AQA A-Level Physics Paper 2, June 2023: Question 6

10 marks · Medium difficulty · Short Answer

State advantages of high-energy electron scattering over alpha particles, sketch diffraction intensity, and calculate nuclear radius constant and density.

Practise this question

Question

$Question 6 comprises five sub-questions. 06.1 asks for two advantages of using high-energy electrons rather than alpha particles to estimate nuclear radii (2 marks). 06.2 describes electron diffraction by target gas in an evacuated chamber (Figure 11) and asks to sketch a graph of electron intensity against scattering angle theta on a blank grid (Figure 12) (2 marks). 06.3 asks to show that R = R_0 A^(1/3) is consistent with nuclei having the same density (2 marks). 06.4 asks for one reason why the constant density derived is only approximate (1 mark). 06.5 gives the radius of chlorine-35 as 4.02 x 10^-15 m and asks to calculate R_0 and the nuclear density (3 marks).$

Question text

06.1 Nuclear radii can be estimated using either alpha particles or high-energy electrons.

State two advantages of using high-energy electrons rather than alpha particles

for this estimate.

[2 marks]

06.2 Figure 11 shows a beam of electrons, each with the same high energy, incident on

a target gas.

The electrons are diffracted by the nuclei in the gas.

The intensities of these diffracted electrons are measured at various angles θ.

The data are used to determine the nuclear radius R of the atoms in the gas.

Figure 11

Sketch on Figure 12 a graph showing how the electron intensity varies with θ.

[2 marks]

Figure 12

06.3 The radius R of a nucleus is related to its nucleon number by R = R A3 .

Show that this equation is consistent with the idea that all nuclei have the same

density.

[2 marks]

06.4 The equation R = R A3 is derived from experimental data.

Suggest one reason why the constant density of nuclear material derived from this

equation is only approximate.

[1 mark]

06.5 The measured radius R of 35Cl is 4.02 × 10−15 m.

Calculate an estimate of

• the constant R0

• the density of nuclear material.

[3 marks]

R = m density = kg m−3

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark AO

06.1 Any two from: 2 AO1

• using electrons gives greater resolution (as the AO2

wavelength can be made very small)

• electrons can get closer to the nuclei (as there is no

electrostatic repulsion)

• electrons have less recoil (as their mass is small

compared to the nucleus) OWTTE on each advantage

• free electrons are easier to accelerate Allow reverse arguments

OR give energy to (as charge-to-mass ratio is higher)

• electrons are easier to produce

• scattering distributions are easier to interpret

OR strong nuclear interaction is not involved

• using alpha particles only gives the distance of closest

approach/upper limit to the radius.

Do not allow U-shaped graphs.

06.2 A curved line showing an decrease in intensity with 2 AO1

increase in θ. The initial part of the curve may be absent. ×2

There is a single non-zero minimum.

Award MAX 1 for a line that covers less than

half the θ axis.

06.3 Amnucleon 1 Do not accept M unlabelled. 2 AO1

Density (= mass÷volume) = 4 1

πR3 Accept m (ie lowercase) unlabelled. AO2

Amnucleon Condone mn (mass of neutron in Data sheet)

(substituting density = 3 )

41 for mnucleon.

π�R0A3�

3 Accept only 1.67 ×10-27 (kg) for mass of

nucleon (ie to 3 sf).

3mnucleon Allow ecf to MP2 for any misrepresentation of

Density = 3 (in which) all (terms are) constant OR m provided it is clear that it signifies mass of

4π(R0)

the expression does not depend on A 2 a single nucleon.

2 If the constants are identified the equation

may be converted into a ratio. The equation

may be rearranged to have R0 as the subject

with work to show that this is a constant when

the density is constant.

06.4 Any one of: 1 AO3

The mass of the nucleus is not exactly A × mnucleon Do not accept “density of individual nucleons

(because this ignores the binding energy) can be different from each other”. It is not

allowed as it does not occur in the working

OR equation.

