AQA A-Level Physics Paper 2, June 2023: Question 7

10 marks · Medium difficulty · Short Answer

Explain the role of moderators in nuclear fission, calculate energy loss in neutron collisions, and determine the number of collisions required to thermalise a neutron.

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Question

Question 07 focuses on nuclear reactor moderators. Part 07.1 asks for another moderator material besides carbon. Part 07.2 asks for two benefits of slowing down neutrons. Part 07.3 models a neutron-nucleus collision using gliders on an air track, providing a graph of v/u against mM/mN, and asks to show that when mM/mN is 12, glider N loses about 30% of its kinetic energy. Part 07.4 gives an exponential equation Ef = E0 * e^(-by) with b = 0.73, and asks for the minimum collisions y needed to slow a 1.0 MeV neutron to the thermal energy of an ideal gas at 350 K. Part 07.5 asks to explain why elements with a small nucleon number are preferred as moderators.
Question text

07.1 Carbon is used as the moderator in some thermal nuclear reactors.

Identify one other material commonly used as a moderator.

[1 mark]

07.2 State two benefits of slowing down the neutrons released during fission.

[2 marks]

07.3 The collision of a neutron with the nucleus of a moderator atom is modelled using two

gliders on a horizontal frictionless air track.

In Figures 13 and 14 the glider N of mass mN represents the neutron and the

glider M of mass mM represents the moderator nucleus.

Figure 13 shows glider N travelling with initial speed u towards the stationary

glider M.

Figure 13

The gliders collide. N rebounds with speed v as shown in Figure 14.

Figure 14

v mM

Figure 15 shows the variation of the ratio with the ratio .

u mN

Figure 15

mM

Show that when is 12, N loses about 30% of its initial kinetic energy

mN

in the collision.

[2 marks]

07.4 In a reactor, the speed of a fast-moving neutron is reduced by a series of y random

collisions with carbon-12 nuclei.

The final kinetic energy Ef of the neutron is

E = E e−by

f 0

where E0 is the initial kinetic energy of the neutron and b = 0.73

A thermal neutron has kinetic energy equivalent to that of the average particle of an

ideal gas with a temperature of 350 K.

One neutron has an initial kinetic energy of 1.0 MeV.

Calculate the minimum value of y required so that this neutron becomes a thermal

neutron.

[3 marks]

25 y =

07.5 Explain, with reference to Figure 15, why elements with a small nucleon number are

preferred as moderator materials.

[2 marks]

END OF SECTION A

Section B

Each of Questions 08 to 32 is followed by four responses, A, B, C and D.

For each question select the best response.

Only one answer per question is allowed.

For each question, completely fill in the circle alongside the appropriate answer.

CORRECT METHOD WRONG METHODS

If you want to change your answer you must cross out your original answer as shown.

If you wish to return to an answer previously crossed out, ring the answer you now wish to select

as shown.

You may do your working in the blank space around each question but this will not be marked.

Do not use additional sheets for this working.

Mark scheme

Show the mark scheme Mark scheme for Question 07 detailing answers and marking points for 07.1 to 07.5. 07.1 accepts heavy water, beryllium, or water (1 mark). 07.2 accepts points on increased probability of fission/absorption by U-235 and less structural damage or heat extraction (2 marks). 07.3 shows (v/u)^2 = 0.85^2 = 0.72, giving ~28% kinetic energy loss (2 marks). 07.4 shows calculation of final KE as 3/2 kT = 7.2 x 10^-21 J (or 0.045 eV), equating Ef = E0 e^(-by) to obtain y = 23.2 to 24 (3 marks). 07.5 awards marks for identifying that lower mass causes a greater reduction in speed per collision, meaning fewer collisions and a thinner moderator are needed (2 marks). Total: 10 marks.

Question Answers Additional comments/Guidelines Mark AO

07.1 Heavy water 1 AO1

OR

Beryllium / Be Accept D2O and H2O

OR

(normal) Water

Any two points from:

07.2 Condone the answer: 2 AO1

• U-235/Uranium fuel will (be more likely to) absorb the As an alternative to the first point AO2

neutron Fission of U-236 is much more likely.

• slow neutrons are less damaging OR cause less fatigue Condone the answer:

to the structure of the reactor/shielding/etc Absorption by U-238 is less likely.

• slow neutrons (spend longer within the fissionable

material and) increase the chance of causing fission

• slowing neutrons transfers heat energy to the moderator

(which can make heat easier to extract)

all points OWTTE

07.3 1 2 1 can be for any of the terms shown 2 AO2

m v 2 equating to the kinetic energy ratio.

final kinetic energy N v

= 2 = = 0.852 = 72% AO3

initial kinetic energy 1 2 u

26 m uN 2 can be an ecf but only for an arithmetic

error.

