Thermionic Emission and Electron Acceleration Revision Notes | A Level AQA Physics

1. Why Thermionic Emission Changed Physics

At the end of the nineteenth century, physicists had no direct evidence that atoms contained smaller charged particles. Then J.J. Thomson used a heated wire and a high voltage to produce a beam of particles that bent in electric and magnetic fields — and the electron was discovered. The two ideas at the heart of that experiment are exactly what this lesson covers: thermionic emission and electron acceleration.

These concepts are not just historical curiosities. Every cathode-ray tube, electron microscope, and X-ray machine relies on the same physics. Understanding them also gives you a clean, concrete context for applying the work-energy theorem at the sub-atomic scale — which is precisely what AQA tests in this topic.

2. Thermionic Emission — The Principle

Picture a metal filament — essentially a very fine wire — connected to a low-voltage supply so that a current heats it. Inside the metal, conduction electrons are already free to move between atoms. As the filament temperature rises, those electrons gain more and more thermal (kinetic) energy.

At the surface of the metal, electrons are held in by electrostatic attraction to the positive ion lattice. This surface barrier has a specific energy threshold: the work function of the metal. When an electron's thermal energy exceeds the work function, it can break free from the surface entirely. This process — the emission of electrons from a heated metal surface — is called thermionic emission.

The rate of emission increases sharply with temperature: more electrons have enough energy to escape as the filament gets hotter. In practice, the filament (cathode) is enclosed in a vacuum so that emitted electrons are not scattered by air molecules before they can be accelerated.

• The metal filament is heated by passing a current through it (the heater supply).
• Electrons near the surface gain thermal energy from lattice vibrations.
• Electrons with energy greater than the work function escape from the surface.
• A vacuum surrounds the filament so emitted electrons travel freely.
• The emitted electrons form a cloud (space charge) just outside the cathode surface.

A useful way to picture this: the work function acts like a cliff edge. Electrons inside the metal are at the base; the cliff height is the work function energy. Heating the metal gives electrons a running jump. Only those with enough energy to clear the top of the cliff actually escape. The rest fall back into the metal.

This spec point is fully covered once you can describe the heating mechanism, name the work function as the energy barrier, and state that a vacuum is required. Those three elements together earn full marks on a 'describe thermionic emission' question.

3. Accelerating Electrons Through a Potential Difference

Once electrons have been emitted from the cathode, they need to be given a controlled, measurable amount of kinetic energy. This is where the accelerating voltage comes in. A positive anode is placed a short distance from the cathode, and a potential difference $V$ is applied between them.

An electron (charge $-e$, where $e = 1.6 \times 10^{-19}$ C) is attracted towards the positive anode. As it moves through the potential difference, the electric field does work on it. The work done on the electron equals the charge multiplied by the potential difference: $W = eV$. All of that work goes into the kinetic energy of the electron, provided it starts from rest at the cathode.

The equation $\frac{1}{2}mv^2 = eV$ is a direct application of the work-energy theorem. The left side is the kinetic energy the electron ends up with; the right side is the energy transferred to it by the electric field. Because the electron starts from rest, there is no initial kinetic energy term to subtract.

To find the speed $v$, rearrange for $v^2$ first, then take the square root. This two-step rearrangement is where marks are most often dropped — rushing straight to $v$ without writing $v^2$ as an intermediate line is a reliable way to make an arithmetic error under exam pressure.

4. Worked Calculation — Finding Electron Speed

Theory is only useful when you can apply it under exam conditions. The calculation below models the exact method AQA expects: state the equation, substitute with units, rearrange, and give a final answer with correct units and significant figures. Work through each step yourself before revealing the next one.

Worked example: Calculating the speed of an electron accelerated through a potential difference

An electron is accelerated from rest through a potential difference of 3.0 kV. Calculate the speed of the electron after acceleration. (Mass of electron $m = 9.11 \times 10^{-31}$ kg; charge of electron $e = 1.6 \times 10^{-19}$ C.)