The volume equation assumes that the nucleus is a perfect

sphere (which is not true) OWTTE

The density equation that uses the nuclear radius formula 23

implies that the density is uniform within a nucleus which is

not true. OWTTE

Protons have a slightly different mass to a neutron

AO2

06.5 3

R 4.02 ×10−15 ×3

R = = = 1.2(3) ×10−15 (m)

01 1 1

A3 (35)3

mass Am

Substitutes values into density equation 2 2e.g.Density=� = 4 3� =

volume πR

35 × 1.67 × 10−27

4 −15 3

π�4.02 × 10 �

24 3

3 Evidence of a calculation must be given to

Density = 2.1 × 1017 (kg m−3)

3 gain this mark

Accept 2.15 but not 2.2

Total 10

How to answer it

Nuclear Radius, Electron Diffraction & Nuclear Density

📋 What This Question Tests

This question assesses your understanding of nuclear structure: comparing high-energy electron diffraction with alpha particle scattering, sketching and interpreting electron diffraction intensity curves, mathematically deriving why nuclear density is independent of nucleon number (constant density model), explaining the physical approximations behind this model, and calculating the constant R₀ and nuclear density from experimental data.

Question 06.1 · 2 Marks

Advantages of High-Energy Electron Diffraction

Comparing probes of the nucleus: Electrons vs Alpha Particles

✅ Acceptable Answers (Any Two)

Greater resolution: High-energy electrons have a very short de Broglie wavelength (comparable to nuclear diameter).
No strong nuclear force: Electrons are leptons, so they interact via the electromagnetic/weak interaction only, not the strong interaction (easier interpretation).
No Coulomb repulsion: Unlike alpha particles, negative electrons are attracted rather than repelled by positive nuclei, so they can penetrate much closer.
Direct nuclear measurement: Alpha scattering gives only the distance of closest approach (an upper limit), not the actual nuclear radius.
Less recoil: Electrons have negligible mass compared to the nucleus, minimising recoil energy loss.
Easier to produce & accelerate: Electrons have a much higher charge-to-mass ratio.

❌ Common Errors & Pitfalls

Vaguely stating "electrons are smaller" without linking to de Broglie wavelength or resolution.
Saying "electrons don't experience electric forces" (they do—they are attracted to the nucleus, not repelled!).
Forgetting that alpha particles are subject to the strong nuclear force at very close approach, distorting pure Coulomb scattering.

Mark Scheme Breakdown: 1 mark per distinct valid advantage (up to 2 marks). Reverse arguments comparing alpha limitations are fully accepted.

Question 06.2 · 2 Marks

Electron Diffraction Intensity Curve

Sketching Intensity vs Scattering Angle θ

✅ Graph Characteristics

Mark 1: A continuous curved line showing an overall decrease in intensity as scattering angle θ increases.
Mark 2: Shows a single, distinct non-zero minimum (the first diffraction minimum), followed by a lower secondary maximum that continues to fall.

Visual Appearance: Starts high on the intensity axis near θ = 0, curves downward steeply, bottoms out at a dip above the horizontal axis (must not touch zero), rises slightly into a small hump, then continues decreasing towards larger θ.

🧠 Exam Technique & Diagram Rules

Do NOT touch the axis: The intensity at the first diffraction minimum is non-zero due to nuclear shape factors and background scattering. Touching zero loses the second mark!
Extend past halfway: Ensure your curve spans across more than half the θ-axis. Truncated curves receive a maximum of 1 mark.
No U-shapes: Avoid symmetrical parabolic or U-shaped curves; it must be an asymmetric diffraction pattern.

Mark Scheme Breakdown: 1 mark for overall decreasing trend with angle; 1 mark for a single non-zero first minimum.

