A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Thermionic Emission and Electron Acceleration — mark scheme explained
The short answer
The principle of thermionic emission is a fundamental concept in physics that describes the process by which electrons are emitted from a heated metal surface. This phenomenon is crucial in various applications, including cathode-ray tubes (CRTs), electron microscopes, and X-ray tubes.
The question
An electron is accelerated through a potential difference of 500 V. Calculate the final velocity of the electron. [Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
1. Use the equation 1/2 mv 2 = eV to find the kinetic energy gained by the electron.
- S2
2. Substitute the values: m = 9.11 × 10 -31 kg, e = 1.602 × 10 -19 C, and V = 500 V.
- S3
3. Calculate the kinetic energy: 1/2 mv 2 = (1.602 × 10 -19 C) × 500 V = 8.01 × 10 -17 J.
- S4
4. Solve for v: v 2 = (2 × 8.01 × 10 -17 J) / (9.11 × 10 -31 kg).
- S5
5. Simplify and take the square root: v ≈ √(1.76 × 10 14 ) = 1.32 × 10 7 m/s.
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
1. Use the equation 1/2 mv 2 = eV to find the kinetic energy gained by the electron.
- S2
2. Substitute the values: m = 9.11 × 10 -31 kg, e = 1.602 × 10 -19 C, and V = 500 V.
- S3
3. Calculate the kinetic energy: 1/2 mv 2 = (1.602 × 10 -19 C) × 500 V = 8.01 × 10 -17 J.
- S4
4. Solve for v: v 2 = (2 × 8.01 × 10 -17 J) / (9.11 × 10 -31 kg).
- S5
5. Simplify and take the square root: v ≈ √(1.76 × 10 14 ) = 1.32 × 10 7 m/s.
Final answer: The final velocity of the electron is approximately 1.32 × 10 7 m/s.
Common mistakes
- Forgetting to convert units, such as kV to V, before using them in calculations. — Always check and convert units to their base form (e.g., from kV to V) before substituting them into equations.
- Using the wrong value for the charge of an electron (e). — Memorize the correct value for the charge of an electron (e = 1.602 × 10 -19 C) and double-check it before using it in equations.
- Confusing kinetic energy with potential energy. — Clearly understand the difference between kinetic energy and potential energy. Kinetic energy is associated with motion, while potential energy is associated with position or configuration.
- Forgetting to take the square root when solving for velocity (v). — Always remember to take the square root of both sides of the equation when solving for velocity (v) from v 2 .
- Using the wrong formula for kinetic energy. — Memorize and use the correct formula for kinetic energy: 1/2 mv 2 .
- Not understanding the relationship between potential difference and kinetic energy. — Understand and remember that 1/2 mv 2 = eV, where the kinetic energy (1/2 mv 2 ) is equal to the work done by the electric field (eV).
Where the marks go
- Full worked solution (all marking points)5 marks