AQA A-Level Physics Paper 2, June 2023: Question 4

8 marks · Medium difficulty · Short Answer

Analyze the electric field and particle motion in a spark detector, and evaluate relationships between spark rate and distance from an alpha source.

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Question

Question 4 shows a spark counter consisting of a metal mesh at 0 V placed 5.0 mm above a wire at 4000 V. Figure 5 illustrates the equipotential surfaces between the mesh and wire, with equipotentials evenly spaced near the mesh and curving and crowding closely near the wire, with point P located on the 2000 V line. Part 04.1 asks to sketch electric field strength E versus distance d on Figure 6. Part 04.2 asks for the ratio of the speeds of the electron and argon ion when they reach the wire and mesh respectively. Part 04.3 asks how air affects this ratio. Part 04.4 provides Figure 7, a graph of the number of sparks N against height h of the alpha source, and asks to test two hypotheses: an inverse-square law and an exponential relationship.
Question text

04 Figure 4 shows a spark detector used to detect alpha particles.

Figure 4

The detector consists of a metal mesh placed 5.0 mm above a wire.

A potential difference of 4000 V is applied between the mesh and the wire.

Molecules in the air between the mesh and the wire are ionised by an alpha particle

and a spark is produced.

Figure 5 shows equipotentials between the mesh and the wire.

Figure 5

04.1 Figure 5 shows a dashed line between the mesh and the wire.

Sketch on Figure 6 a graph to show how the magnitude E of the electric field strength

varies with the distance d from the mesh along this dashed line.

No values are required on the E axis.

[2 marks]

Figure 6

An alpha particle passes through the mesh.

The alpha particle ionises an argon atom at P on Figure 5, releasing one electron.

The electron and the argon ion have no kinetic energy at P.

The electron then travels to the wire and the argon ion travels to the mesh.

04.2 speed of electron when it reaches the wire

Calculate the ratio .

speed of argon ion when it reaches the mesh

Assume that the air has no effect on the motion of the electron or on the motion of

the argon ion.

mass of argon ion = 6.64 × 10−26 kg

[2 marks]

ratio =

04.3 In practice, the air does affect the motion of the electron and the motion of the

argon ion.

Suggest how the presence of air between the mesh and the wire changes the ratio in

*11* Question 04.2.

No numerical detail is required.

[1 mark]

04.4 The alpha source in Figure 4 is moved to different heights h above the mesh.

Figure 7 shows how the number of sparks N produced in 10 minutes varies with h.

No sparks are produced when the source is not present.

Figure 7

Student A suggests that the spark rate obeys an inverse-square law.

Student B suggests that the spark rate decreases exponentially with h.

Determine whether either student is correct.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 04. 04.1 awards 2 marks for a horizontal line above zero for over half the distance, then curving upwards. 04.2 awards 2 marks for showing equal kinetic energies and calculating the ratio of speeds as sqrt(6.64e-26 / 9.11e-31) = 270. 04.3 awards 1 mark for explaining that collisions cause the argon ion to lose more energy/momentum, making the ratio larger. 04.4 awards 3 marks: 1 for testing the inverse-square law using two data points (e.g. N*h^2), 1 for testing exponential decay (e.g. constant half-distance or N*e^h), and 1 for concluding that both students are incorrect with valid reasoning.

Question Answers Additional comments/Guidelines Mark AO

04.1 Horizontal line above zero for > half the distance 2 AO2

×2

(then) curves upwards

Marks are independent.

statement that (kinetic) energy is the same

04.2 2 AO3

×2

OR

Some factors (e.g. ½) that cancel may be

correct substitution of data into equation/ratio absent.

6.64 ×10–26

ratio = –31 = 270

9.11 10×

04.3 Due to collisions argon loses more energy / speed / Mark is for the explanation 1 AO3

momentum (than electron) 15

Do not allow explanations that suggests the

OR particles have different distances to travel.

Electron is less ionising (than argon) / Do not allow references to air resistance.

Argon is more ionising (than electron)

Accept idea that “argon has higher probability

of collision (than electron)”

so ratio is larger.

Accept reverse arguments

04.4 1 test inverse square: 3 AO3

Evidence of suitable test for student A suggestion involving 2

e.g. Nh = constant ×3

2+ data points 1

2 test exponential:

Evidence of suitable test for student B suggestion involving

e.g. N should decrease by half in equal

2+ sets of data 2

intervals of h

OR test N = k e-h : show that N eh is not

constant

Accept log/ln interpretations of the test.

Both tests performed AND rejects both suggestions 3

Expect to see for student A 10 × 1.02 = 10

and 4 × 1.342 = 7.18 or similar

Expect to see for student B N = 10 to N = 5

changes h by 0.25 and N = 5 to N = 2.5

16 changes h by 0.31 or similar

reject both students’ suggestions with

reasons eg Nh2 is not constant and N does

not fall by the same fraction in equal intervals

of h

Allow answer accepting B if a reference to

experimental error is made and the difference

is small.

Total 8

How to answer it

Spark Detector & Electric Fields

📋 What this question tests

This question assesses your understanding of electric fields and radioactivity across practical and theoretical contexts:

  • Electric Field Strength from Equipotentials: Interpreting spacing of equipotential surfaces to deduce electric field variation ( E = -ΔV/Δx ).
  • Work Done & Particle Kinematics: Applying W = qΔV = ½mv² to charged particles (electrons vs. ions) moving across a potential difference.
  • Microscopic Collision Effects: Explaining the impact of collisions/ionisation on ion vs. electron motion through air.
  • Graphical Hypothesis Testing: Using coordinates from a curve to quantitatively test whether data follows an inverse-square relationship or an exponential decay.

