The Electron Volt and Energy Conversions Revision Notes | A Level AQA Physics

1. Why the Joule Is the Wrong Tool for the Job

Picture the energy needed to lift a textbook off a desk — roughly 10 joules. Now picture the energy of a single electron moving inside an atom. That energy is around $1.6 \times 10^{-19}$ J. Writing that number repeatedly in every calculation is cumbersome and error-prone.

Physicists solved this by inventing a unit scaled to the atomic world: the electron volt. At A Level, you will meet it constantly in topics covering photons, photoelectric effect, atomic energy levels, and particle physics. Getting comfortable with it now pays dividends across the whole course.

2. Defining the Electron Volt

Here is the physical picture. Place an electron between two charged plates with a potential difference of exactly 1 V across them. The electric field does work on the electron as it accelerates from one plate to the other. The amount of work done — and therefore the kinetic energy gained — is defined as one electron volt.

The connection to the elementary charge is not a coincidence — it is the whole point. Work done by an electric field equals charge multiplied by potential difference: $W = QV$. For one electron ($Q = e = 1.6 \times 10^{-19}$ C) accelerated through $V = 1$ V, the work done is $W = 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-19}$ J. That result is exactly 1 eV.

This means the conversion factor between eV and J is the elementary charge $e$. Multiply by $e$ to go from eV to J; divide by $e$ to go from J to eV. Keep that relationship in mind and the conversions become mechanical.

3. Converting eV to J — The Method

Converting from eV to J is the more common direction in AQA questions. You are usually given a photon energy or ionisation energy in eV and need to substitute it into an equation that requires joules. The rule is always the same: multiply the eV value by $1.6 \times 10^{-19}$.

Worked example: Converting eV to J

A photon has energy 2.5 eV. Calculate this energy in joules.

1. $E \text{ (J)} = E \text{ (eV)} \times e$ — The elementary charge e = 1.6 × 10⁻¹⁹ C is the conversion factor. [M1]
2. $E = 2.5 \times 1.6 \times 10^{-19}$ — Replace E(eV) with 2.5. Keep the power of ten separate to avoid arithmetic errors. [M1]
3. $2.5 \times 1.6 = 4.0$ — Verify: 2.5 × 1.6 = 2.5 × 1 + 2.5 × 0.6 = 2.5 + 1.5 = 4.0. Correct.
4. $E = 4.0 \times 10^{-19}$ J — The power of ten stays as 10⁻¹⁹ since we only multiplied the coefficient. [A1]
5. $E = 4.0 \times 10^{-19}$ J — Always include the unit J. Two significant figures match the data given.

Answer: $E = 4.0 \times 10^{-19}$ J

4. Converting J to eV — The Reverse Direction

The reverse conversion comes up when a calculation gives you an answer in joules and the question asks you to express it in eV, or when you need to compare an energy to a known threshold given in eV. Dividing by $e$ undoes the multiplication, so the rule is: divide the joule value by $1.6 \times 10^{-19}$.

Dividing by a number in standard form can feel awkward. The cleanest approach is to write the division explicitly, then handle the coefficient and the power of ten separately. The worked example below shows this method step by step.

Worked example: Converting J to eV

An electron has kinetic energy $3.2 \times 10^{-19}$ J. Express this energy in electron volts.

1. $E \text{ (eV)} = \frac{E \text{ (J)}}{e}$ — Rearrange E(J) = E(eV) × e by dividing both sides by e. [M1]
2. $E \text{ (eV)} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}$ — Place the energy in J in the numerator and e in the denominator. [M1]
3. $\frac{3.2}{1.6} = 2.0$ — Verify: 1.6 × 2 = 3.2. Correct.
4. $\frac{10^{-19}}{10^{-19}} = 10^{0} = 1$ — When the powers of ten are identical, they cancel to give 1. This is why energies close to 1.6 × 10⁻¹⁹ J give answers close to 1 eV.
5. $E = 2.0 \times 1 = 2.0$ eV — The unit is now eV. [A1] Two significant figures match the data.

Answer: $E = 2.0$ eV

5. Handling Larger Multiples — keV and MeV

In particle physics, energies can be thousands or millions of electron volts. AQA uses the prefixes kilo- and mega- in this context: 1 keV = $10^3$ eV and 1 MeV = $10^6$ eV. Before converting to joules, always convert to plain eV first.

Notice the pattern: each step up in prefix shifts the joule value by a factor of $10^3$. If you see 0.51 MeV (the rest-mass energy of an electron), that is $0.51 \times 10^6$ eV, which converts to $0.51 \times 10^6 \times 1.6 \times 10^{-19}$ J $= 8.16 \times 10^{-14}$ J.

Worked example: Converting MeV to J

A proton has kinetic energy 150 MeV. Convert this to joules.

1. $150 \text{ MeV} = 150 \times 10^{6} \text{ eV} = 1.50 \times 10^{8} \text{ eV}$ — 1 MeV = 10⁶ eV. Always resolve the prefix before applying the eV-to-J conversion. [M1]
2. $E = 1.50 \times 10^{8} \times 1.6 \times 10^{-19}$ — Multiply by e = 1.6 × 10⁻¹⁹ J eV⁻¹. [M1]
3. $1.50 \times 1.6 = 2.4$ — Verify: 1.5 × 1.6 = 1.5 × 1 + 1.5 × 0.6 = 1.5 + 0.9 = 2.4. Correct.
4. $10^{8} \times 10^{-19} = 10^{8+(-19)} = 10^{-11}$ — Add the exponents: 8 + (−19) = −11.
5. $E = 2.4 \times 10^{-11}$ J — Two significant figures. Unit is joules. [A1]

Answer: $E = 2.4 \times 10^{-11}$ J

6. Where the Electron Volt Appears in AQA Physics

Recall from the definition: the eV is built on the relationship $W = QV$. That same relationship underpins several other AQA topics. When you calculate photon energies using $E = hf$, the answer often comes out in joules — and you may need to convert to eV to compare it with an ionisation energy or work function given in eV.

In the photoelectric effect, the work function $\phi$ is the minimum energy needed to release an electron from a metal surface. It is almost always quoted in eV. If you calculate the photon energy $E = hf$ in joules, you must convert $\phi$ to joules before subtracting: $E_k = hf - \phi$. Mixing units in that equation is one of the most common errors in this topic.

Energy levels in atoms are also expressed in eV. The ground state of hydrogen is at $-13.6$ eV. Transitions between levels produce photons whose energies equal the difference between levels — again in eV. Converting those differences to joules lets you calculate the frequency or wavelength of the emitted photon using $E = hf$.

7. Exam Technique and Common Errors

AQA mark schemes for conversion questions award marks for the method (correct formula or conversion factor) and the final answer with correct units. You can lose the accuracy mark through a power-of-ten slip while still earning the method mark — so always show your working clearly.

The electron volt is a compact, practical unit that removes the need to write $10^{-19}$ in every atomic-scale energy calculation — and mastering the two-direction conversion between eV and J is a core skill that underpins the photoelectric effect, atomic energy levels, and particle physics throughout the AQA specification.