Question 06.3 · 2 Marks

Derivation: Constant Nuclear Density

Proving that R = R₀A¹/³ implies constant density

📐 Step-by-Step Proof

Express density in terms of mass and volume:
Assuming a spherical nucleus of radius R:
V = ⁴⁄₃ π R³
Total mass: M = A × m_nucleon
ρ = (A × m_nucleon) / (⁴⁄₃ π R³)
Substitute R = R₀A¹/³:
R³ = (R₀ A¹/³ )³ = R₀³ A
ρ = (A × m_nucleon) / (⁴⁄₃ π R₀³ A)
Cancel A and conclude:
ρ = (3 m_nucleon) / (4 π R₀³)
Because m_nucleon and R₀ are constants, the expression does not depend on A. Therefore, all nuclei have approximately constant density.

💡 Examiner Notes

Define your symbols: Do not write mass simply as M without linking it to A × m_nucleon .
Alternative method: You can rearrange for R₀ in terms of ρ, showing that if ρ is constant, R₀ is also constant.
Standard value of nucleon mass: m_nucleon ≈ 1.67 × 10⁻²⁷ kg .

Mark Scheme Breakdown: [1] Correct substitution of V and M = A m into density formula; [1] Algebraic cancellation showing ρ is independent of A.

Question 06.4 · 1 Mark

Why Nuclear Density is Only Approximate

Physical limitations and simplifications in the model

✅ Any One Valid Reason

Mass defect / Binding energy: Total nuclear mass is not exactly A × m_nucleon because binding energy lowers the total nuclear mass.
Non-spherical nuclei: The formula assumes nuclei are perfect spheres; many nuclei have deformed/ellipsoidal shapes.
Non-uniform density profile: Density is not uniform throughout; it has a fuzzy/diffuse boundary where density tails off towards the surface.
Mass difference: Protons and neutrons have slightly different masses ( m_n > m_p ).

❌ Rejected Answers

Do NOT say: "The density of individual nucleons is different from each other." This does not address the assumptions in the working equation.
Avoid vague answers like "experimental error" or "inaccurate measuring instruments."

Mark Scheme Breakdown: 1 mark for any single correct physical reason why nuclear density is an approximation.

Question 06.5 · 3 Marks

Calculation: R₀ and Nuclear Density

Data: Measured radius R for ³⁵₁₇Cl = 4.02 × 10⁻¹⁵ m

📐 Step-by-Step Calculation

Calculate R₀:
R = R₀ A¹/³ ⇒ R₀ = R / A¹/³
For Chlorine-35, A = 35 :
R₀ = (4.02 × 10⁻¹⁵) / (35¹/³)
35¹/³ ≈ 3.271
R₀ = 1.23 × 10⁻¹⁵ m (accepts 1.2 × 10⁻¹⁵ m)
Substitute into the Nuclear Density Formula:
Total mass: M = 35 × (1.67 × 10⁻²⁷ kg) = 5.845 × 10⁻²⁶ kg
Volume: V = ⁴⁄₃ π R³ = ⁴⁄₃ π (4.02 × 10⁻¹⁵)³ ≈ 2.72 × 10⁻⁴³ m³
ρ = (5.845 × 10⁻²⁶) / (2.72 × 10⁻⁴³)
Final Density Value:
Density = 2.1 × 10¹⁷ kg m⁻³ (or 2.15 × 10¹⁷ kg m⁻³)

❌ Common Traps

Using Z instead of A: Do not use 17 (atomic number) for the cubic root! Chlorine-35 has nucleon number A = 35 .
Cube Error: Forgetting to cube the whole radius: (4.02 × 10⁻¹⁵)³ gives order of magnitude 10⁻⁴⁴ .
Rounding Trap: The mark scheme specifically accepts 2.1 × 10¹⁷ or 2.15 × 10¹⁷ , but rejects rounding prematurely to 2.2 × 10¹⁷ without shown working!

Mark Scheme Breakdown: [1] Calculating R₀ = 1.2(3) × 10⁻¹⁵ m; [1] Correct substitution into density formula; [1] Final density = 2.1 × 10¹⁷ kg m⁻³ (must show evidence of calculation).

Topics

Physics · 3.8 Nuclear physics (A-level only)

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.