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NBPhO 2025, Теория и експеримент — Задача 3. Nuclear Reactors

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Nordic-Baltic Physics Olympiad 2025 · 6 т.

Условие

— Topi Lind. In order to maintain a chain reaction in a modern thermal-neutron nuclear reactor one needs three things: 1. nuclear fuel (e. g. $\mathrm{U}^{235}$ ), 2. moderator (e. g. water) and 3. coolant. In most cases the moderator acts as the coolant as well. Neutrons released from a thermal fission of $\mathrm{U}^{235}$ have a mean kinetic energy of approximately $E_0 = 2\ \mathrm{MeV}$ . However, neutrons which are that fast are inefficient in triggering fission of $\mathrm{U}^{235}$ : neutrons need to be slowed down to an average kinetic energy of $E_f = 0.025\ \mathrm{eV}$ . In what follows, justify why non-relativistic approximations can be used unless explicitly instructed otherwise.

i) The rest energy of neutrons $m_{\mathrm{n}}c^2 = 940\ \mathrm{MeV}$ , the Boltzmann constant $k_{\mathrm{B}} = 1.38\times 10^{-23}\ \mathrm{J\,K^{-1}}$ , and the elementary charge $e = 1.602\times 10^{-19}\ \mathrm{C}$ . What is the required speed of neutrons, i. e. the speed of neutrons with kinetic energy $E_f$ ? What is the temperature $T_f$ of a neutron gas where the average kinetic energy of neutrons is $E_f$ ? [1 т.]

ii) What is the initial speed of neutrons, i. e. the speed of neutrons with energy $E_0$ ? [1 т.]

iii) From a completely non-relativistic point of view, what should be the mass of the moderator’s atoms to slow down the fast neutrons as efficiently as possible? If the mass of the moderator’s atoms were to be $M = 135m_{\mathrm{n}}$ , how many collisions with such atoms at a temperature much lower than $T_f$ would a fast neutron need to experience to slow down from $E_0$ to $E_f$ ? Assume that all collisions are elastic and central. [2,5 т.]

iv) Nuclear fuel, i.e. $\mathrm{U}^{235}$ , is placed inside metal rods and pressurized with helium gas to $p_0 = 2.5\ \mathrm{MPa}$ . During operation, as $\mathrm{U}^{235}$ keeps on fissioning inside the fuel rods, there is a build up of gas inside the rods. With a non-invasive ultrasound measurement we can measure that the gas pressure inside the rod after it is finally picked out from the core is $p = 6.5\ \mathrm{MPa}$ . Assuming that the gas released inside the rods is completely made of xenon isotope $^{135}_{54}\mathrm{Xe}$ and that the initial gas volume drops from $V_0 = 18\ \mathrm{cm^3}$ to $V = 9\ \mathrm{cm^3}$ due to the swelling of the fuel pellets, how many moles of xenon are released from fission? What is the ratio of helium to xenon inside the rod? The measurements are done at $T_0 = 20\ ^{\circ}\mathrm{C}$ ; the universal gas constant $R = 8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$ . [1,5 т.]

Решение

— Solution and grading scheme by Topi Lind, Melvin Storbacka, Oleg Kosik, Aleksandr Sorokin and Ludmila Belogrudova.

i) (1 т.)

If the speed of the particle is much less than the speed of light, we can use non-relativistic approach. For non-relativistic particles we know $v = \sqrt{2E_k/m}$ . Substituting values gives us $v_f = 2.2\times 10^3\ \mathrm{m\,s^{-1}}$ . This is much less than the speed of light and thus justified. Another way to justify the applicability of the non-relativistic approach would be to say that kinetic energy is significantly less than the rest energy ( $E_f \ll m_{\mathrm{n}}c^2$ ).

Typo in problem description. $E_f = 0.025\ \mathrm{eV}$ is the mode of the Maxwell–Boltzmann distribution which gives $E = k_bT$ . Average gives $E = (3/2)k_bT$ .

Using the mode of the Maxwell–Boltzmann distribution, $E = k_{\mathrm{B}}T$ , and remembering to convert from eV to J correctly, we find

T_f = \frac{0.025\cdot 1.602\times 10^{-19}}{1.38\times 10^{-23}} = 290\ \mathrm{K}.

With $E = (3/2)k_{\mathrm{B}}T$ we find $T_f \approx 193\ \mathrm{K}$ .

ii) (1 т.)

The non-relativistic approach justified as in previous task. Same approach gives us $v_0 = 2.0\times 10^7\ \mathrm{m\,s^{-1}}$ .

Grading i)+ii): (preliminary)

Expresses $v_f = \sqrt{2E_f/m}$ and/or $v_0 = \sqrt{2E_0/m}$ 0.3 pts
Calculates $v_f = 2.2\times 10^3\ \mathrm{m\,s^{-1}}$ and/or $v_0 = 2.0\times 10^7\ \mathrm{m\,s^{-1}}$ for two correct numerical values; 0.5 pts; if only one value is correct 0.3 pts.
Uses $E_f = \frac{3}{2}k_BT$ 0.3 pts
Using this formula calculates $T = 193\ \mathrm{K}$ 0.3 pts
Justifies the validity of the classical approach in both cases 0.3 + 0.3 pts Remark. Using $E_f = k_BT$ without justification and finding $T = 290\ \mathrm{K}$ gives 0 pts for the formula and 0.3 pts for the numerical calculation.

iii) (2,5 т.)