(Hence) proportion of kinetic energy lost = 28% 2

07.4 3 −21 1+2 3 AO2

final kinetic energy = ( kT =) 7.2 × 10 (J) 1 Both marks must come from the same

2 alternative route and have consistent units ×3

initial kinetic energy = (W = QV) = 1.6 × 10−13 (J) 2 (which may not be seen).

1 Initial kinetic energy =

OR

3 × 1.38 × 10−23 × 350 = 7.245 × 10−21 J

final kinetic energy = 0.045 (eV) 1 2

6 7.245×10−21

initial kinetic energy = 1.0 × 10 (eV) 2

= −13 = 0.045 eV

1.60×10

E0 2 Using the eV unit alternative the second

1n

(Rearranging equation y = E ) mark cannot be given without an attempt at

f

b the first mark.

The 1.0 × 106 eV can be seen in a later

substitution provided eV is used throughout.

1.0 ×106

1n

3 0.045

y = 23.(2) 3 y = = 23.2

0.73

Condone answer 24 provided it is given as an

integer.

07.5 Idea that the model/Figure 15 shows that low nucleon Condone the use of nuclear mass instead of 2 AO2

number (and so low mass) gives a greater mass number. ×2

change/reduction in speed/KE (in a collision)

Idea that fewer collisions needed (with a low mass number

so moderator can be thinner)

Total 10

How to answer it

Nuclear Reactor Moderation & Neutron Collisions

📌 Overview

What This Question Tests

This question assesses your understanding of thermal nuclear reactor physics and particle collisions. Specifically, it tests:

  • Reactor Engineering: Identification of moderator materials and the physical necessity of moderating fission neutrons to thermal energies.
  • Graph & Kinetic Energy Analysis: Relating kinetic energy ratios to velocity ratios ( Ek ∝ v² ) using collision mechanics graphs.
  • Thermal Energy & Logarithmic Decay: Linking ideal gas thermal kinetic energy ( E = ³⁄₂ kT ) to exponential energy loss models ( Ef = E₀ e-by ) across units (MeV to Joules/eV).
  • Scientific Evaluation: Explaining why low nucleon number moderators are vastly more efficient in reactor cores.
Question 07.1

Moderator Materials

Identification of alternative moderator materials [1 mark]

✅ Acceptable Answers

  • Water (or normal water / H₂O)
  • Heavy water (or D₂O / deuterium oxide)
  • Beryllium (or Be)

❌ Common Errors

  • Boron or Cadmium: These are used in control rods to absorb neutrons, NOT as moderators. Confusing moderators with control rods is a classic mistake.
  • Uranium: This is the reactor fuel, not a moderator.
Mark Breakdown: 1 mark awarded for identifying any one suitable substance. Chemical formulas ( H₂O , D₂O , Be ) are fully credited.
Question 07.2

Benefits of Moderation

State two benefits of slowing down fission neutrons [2 marks]

✅ Mark Scheme Points (Any 2)

  • Increased absorption/fission: U-235 fuel is far more likely to absorb slow (thermal) neutrons, increasing the probability of inducing further fission.
  • Decreased capture by U-238: Absorption of neutrons by non-fissionable U-238 is reduced.
  • Less structural damage: Slower neutrons cause less radiation damage or material fatigue to the reactor vessel/shielding.
  • Heat transfer: Slowing down transfers kinetic energy as thermal energy to the moderator/coolant, aiding heat extraction.

🧠 Exam Technique

Always specify which isotope is involved. Fast neutrons tend to bounce off or get captured parasitically by U-238 without causing fission. U-235 has a much larger "cross-section" (higher capture probability) for slow/thermal neutrons.

Mark Breakdown: 1 mark per distinct benefit (maximum 2 marks). Phrases like "increases chance of chain reaction" need to be linked to neutron absorption by U-235.
Question 07.3

Collision Energy Loss (Show That)

Verify that glider N loses about 30% of its kinetic energy when mM / mN = 12 [2 marks]

📐 Step-by-Step Calculation

  1. Read the graph (Figure 15):
    At mM / mN = 12 , read the value on the vertical axis:
    v / u = 0.85
  2. Relate velocity to Kinetic Energy:
    Ek, final / Ek, initial = (½ mN v²) / (½ mN u²) = (v / u)²
  3. Calculate retained kinetic energy fraction:
    (0.85)² = 0.7225 ≈ 72%
  4. Calculate percentage energy lost:
    Percentage lost = 100% - 72.25% = 27.75% ≈ 28% (which is "about 30%").

