NBPhO 2025

Part i (1 point)

If the speed of the particle is much less than the speed of light, we can use the non-relativistic approach. For non-relativistic particles we know $v = \sqrt{2E_k/m}$. Substituting values gives us $v_f = 2.2 \times 10^3\,\text{m\,s}^{-1}$. 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 ($E_f \ll m_n c^2$).

Typo in problem description. $E_f = 0.025\,\text{eV}$ is the mode of the Maxwell–Boltzmann distribution which gives $E = k_BT$. The average gives $E = \frac{3}{2}k_BT$.

Using the mode of the Maxwell–Boltzmann distribution, $E = k_BT$, and converting from eV to J:

$$T_f = \frac{0.025 \times 1.602 \times 10^{-19}}{1.38 \times 10^{-23}} = 290\,\text{K}.$$

With $E = \frac{3}{2}k_BT$ we find $T_f \approx 193\,\text{K}$.

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Part ii (1 point)

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

Grading for parts i and ii (preliminary)

• Expresses $v_f = \sqrt{2E_f/m}$ and/or $v_0 = \sqrt{2E_0/m}$: 0.3 pts
• Calculates $v_f = 2.2\times 10^3\,\text{m\,s}^{-1}$ and/or $v_0 = 2.0\times 10^7\,\text{m\,s}^{-1}$ 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 $T = 193\,\text{K}$: 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 $T = 290\,\text{K}$ gives 0 pts for the formula and 0.3 pts for the numerical calculation.

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Part iii (2.5 points)

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 into 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 moderator atoms should be the same as the neutron’s.

In a single collision with a stationary atom of the moderator, the speed of the neutron decreases by the factor $\frac{m_1 - m_2}{m_1 + m_2}$, 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,

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

Grading (preliminary)

• $T \ll T_f$, so moderator atoms are essentially at rest: 0.3 pts
• Justifies that maximum momentum transfer is when $m_n = M$: 0.4 pts
• Applies energy and momentum conservation: 0.3 + 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

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Part iv (1.5 points)

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 MPa to 5 MPa as we know from Boyle’s law $P_1V_1 = P_2V_2 \Rightarrow P_2 = P_1V_1/V_2$. Thus, the release of xenon must contribute 1.5 MPa’s worth of pressure due to Dalton’s law $p_\text{tot} = p_\text{He} + p_\text{Xe}$. From the ideal gas law we find the amount of xenon moles as

$$n_\text{Xe} = \frac{p_\text{Xe} V}{RT_0} = 5.5\times 10^{-3}\,\text{mol},$$

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

$$n_\text{He} = \frac{p_\text{He} V_0}{RT_0} = 1.8\times 10^{-2}\,\text{mol},$$

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

$$\frac{n_\text{He}}{n_\text{Xe}} \approx 3.3.$$

Grading (preliminary)

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