Energy Levels and Photon Emission Revision Notes | A Level AQA Physics

1. The Mystery of Coloured Light

Heat a gas of hydrogen atoms and something remarkable happens: instead of producing a smooth rainbow of colours, the light breaks into a handful of sharp, precise lines. Each line sits at an exact frequency — no more, no less. This is not what classical physics predicted, and for decades it was a genuine puzzle.

The answer lies inside the atom itself. Electrons are not free to orbit at any energy they choose. They are confined to specific, quantised energy levels — and the light you see is the fingerprint of electrons jumping between them.

2. Discrete Energy Levels and Line Spectra

Picture the energy levels of a hydrogen atom as a ladder with unevenly spaced rungs. The lowest rung is the ground state ($n = 1$), and higher rungs are called excited states ($n = 2, 3, 4, \ldots$). Crucially, an electron can only stand on a rung — never between them.

When an electron drops from a higher rung to a lower one, it releases exactly the energy difference between those two levels — and it does so as a single photon. Because the energy levels are fixed, the photon has a fixed frequency. That fixed frequency appears as a sharp line in the emission spectrum.

For atomic hydrogen, the visible lines form the Balmer series — transitions that end on the $n = 2$ level. The red line (H-alpha) arises from the $n = 3 \to n = 2$ transition; the blue-violet lines come from higher levels dropping to $n = 2$. Transitions ending on $n = 1$ produce ultraviolet photons (the Lyman series), and those ending on $n = 3$ produce infrared (the Paschen series).

The key point for your exam: line spectra are direct experimental evidence that energy levels in atoms are discrete. A continuous spectrum would imply electrons could lose any amount of energy — but they cannot. The sharp lines prove otherwise.

3. The Photon Emission Equation

Now that the physical picture is clear, the equation that links it all together is straightforward. When an electron drops from energy level $E_1$ to a lower level $E_2$, the energy of the emitted photon equals the difference between those two levels.

Notice that $E_1 > E_2$ by definition — the electron is falling to a lower energy. The right-hand side is always positive, which means $hf$ is always positive, consistent with a real photon being emitted. If you ever get a negative value, you have subtracted in the wrong order.

You can also combine this with $c = f\lambda$ to find the wavelength of the emitted light: rearrange to $f = c / \lambda$ and substitute into $hf = E_1 - E_2$, giving $hc/\lambda = E_1 - E_2$. This is particularly useful when a question gives you a wavelength and asks for the energy difference.

4. Working with Electronvolts and Joules

Energy levels in atoms are tiny on the joule scale. The ground state of hydrogen sits at about $-2.18 \times 10^{-18}$ J — an awkward number to work with. The electronvolt (eV) was introduced precisely to make atomic energies manageable.

In AQA questions, energy levels may be quoted in either joules or electronvolts. Your job is to check the units before substituting into $hf = E_1 - E_2$. Planck's constant $h = 6.63 \times 10^{-34}$ J s works with joules — so if the levels are given in eV, convert first.

For hydrogen, the energy levels are often quoted as negative values: the ground state is $-13.6$ eV, the first excited state is $-3.4$ eV, the second is $-1.51$ eV, and so on, approaching zero (the ionisation level) as $n$ increases. The negative sign indicates that energy must be supplied to free the electron from the atom.

5. Worked Calculations

Theory is only useful when you can apply it under exam conditions. The following worked examples take you through the two most common question types: finding the frequency of an emitted photon, and finding the wavelength. Work through each step yourself before revealing the answer.

Worked example: Calculating the frequency of an emitted photon

An electron in a hydrogen atom falls from the $n = 3$ level ($E_1 = -1.51$ eV) to the $n = 2$ level ($E_2 = -3.40$ eV). Calculate the frequency of the photon emitted. ($h = 6.63 \times 10^{-34}$ J s, $1$ eV $= 1.6 \times 10^{-19}$ J)

1. $E_1 - E_2 = -1.51 - (-3.40) = 1.89$ eV — Always subtract the lower level from the higher level. Both values are negative, so subtracting a more negative number gives a positive result. [M1]
2. $\Delta E = 1.89 \times 1.6 \times 10^{-19} = 3.024 \times 10^{-19}$ J — Multiply by $1.6 \times 10^{-19}$ to convert eV to J. Planck's constant is in J s, so the units must match. [M1]
3. $f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}}{6.63 \times 10^{-34}}$ — Divide the energy difference by Planck's constant to isolate $f$. [M1]
4. $f = \frac{3.024 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.56 \times 10^{14}$ Hz — This frequency lies in the visible red region of the spectrum — consistent with the H-alpha line of hydrogen. [A1]
5. Visible light: $\approx 4 \times 10^{14}$ to $7 \times 10^{14}$ Hz — $4.56 \times 10^{14}$ Hz sits within the visible range, confirming the answer is physically reasonable.

