AQA A-Level Physics Paper 3 (3BD), June 2025: Question 1

✅ Mark Scheme Requirements

Any 2 from (Mark 1 & 2):

• Glass tube evacuated / containing low-pressure gas.
• Heater/filament/thin wire connected to low-voltage supply (AC or DC) / has current running through it to emit electrons thermionically.
• Anode with a central hole connected to the positive terminal of a high-voltage supply to accelerate and collimate electrons.

Mark 3 (Voltage Specifications):

• Heater power supply: ≤ 20 V (typically 6 V or 12 V).
• Accelerating anode potential difference: > 200 V and < 1000 kV (typically 1 kV – 5 kV; must NOT be AC).

🧠 Diagram Instructions

💡 Key Knowledge: Velocity Selector

When electric and magnetic fields are crossed such that magnetic force balances electric force:

eE = Bev ⇒ e(V / d) = Bev ⇒ v = V / (B × d)

Given typical values: V = 750 V , d = 0.050 m , B = 0.64 × 10⁻³ T :

v = 750 / (0.05 × 0.64 × 10⁻³) = 2.34 × 10⁷ m s⁻¹

💡 Key Knowledge: Vertical Velocity Component

Upon exiting the electric field after deflecting by angle θ = 25° :

tan 25° = vᵥ / v

vᵥ = 2.34 × 10⁷ × tan 25° = 1.09 × 10⁷ m s⁻¹

📐 Step-by-Step Calculation

1. Time spent between plates:
   t = L / v = 0.10 / 2.34 × 10⁷ = 4.27 × 10⁻⁹ s
   (Plate length L = 0.10 m)
2. Calculate acceleration:
   a = (vᵥ − uᵥ) / t = (1.09 × 10⁷ − 0) / 4.27 × 10⁻⁹
   a = 2.56 × 10¹⁵ m s⁻² ≈ 2.5 × 10¹⁵ m s⁻²

❌ Critical Exam Restriction

DO NOT USE DATA BOOKLET VALUES!

The prompt strictly states: "You should not use the values of the electron charge, mass or specific charge from the data and formulae booklet."

Using a = eE / m = (1.6 × 10⁻¹⁹ × E) / 9.11 × 10⁻³¹ scores 0 marks because you are using the very value Thomson set out to discover!

📐 Step-by-Step Working

1. Relate electric force to acceleration:
   F = ma = eE = e(V / d)
2. Rearrange for specific charge (e/m):
   e / m = (a × d) / V
3. Substitute experimental values:
   e / m = (2.56 × 10¹⁵ × 0.050) / 750
4. Final calculated value:
   e / m = 1.7 × 10¹¹ C kg⁻¹

🧠 Alternative Valid Approaches

You can also express specific charge directly from the combined field parameters:

All valid paths give 1.7 × 10¹¹ C kg⁻¹ .

✅ Model Answer

• The specific charge of the electron was much larger (around 1800 to 2000 times larger) than that of the largest known specific charge at the time: the hydrogen ion / proton. [1 mark]
• Since charge was assumed to be comparable, this proved that the electron has a much smaller mass than a hydrogen ion/atom, showing it is a subatomic particle. [1 mark]

❌ Examiner Pitfalls to Avoid

• Vague comparisons: Writing simply "it was larger" loses the mark. It must be stated as "much larger" or quantitatively "~2000 times larger".
• Hydrogen Atom vs Ion: Do NOT refer merely to "hydrogen" or a "hydrogen atom". It must be compared to a hydrogen ion (or proton).
• Confusing size with mass: Do not say "electrons are smaller in size"; the conclusion relates strictly to mass.

💡 The Two Experimental Methods

Method 1: Fix B → adjust V until horizontal → turn off B → measure resulting angle θ on screen.

Method 2: Choose angle θ on scale → set V so beam hits θ → adjust B until beam is horizontal.

✅ Which Method is Best & Why? (3 Marks)

Method 2 is preferred. (Required to access full marks)

Mark Breakdown (Any 3 points):

• In Method 1, the deflected spot is unlikely to align exactly on a scale division mark, causing higher reading uncertainty.
• In Method 2, the spot can be precisely adjusted using the fine p.d. control to sit exactly on a prominent scale division line.
• Method 2 allows the experimenter to select a large angle θ (e.g. 15°), which automatically minimises the percentage uncertainty in θ ( % uncertainty = Δθ / θ × 100 ).
• The percentage uncertainty / resolution in the B-field reading (or ammeter current) is much smaller than the angular scale resolution.