❌ Common Trap: Forgetting the Square

A common student error is subtracting the velocity ratio directly: 1 - 0.85 = 0.15 (15%) .

Remember: Kinetic energy is proportional to the square of speed ( Ek ∝ v² ). You MUST square the ratio v / u before finding the remaining energy fraction!

Mark Breakdown:
• Mark 1: Equating kinetic energy ratio to (v/u)² and calculating 0.85² = 0.72 (or 72%).
• Mark 2: Showing that the lost proportion is 1 - 0.72 = 0.28 (or 28%), confirming it is "about 30%".
Question 07.4

Number of Collisions for Thermalisation

Calculate the minimum number of collisions y needed to moderate a 1.0 MeV neutron [3 marks]

📐 Step-by-Step Calculation

  1. Calculate thermal kinetic energy Ef:
    Thermal neutron energy equals mean gas particle kinetic energy at T = 350 K :
    Ef = ³⁄₂ k T = 1.5 × (1.38 × 10⁻²³ J K⁻¹) × 350 K
    Ef = 7.245 × 10⁻²¹ J
  2. Convert initial energy E₀ to Joules (or Ef to eV):
    E₀ = 1.0 MeV = 1.0 × 10⁶ eV
    E₀ = 1.0 × 10⁶ × (1.60 × 10⁻¹⁹ C) = 1.60 × 10⁻¹³ J
    (Alternatively, Ef in eV = (7.245 × 10⁻²¹) / (1.60 × 10⁻¹⁹) = 0.0453 eV)
  3. Rearrange exponential decay formula:
    Ef = E₀ e-by ⇒ e-by = Ef / E₀ ⇒ eby = E₀ / Ef
    by = ln(E₀ / Ef) ⇒ y = ln(E₀ / Ef) / b
  4. Substitute values:
    y = ln((1.60 × 10⁻¹³) / (7.245 × 10⁻²¹)) / 0.73
    y = ln(2.208 × 10⁷) / 0.73 = 16.91 / 0.73 = 23.16 ≈ 23.2
    Since y must be collisions, y = 23 (or rounded up to 24).

💡 Key Knowledge & Traps

  • Unit Consistency: You cannot divide MeV by Joules inside a natural log. Both energies MUST be in Joules, or both in eV.
  • Ideal Gas Kinetic Theory: Mean translational kinetic energy of an ideal gas particle is given on the formula sheet as Ek = ³⁄₂ k T .
  • Constants Used:
    Boltzmann constant: k = 1.38 × 10⁻²³ J K⁻¹
    Elementary charge: e = 1.60 × 10⁻¹⁹ C
Mark Breakdown:
• Mark 1: Correct calculation of thermal energy Ef = ³⁄₂ kT = 7.2 × 10⁻²¹ J (or 0.045 eV ).
• Mark 2: Initial energy converted correctly to 1.6 × 10⁻¹³ J (or matched unit basis 1.0 × 10⁶ eV ).
• Mark 3: Final answer of y = 23.2 (accepts 23 or 24 integer).
Question 07.5

Moderator Nucleon Number Evaluation

Explain, using Figure 15, why elements with a small nucleon number are preferred [2 marks]

✅ Ideal Explanation

  • Mark 1 (Graph Link): Figure 15 shows that when the mass ratio mM / mN is smaller, the rebound speed ratio v / u is much smaller. This means a low nucleon number moderator produces a much greater reduction in neutron speed/kinetic energy per collision.
  • Mark 2 (Reactor Consequence): Because each collision removes a larger fraction of energy, fewer collisions are required to reach thermal energies, meaning the moderator can be smaller/thinner/more compact.

🧠 Exam Insight

Notice how Figure 15 approaches v/u = 0 as mM / mN → 1 ! When a neutron hits a particle of nearly identical mass (like a proton in ¹H), almost all kinetic energy can be transferred in a single collision (like billiard balls).

Examiners strictly require you to reference Figure 15 (link low ratio to lower rebound speed / larger energy loss) AND draw the operational conclusion (fewer collisions needed).

Mark Breakdown:
• Mark 1: Stating that low mass/nucleon number gives greater reduction in speed / KE per collision.
• Mark 2: Concluding that fewer collisions are needed (or moderator can be thinner/less bulky).

Topics

Physics · 3.8 Nuclear physics (A-level only) · 3.6 Further mechanics and thermal physics (A-level only) · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.