Answer: $f = 4.56 \times 10^{14}$ Hz

Worked example: Calculating the wavelength of an emitted photon

An electron transitions between two energy levels separated by $4.09 \times 10^{-19}$ J. Calculate the wavelength of the emitted photon. ($h = 6.63 \times 10^{-34}$ J s, $c = 3.0 \times 10^{8}$ m s$^{-1}$)

1. $\frac{hc}{\lambda} = E_1 - E_2$ — Substitute $f = c/\lambda$ into $hf = E_1 - E_2$. This links wavelength directly to the energy difference. [M1]
2. $\lambda = \frac{hc}{E_1 - E_2}$ — Multiply both sides by $\lambda$ and divide both sides by $(E_1 - E_2)$. [M1]
3. $\lambda = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{4.09 \times 10^{-19}}$ — The energy difference is already in joules, so no conversion is needed here.
4. $6.63 \times 10^{-34} \times 3.0 \times 10^{8} = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$ J m — $6.63 \times 3.0 = 19.89$, and $10^{-34} \times 10^{8} = 10^{-26}$.
5. $\lambda = \frac{1.989 \times 10^{-25}}{4.09 \times 10^{-19}} = 4.86 \times 10^{-7}$ m — $1.989 / 4.09 \approx 0.486$, and $10^{-25} / 10^{-19} = 10^{-6}$, giving $0.486 \times 10^{-6} = 4.86 \times 10^{-7}$ m. This is 486 nm — the blue-green Balmer line of hydrogen. [A1]

Answer: $\lambda = 4.86 \times 10^{-7}$ m (486 nm)

6. Reading Energy Level Diagrams

AQA questions frequently present an energy level diagram and ask you to identify which transition produces a given spectral line, or to calculate the frequency for a specified arrow. The diagram shows horizontal lines at fixed energies, with downward arrows representing photon emission and upward arrows representing photon absorption.

A longer arrow on the diagram means a larger energy difference, which means a higher-frequency (shorter-wavelength) photon. The longest possible arrow — from the highest excited state down to the ground state — produces the highest-frequency photon. Transitions between adjacent levels near the top of the diagram produce the lowest-frequency photons.

When a question asks you to identify which transition corresponds to a visible photon, recall that visible light spans roughly 400 nm to 700 nm, corresponding to frequencies of about $4.3 \times 10^{14}$ Hz to $7.5 \times 10^{14}$ Hz. Calculate the energy difference for each candidate transition and check whether $hf$ falls in that range.

7. Exam Technique and Common Mistakes

The calculation questions on this topic are very structured, and marks follow a predictable pattern. Examiners award one method mark for selecting the correct equation, one for correct substitution with units, and one accuracy mark for the final answer. Losing the accuracy mark through a unit error is the most common avoidable mistake at A level.

For extended-response questions asking you to explain how line spectra provide evidence for discrete energy levels, structure your answer in four beats: (1) electrons in atoms occupy discrete energy levels; (2) when an electron drops from a higher to a lower level, a photon is emitted with energy $hf = E_1 - E_2$; (3) because the energy levels are fixed, the photon frequency is fixed; (4) each fixed frequency appears as a sharp line in the spectrum, and the existence of these discrete lines — rather than a continuous spread — confirms that energy levels are quantised.

This lesson has covered all three AQA specification requirements for this topic: line spectra as evidence for discrete energy levels, the equation $hf = E_1 - E_2$ with full worked application, and the conversion between joules and electronvolts. These ideas connect directly back to the photon model introduced earlier in the course — every emitted photon carries exactly one quantum of energy, and that quantum is determined by the atomic structure of the emitting element.