Question 04.1

Sketching Electric Field Strength (E) vs Distance (d) from Mesh to Wire [2 Marks]

✅ Required Sketch Features

  • Mark 1: A horizontal line above zero ( E > 0 ) starting at d = 0 and continuing for more than half the distance towards the wire.
  • Mark 2: The line curves steadily upwards as it approaches d = 5.0 mm (the wire).

💡 Key Physics

Electric field strength is the potential gradient: E = -ΔV / Δx .

  • Near the mesh, equipotentials are evenly spaced horizontal lines → uniform field → constant E .
  • Near the wire, equipotentials become concentric cylindrical surfaces and bunch tightly together → field gradient increases dramatically → E rises sharply.

❌ Common Errors

  • Starting the curve right from d = 0 (ignoring that the equipotentials remain parallel and uniformly spaced for the top half).
  • Starting E at zero at d = 0 (the mesh is at 0 V , but the gradient ΔV/Δx is definitely non-zero!).
  • Drawing a straight diagonal line from the mesh to the wire.

🧠 Exam Technique

Always relate the spacing between equipotentials in the provided diagram directly to the value plotted on the E axis: Equal spacing = Constant E; Closer spacing = Increasing E.

Question 04.2

Calculating the Speed Ratio of Electron to Argon Ion [2 Marks]

📐 Step-by-Step Calculation

  1. Identify Work Done: Both the electron and the singly charged argon ion carry equal magnitude of charge ( e = 1.60 × 10⁻¹⁹ C ). Moving to their respective electrodes across the same potential gap from point P gives them equal kinetic energy:
    Ek = ½ me ve² = ½ mion vion²
  2. Rearrange for Velocity Ratio:
    me ve² = mion vion²
    (ve / vion)² = mion / me
    ve / vion = √(mion / me)
  3. Substitute Mass Values:
    mion = 6.64 × 10⁻²⁶ kg
    me = 9.11 × 10⁻³¹ kg (from data sheet)
    ratio = √( (6.64 × 10⁻²⁶) / (9.11 × 10⁻³¹) ) = √(72887) ≈ 270

✅ Mark Scheme Breakdown

Mark 1: Stating kinetic energies are equal ( Ek(e) = Ek(ion) ) OR correct substitution into ratio/equation.

Mark 2: Final calculated ratio = 270 (accept 269 – 270).

💡 Why the potential at P doesn't matter:

Whether P is at 2000 V or any intermediate potential, the work done on both particles originates from traversing complementary parts of the field with identical charges; or assuming both accelerate across the full potential difference.

❌ Common Errors

  • Forgetting to take the square root of the mass ratio (giving 7.29 × 10⁴ instead of 270).
  • Inverting the ratio ( √(me/mion) = 3.7 × 10⁻³ ) by putting electron mass in the numerator.
  • Looking up the mass of a proton or neutron instead of electron mass from the data sheet.

Question 04.3

Effect of Air on the Speed Ratio [1 Mark]

✅ Acceptable Responses

The ratio is larger, because:

  • Due to collisions, the argon ion loses more energy / speed / momentum than the electron.
  • OR The argon ion is larger / has a higher probability of collision with air molecules than the electron.
  • OR The electron is less ionising than the argon ion.

❌ What the Examiner Rejected

  • "Air resistance": Do NOT refer to macroscopic fluid drag/air resistance. Must be discussed in terms of particle collisions or ionisation.
  • Saying particles have different distances to travel (the question asks about speed reduction due to presence of air).

Question 04.4

Evaluating Student A (Inverse-Square) & Student B (Exponential) [3 Marks]

📐 Testing the Two Hypotheses

Test for Student A (Inverse Square: N ∝ 1/h² → N × h² = constant):
  • At h = 1.0 cm , N = 10.0 × 10³ → N × h² = 10.0 × 1.0² = 10.0
  • At h = 1.34 cm , N = 4.0 × 10³ → N × h² = 4.0 × 1.34² = 7.18
  • Since 10.0 ≠ 7.18 , the product is not constant.
Test for Student B (Exponential Decay: constant halving distance Δh):
  • Halving from N = 10.0 to N = 5.0 :
    h goes from 1.00 cm to 1.25 cm → Δh₁ = 0.25 cm
  • Halving from N = 5.0 to N = 2.5 :
    h goes from 1.25 cm to 1.56 cm → Δh₂ = 0.31 cm
  • Since Δh₁ ≠ Δh₂ (0.25 cm ≠ 0.31 cm), it does not fall by equal fractions in equal intervals.
    Alternative test: N ekh is not constant.

✅ Mark Scheme Breakdown

Mark 1: Suitable quantitative test for Student A using 2+ data points (e.g. testing whether N h² = constant ).

Mark 2: Suitable quantitative test for Student B using 2+ data pairs (e.g. checking if half-distance is constant, or testing N eh / log values).

Mark 3: Clear conclusion rejecting both students with numerical justification.
(Note: MS allows accepting B only if experimental error is explicitly argued for the slight variation).

🧠 Exam Technique: Testing Relationships from Graphs

  • Always state the test condition explicitly: Write "If inverse square, N × h² = constant" before writing numbers.
  • Read easy grid points: Choose points where the curve crosses exact grid intersections (e.g., h = 1.0, N = 10 ).
  • Show working for at least two points for each hypothesis, state the values calculated, and compare them explicitly.

Topics

Physics · Practical skills · 3.7 Fields and their consequences (A-level only) · 3.8 Nuclear physics (A-level only) · Data analysis

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.