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
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct substitution of values into the equation, showing working, and providing the final answer with appropriate units. For conceptual questions, marks are given for clear and accurate explanations that demonstrate understanding of the principles involved.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Determine
- Find or derive a specific value or result from given information.
- Explain
- Provide a clear and detailed account of the principles or processes involved.
- Describe
- Give a detailed account of the characteristics, features, or steps in a process.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 2-3 minutes per mark for calculation questions and 1-2 minutes per mark for conceptual questions to ensure you have enough time to show all working and provide detailed answers.
- 1. Use the equation 1/2 mv 2 = eV to find the kinetic energy gained by the electron.1 mark
- 2. Substitute the values: m = 9.11 × 10 -31 kg, e = 1.602 × 10 -19 C, and V = 500 V.1 mark
- 3. Calculate the kinetic energy: 1/2 mv 2 = (1.602 × 10 -19 C) × 500 V = 8.01 × 10 -17 J.1 mark
- 4. Solve for v: v 2 = (2 × 8.01 × 10 -17 J) / (9.11 × 10 -31 kg).1 mark
- 5. Simplify and take the square root: v ≈ √(1.76 × 10 14 ) = 1.32 × 10 7 m/s.1 mark
Final answer: The final velocity of the electron is approximately 1.32 × 10 7 m/s.
Work through every step correctly and you earn all 5 marks.
Another worked example
An electron gains a kinetic energy of 4.806 × 10 -17 J when accelerated through a potential difference. Calculate the potential difference V.
- 1. Use the equation 1/2 mv 2 = eV to find the potential difference V.1 mark
- 2. Substitute the values: m = 9.11 × 10 -31 kg, e = 1.602 × 10 -19 C, and kinetic energy = 4.806 × 10 -17 J.1 mark
- 3. Rearrange to solve for V: the kinetic energy gained equals eV, so V = KE / e = (4.806 × 10 -17 J) / (1.602 × 10 -19 C).1 mark
- 4. Simplify and calculate: V = (4.806 × 10 -17 J) / (1.602 × 10 -19 C) = 300 V.1 mark
Final answer: V ≈ 300 V
Work through every step correctly and you earn all 4 marks.
Common mistakes
Forgetting to convert units, such as kV to V, before using them in calculations.
Why it happens: Students often overlook the need to ensure all units are consistent before performing calculations. This can lead to incorrect answers and lost marks.
Fix: 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).
Why it happens: Students might use incorrect values, such as 1.602 × 10 -18 C instead of 1.602 × 10 -19 C, leading to significant errors in calculations.
Fix: 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.
Why it happens: Students might mix up the concepts of kinetic energy (1/2 mv 2 ) and potential energy (eV), leading to incorrect application of the equation 1/2 mv 2 = eV.
Fix: 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).
Why it happens: Students might solve for v 2 and forget to take the square root, leading to an incorrect final answer.
Fix: 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.
Why it happens: Students might use incorrect formulas, such as mv 2 instead of 1/2 mv 2 , leading to errors in calculations.
Fix: Memorize and use the correct formula for kinetic energy: 1/2 mv 2 .
Not understanding the relationship between potential difference and kinetic energy.
Why it happens: Students might not grasp that the kinetic energy gained by an electron is directly proportional to the potential difference through which it is accelerated, leading to conceptual errors.
Fix: 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
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Electron Velocity | Use energy conservation to find an electron's speed after acceleration through a potential difference. | 5 |
| Calculate Potential Difference | Use the work-energy relation KE = eV to find the accelerating voltage. | 4 |
| Calculate Kinetic Energy | Find the kinetic energy an electron gains when accelerated through a given potential difference. | 3 |
| Calculate Electron Velocity | Use energy conservation to find electron speed after acceleration through a potential difference. | 5 |
| Total across these question types | 17 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.