1. $V = 3.0 \text{ kV} = 3.0 \times 10^3 \text{ V} = 3000 \text{ V}$ — Always convert kV to V before substituting. Leaving the value in kV would give a speed that is wrong by a factor of $\sqrt{1000}$. [M1]
2. $\frac{1}{2}mv^2 = eV$ — This is the core equation from the specification. Writing it first earns the method mark even if arithmetic errors follow. [M1]
3. $v^2 = \frac{2eV}{m}$ — Multiply both sides by 2, then divide both sides by $m$. Keep this as a separate line — it prevents sign and factor errors.
4. $v^2 = \frac{2 \times 1.6 \times 10^{-19} \times 3000}{9.11 \times 10^{-31}}$ — Substitute $e = 1.6 \times 10^{-19}$ C, $V = 3000$ V, $m = 9.11 \times 10^{-31}$ kg. Include units at this stage. [M1]
5. $2 \times 1.6 \times 10^{-19} \times 3000 = 2 \times 4.8 \times 10^{-16} = 9.6 \times 10^{-16}$ J — $1.6 \times 10^{-19} \times 3000 = 4.8 \times 10^{-16}$; then multiply by 2 to get $9.6 \times 10^{-16}$.
6. $v^2 = \frac{9.6 \times 10^{-16}}{9.11 \times 10^{-31}} = 1.054 \times 10^{15} \text{ m}^2\text{s}^{-2}$ — $\frac{9.6}{9.11} \approx 1.054$ and $10^{-16} \div 10^{-31} = 10^{15}$. Keep at least 3 significant figures here to avoid rounding errors in the final step.
7. $v = \sqrt{1.054 \times 10^{15}} = \sqrt{1.054} \times \sqrt{10^{15}} \approx 1.027 \times 10^{7.5} = 1.027 \times 3.162 \times 10^{7} \approx 3.2 \times 10^{7} \text{ m s}^{-1}$ — $\sqrt{10^{15}} = 10^{7.5} = 10^7 \times \sqrt{10} \approx 10^7 \times 3.162$. So $v \approx 1.027 \times 3.162 \times 10^7 \approx 3.2 \times 10^7$ m/s. [A1]

Answer: $v \approx 3.2 \times 10^7 \text{ m s}^{-1}$

Notice that the electron reaches roughly one-tenth of the speed of light. At speeds approaching $c$, relativistic effects become significant, but at A-level you are not required to apply relativistic corrections — the classical equation $\frac{1}{2}mv^2 = eV$ is sufficient for all AQA questions on this topic.

5. The Electron Gun — Putting It All Together

An electron gun combines both processes in a single device. The cathode is heated by a separate low-voltage heater supply, causing thermionic emission. The emitted electrons are then drawn towards the anode by the accelerating voltage $V$. A small hole in the anode allows a narrow beam of electrons to pass through and continue in a straight line.

The two voltage supplies are independent and serve different purposes. The heater voltage controls the temperature of the filament and therefore the rate of electron emission. The accelerating voltage $V$ controls the kinetic energy — and hence the speed — of the electrons in the beam. Increasing $V$ increases the speed of each electron; it does not change the rate of emission.

The chart shows that electron speed increases with accelerating voltage, but the relationship is not linear — speed is proportional to $\sqrt{V}$, because $v = \sqrt{\frac{2eV}{m}}$. Doubling the voltage increases the speed by a factor of $\sqrt{2} \approx 1.41$, not by a factor of 2. This is a classic exam analysis point.

6. Exam Technique and Common Mistakes

AQA questions on this topic tend to fall into three categories: describe thermionic emission (3–4 marks), calculate electron speed (3 marks), and analyse or evaluate a scenario involving a change in voltage or temperature. Each type has its own mark-winning pattern.

For 'describe thermionic emission' questions, the mark scheme typically awards one mark each for: the filament is heated; electrons gain thermal energy; electrons with energy greater than the work function escape the surface. Mentioning the vacuum is often credited as a fourth mark. Avoid vague phrases like 'electrons are released by heat' — name the work function explicitly.

For calculation questions, the method marks are awarded for writing $\frac{1}{2}mv^2 = eV$ and for a correct substitution. The accuracy mark requires the correct final value with appropriate units. If you make an arithmetic slip but your method is correct, you can still earn 2 out of 3 marks — so always show the equation and substitution clearly.

Together, thermionic emission and the equation $\frac{1}{2}mv^2 = eV$ cover both specification requirements for this section of the Turning Points topic. Thermionic emission explains how a controlled source of free electrons is produced; the energy equation quantifies exactly how much kinetic energy — and therefore speed — those electrons acquire when accelerated through a known potential difference.