AQA A-Level Physics Paper 1, June 2025: Question 8

📐 Step-by-Step Calculation

1. State the energy conservation equation:
   Electrical work done = Maximum kinetic energy gained
   QV = ½mv²  →  v = √(2QV / m)
2. Substitute known physical constants & values:
   Charge of electron, e = 1.60 × 10⁻¹⁹ C
   Mass of electron, m = 9.11 × 10⁻³¹ kg
   Potential difference, V = 130 V
   v = √[(2 × 1.60 × 10⁻¹⁹ × 130) / (9.11 × 10⁻³¹)]
3. Compute the numerical result:
   v = √(4.16 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = √(4.566 × 10¹³)
   v = 6.757 × 10⁶ m s⁻¹
4. Round appropriately (2 s.f.):
   Maximum speed = 6.8 × 10⁶ m s⁻¹

✅ Mark Scheme Breakdown

• Mark 1: Correct equation rearranged or substituted:
  QV = ½mv² or v = √(2QV / m)
• Mark 2: Final answer that rounds to 6.8 × 10⁶ m s⁻¹ .

❌ Common Errors

• Forgetting to square root: Calculating v² = 4.57 × 10¹³ and leaving it as the speed.
• Using wrong mass: Accidentally using the mass of a proton ( 1.67 × 10⁻²⁷ kg ) instead of an electron.
• Premature rounding: Rounding intermediate values too early, leading to out-of-range final answers.

🧠 Exam Technique

Always double-check that your computed speed is less than the speed of light ( c = 3.00 × 10⁸ m s⁻¹ ). 6.8 × 10⁶ m s⁻¹ is roughly 2% of c , confirming that non-relativistic kinetic energy formulas apply.

💡 Key Physics Concepts

Electrons do not travel in a vacuum inside the fluorescent tube; it is filled with vapour (such as mercury atoms). As electrons accelerate through the potential difference, they collide with these atoms.

• These collisions are inelastic.
• Kinetic energy of the free electrons is transferred to excite orbital electrons in the gas atoms.
• After each collision, the electron loses kinetic energy and has to re-accelerate from a much lower speed.

✅ Correct Answer

Electrons undergo inelastic collisions with atoms (mercury/gas particles), transferring and losing their kinetic energy.

❌ Common Errors

• Vague answers: Stating just "they hit things" or "there is resistance" without identifying collisions with gas atoms or loss of KE.
• Missing the energy aspect: Failing to state that kinetic energy is lost/transferred during these collisions.

🧠 Exam Technique

Always specify who collides with what and what happens to the energy. The phrase "inelastic collision with gas atoms transferring kinetic energy" is a high-scoring staple across AQA particle physics questions.

💡 How Fluorescent Tubes Work

1. Accelerated electrons excite gas atoms via inelastic collisions.
2. De-excitation of the gas atoms produces high-energy ultraviolet (UV) photons (wavelengths < 380 nm).
3. These UV photons strike the phosphor coating on the inside wall of the tube.
4. The coating absorbs UV photons and de-excites in multi-step transitions, emitting longer visible wavelengths (white light).

✅ Correct Answer & Marking Points

• Mark 1 (AO3): Yes, AND the graph shows that substance X emits (most) photons in the ultraviolet (UV) region of the spectrum (peak at ~255 nm, which is below 380 nm).
• Mark 2 (AO1): This UV radiation excites the phosphor / fluorescent coating, which then de-excites to emit longer wavelengths across the visible spectrum (white light).

❌ Common Errors

• Concluding "No": Saying "No, because the graph shows line peaks rather than a continuous spectrum" or "No, because 255 nm is not visible light." This shows a total misunderstanding of how fluorescent tubes work—the tube relies on the phosphor coating to produce visible light!
• Omitting the coating: Explaining that UV is emitted, but not mentioning the role of the fluorescent coating/lining in converting UV into visible light.

🧠 Exam Technique

Read the question carefully: it asks if white light can be produced by a fluorescent tube using X, not directly by substance X alone. Always recall the two-stage process: Gas emits UV → Coating absorbs UV and emits visible light.

📐 Step-by-Step Calculation

1. Read the peak wavelength from Figure 12:
   Look at the tallest peak: it sits midway between 250 nm and 260 nm.
   λ_peak = 255 nm = 2.55 × 10⁻⁷ m
   (Acceptable range: 250 nm to 260 nm)
2. Calculate photon energy in Joules:
   E = hc / λ
   h = 6.63 × 10⁻³⁴ J s ,   c = 3.00 × 10⁸ m s⁻¹
   E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (2.55 × 10⁻⁷)
   E = (1.989 × 10⁻²⁵) / (2.55 × 10⁻⁷) = 7.80 × 10⁻¹⁹ J
3. Convert energy from Joules to electron-volts (eV):
   Divide by the elementary charge ( 1.60 × 10⁻¹⁹ C ):
   E (in eV) = (7.80 × 10⁻¹⁹ J) / (1.60 × 10⁻¹⁹ J/eV)
   E = 4.875 eV ≈ 4.9 eV

✅ Mark Scheme Breakdown

• Marks 1 & 2 (Method): Max 2 from:
  • Reading λ accurately in range 250 nm to 260 nm ( 2.55 × 10⁻⁷ m used without power-of-ten error)
  • Using E = hc / λ
  • Converting energy to eV (dividing by 1.60 × 10⁻¹⁹ )
• Mark 3 (Accuracy): Final value of 4.9 eV (allow 4.77 eV to 4.97 eV depending on reading within 250–260 nm; accept 5.0 eV).

❌ Common Errors

• Power of ten error (POT): Forgetting that nanometres mean ×10⁻⁹ m . Note that the mark scheme explicitly states: "Do not condone POT error in MP1 bullet point 1".
• Multiplying by 1.6 × 10⁻¹⁹ instead of dividing: Remember: 1 eV = 1.60 × 10⁻¹⁹ J , so to convert J → eV , you must divide.
• Misreading the scale: Choosing a secondary peak (e.g. at 435 nm or 545 nm) instead of the maximum intensity peak at ~255 nm.

🧠 Exam Technique

Always write out your conversions with units:
E (eV) = E (J) / (1.60 × 10⁻¹⁹ J eV⁻¹) . Notice that atomic transitions typically have energies between 1 eV and 20 eV. If you obtain an answer like 10⁻³⁸ eV or 10¹⁹ eV , stop immediately and check whether you multiplied instead of dividing!