{2}k_B T$, and Claude derives $v_\mathrm{mode}^{2} = 2 k_B T/m$ from $\partial_v[v^{2}\exp(-mv^{2}/(2k_B T))] = 0$ to make the factor $3/2$ precise. Part (ii) adds an exact relativistic cross-check, $\gamma = 1.00213$, $v_0/c = 0.0651$ vs the non-relativistic $0.0652$, and quotes the energy ratio $E_0/E_f = 8\times 10^{7}$ that part (iii) must dissipate. Part (iii) ranks real-world moderators against the $M = m_n$ ideal: water (proton, optimal but absorbs neutrons via $n + p \to d + \gamma$), heavy water (CANDU design, allows natural-uranium fuel), graphite (Chicago Pile-1, 1942), with the logarithmic energy decrement $\xi = 1 + \alpha^{2}\ln\alpha^{2}/(1-\alpha^{2})$ named as the realistic isotropic-CoM-frame analogue of the head-on count, and the $M \gg m_n$ scaling $N \approx (M/4 m_n)\ln(E_0/E_f)$ derived as a closed-form check. Sign and limiting-case checks on $\alpha$ are listed for $M < m_n$, $M = m_n$, and $M > m_n$, with $|\alpha| \to 1$ as $M \to \infty$ tied to the brick-wall reflection limit. Part (iv) compresses to a one-line closed form $n_{\text{Xe}} = (pV - p_0 V_0)/(R T_0)$, checks the $V \to V_0$ limit (xenon overestimated by a factor $8/3$ if pellet swelling is ignored) and the critical-volume limit $V \to V_0/(p/p_0) \approx 6.9\,\text{cm}^{3}$ at which $n_{\text{Xe}} \to 0$, then converts the answer to atom count $N_\text{Xe} \approx 3.3\times 10^{21}$, ties it to the $\sim 6\%$ mass-135 fission yield, and contextualises against $^{135}$Xe as a famous neutron poison ($\sigma \sim 3\times 10^{6}\,\text{barn}$) responsible for the post-shutdown xenon transient that contributed to Chernobyl. The cross-part observation that the $M = 135\,m_n$ moderator of part (iii) is the same nucleus as the xenon-135 of part (iv) — a reactor poison playing the role of a hypothetical bad moderator — is a structural reading the grading scheme does not reward but the problem clearly invites.

Where the official solution is sharper. Two places. (1) The official explicitly flags the problem statement as containing a typo — ”$E_f = 0.025\,\text{eV}$ is the mode of the Maxwell–Boltzmann distribution which gives $E = k_B T$. The average gives $E = \tfrac{3}{2}k_B T$” — and is honest that the literal reading “average” was likely an editing slip. Claude takes the literal-reading-as-intended path, lands on $193\,\text{K}$ for the right reason, and even spells out the mode-vs-mean derivation; but he frames the $290\,\text{K}$ branch as a textbook-folklore shortcut rather than as the value the problem author probably meant to ask for. The official’s “typo” framing is more candid about the problem’s intent. (2) Part (iii), the $T \ll T_f$ argument: the problem statement gives “at a temperature much lower than $T_f$” precisely so the student can justify treating moderator atoms as stationary, and the grading rubric awards 0.3 pts for this single connection. Claude assumes stationarity from the first sentence of the kinematic setup and never references the problem’s temperature condition — the analysis is right, but the explicit justification the rubric calls for is missing. This is exactly the 0.2 pt dock above.