In a collision between particle 1 ( $m_1, v_{1,i}$ and $v_{1,f}$ ) and particle 2 ( $m_2, v_{2,i}$ and $v_{2,f}$ ) momentum is conserved,

m_1(v_{1,f} - v_{1,i}) = m_2(v_{2,i} - v_{2,f})

and since the collisions are elastic, kinetic energy is also conserved:

m_1(v_{1,f}^2 - v_{1,i}^2) = m_2(v_{2,i}^2 - v_{2,f}^2).

Dividing the latter by the former leads to

v_{1,i} + v_{1,f} = v_{2,i} + v_{2,f}.

Substituting this to the conservation of momentum gives for particle 1:

v_{1,f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1,i} + \frac{2m_2}{m_1 + m_2}v_{2,i},

and similarly for particle 2:

v_{2,f} = \frac{2m_1}{m_1 + m_2}v_{1,i} + \frac{m_1 - m_2}{m_1 + m_2}v_{2,i}.

We see that with $m_1 = m_2$ there is a maximum transfer of momentum. Assuming that particle 1 is the neutron and particle 2 is the target, and that the target is at rest for all intents and purposes, the mass of the moderators atoms should be the same as the neutrons.

In a single collision with a stationary atom of the moderator, the speed of the neutron decreases by the factor of $\frac{m_1 - m_2}{m_1 + m_2}$ , so the speed of the neutron after $N$ head-on collisions with stationary atoms of moderator will be

v_N = v\left(\frac{m_1 - m_2}{m_1 + m_2}\right)^N.

Hence,

\begin{aligned} N &= \frac{\ln(v_f/v_0)}{\ln[(m_1 - m_2)/(m_1 + m_2)]} \\ &= \frac{1}{2}\frac{\ln(E_f/E_0)}{\ln[(m_1 - m_2)/(m_1 + m_2)]} = 614. \end{aligned}

Grading: (preliminary)

$T \ll T_f$ , so moderator atoms are essentially at rest 0.3 pts
Justifies that maximum momentum transfer is when $m_{\mathrm{n}} = M$ 0.4 pts
Applies energy and momentum conservation 0.3 pts+ 0.3 pts
Expresses $u = v\frac{m_1 - m_2}{m_1 + m_2}$ 0.4 pts
Expresses $v_f = v_0\left(\frac{m_1 - m_2}{m_1 + m_2}\right)^N$ 0.5 pts
Calculates $N = 614$ 0.3 pts

iv) (1,5 т.)

We can model the gas inside the rod as an ideal gas. Simply due to swelling the pressure inside the rod would increase from $2.5\ \mathrm{MPa}$ to $5\ \mathrm{MPa}$ as we know from Boyle’s law $P_1V_1 = P_2V_2 \rightsquigarrow P_2 = P_1V_1/V_2$ . Thus, the release of xenon must contribute $1.5\ \mathrm{MPa}$ ’s worth of pressure due to Dalton’s law $p_{\mathrm{tot}} = p_{\mathrm{He}} + p_{\mathrm{Xe}}$ . From ideal gas law we find the amount of xenon moles as

n_{\mathrm{Xe}} = p_{\mathrm{Xe}}V/RT_0 = 5.5\times 10^{-3}\ \mathrm{mol},

where $p_{\mathrm{Xe}} = 1.5\ \mathrm{MPa}$ , $V_2 = 9\ \mathrm{cm^3}$ and $T = 293\ \mathrm{K}$ . In a similar manner we find that the amount of helium in the beginning was

n_{\mathrm{He}} = p_{\mathrm{He}}V_0/RT_0 = 1.8\times 10^{-2}\ \mathrm{mol},

where $P_{\mathrm{He}} = 2.5\ \mathrm{MPa}$ , $V_0 = 18\ \mathrm{cm^3}$ and $T_0 = 293\ \mathrm{K}$ . The ratio of the two is

\frac{n_{\mathrm{He}}}{n_{\mathrm{Xe}}} \approx 3.3.

Grading: (preliminary)

Applies Boyle’s law 0.3 pts
Applies Dalton’s law 0.4 pts
Expresses $n_{\mathrm{Xe}} = p_{\mathrm{Xe}}V/RT_0$ 0.2 pts
Calculates $n_{\mathrm{Xe}} = 5.5\times 10^{-3}\ \mathrm{mol}$ 0.2 pts
Expresses $n_{\mathrm{He}} = p_{\mathrm{He}}V_0/RT_0$ 0.2 pts
Calculates $\frac{n_{\mathrm{He}}}{n_{\mathrm{Xe}}} = 3.3$ 0.2 pts

Оригинал в Архива: NBPhO_2025_problems_v1.0.pdf · официални решения: NBPhO_2025_solutions_v1.0.pdf