Ionisation, Excitation & the Fluorescent Tube Revision Notes | A Level AQA Physics

1. Why Atoms Glow: The Central Mystery

A fluorescent tube contains no glowing filament. There is no burning, no incandescence — and yet it produces bright, steady light. The explanation lives entirely at the atomic scale, in the behaviour of electrons absorbing and releasing energy.

To understand the fluorescent tube, you need two key processes: excitation and ionisation. These are not the same thing, and the distinction between them is precisely what examiners test. By the end of this lesson, you will be able to explain both with the precision the mark scheme demands.

2. Energy Levels: The Staircase Model

Picture the electrons in an atom as being confined to a staircase — they can stand on one step or another, but never between steps. Each step corresponds to a discrete energy level. The lowest step is the ground state, and higher steps are called excited states.

Electrons do not drift between levels spontaneously. They only move when they absorb or release exactly the right amount of energy — the energy difference between two specific levels. This quantised nature of energy is the foundation of everything that follows.

Energy levels are measured in electronvolts (eV) or joules (J). For AQA, you will often see them expressed as negative values — the ground state is the most negative (most tightly bound), and energy increases (becomes less negative) as you move to higher levels. The zero of energy corresponds to the electron being completely free of the atom.

The chart above uses hydrogen's energy levels as a concrete reference. Notice that the gap between the ground state and the first excited state is large — this is why it takes a significant photon energy to excite a hydrogen electron from n=1 to n=2. Higher levels are closer together in energy.

3. Excitation and Ionisation: The Critical Distinction

Here is where the two processes diverge. Both involve an electron gaining energy — but what happens next is completely different, and this difference determines whether you score the mark or lose it.

After excitation, the electron is unstable in its higher level. It will fall back down — either directly to the ground state or via intermediate levels — and release a photon each time it drops. The energy of that photon equals the energy difference between the two levels: $E = hf$, where $h$ is the Planck constant and $f$ is the frequency of the emitted photon.

The key threshold: if the energy supplied is less than the ionisation energy, excitation occurs (the electron moves to a higher bound level). If the energy supplied equals or exceeds the ionisation energy, ionisation occurs (the electron escapes entirely). There is no photon emitted during ionisation itself — the energy goes into freeing the electron, not into radiation.

Commit that table to memory. Exam questions frequently ask you to compare excitation and ionisation, and a vague answer that conflates the two will not score. The word 'removed' is the trigger word for ionisation; 'higher energy level' is the trigger phrase for excitation.

4. The Fluorescent Tube: A Step-by-Step Account

Now that excitation and ionisation are clear, you can trace exactly what happens inside a fluorescent tube. The tube is a sealed glass cylinder containing mercury vapour at low pressure and a small amount of inert gas. The inner surface is coated with a phosphor — a material that absorbs ultraviolet radiation and re-emits visible light.

When the tube is switched on, a high voltage is applied across the electrodes at each end. This accelerates free electrons through the mercury vapour. These fast-moving electrons collide with mercury atoms, and this is where excitation and ionisation both play their roles.

Worked example: How a Fluorescent Tube Produces Visible Light

Trace the sequence of physical processes from switching on the tube to producing visible light.

1. A potential difference is applied across the tube. Free electrons (released from the heated cathode) are accelerated through the mercury vapour. — The electric field does work on the electrons, giving them kinetic energy. The higher the voltage, the greater the kinetic energy of the electrons.
2. Fast electrons collide with mercury atoms. If the colliding electron has kinetic energy greater than or equal to the ionisation energy of mercury, it can remove a bound electron from the mercury atom. — This produces mercury ions (Hg⁺) and releases additional free electrons. These extra electrons are then accelerated by the field, sustaining the discharge. This is why the tube needs a starter — initially there are very few free electrons.
3. Electrons with kinetic energy less than the ionisation energy but sufficient to bridge an energy level gap collide with mercury atoms and excite bound electrons to higher energy levels. — The mercury atom absorbs the energy and its electron moves to an excited state. The colliding electron loses exactly the energy corresponding to the level gap — this is an inelastic collision.
4. The excited mercury electrons are unstable. They fall back to lower energy levels (often the ground state) and emit photons. The dominant emission from mercury is in the ultraviolet region, around 254 nm. — The photon energy equals the energy level difference: $E = hf$. Because mercury's energy level gaps correspond to UV frequencies, the emitted photons are ultraviolet — invisible to the human eye.
5. The UV photons strike the phosphor coating on the inner surface of the tube. The phosphor atoms absorb the UV photons, their electrons are excited, and then fall back down in smaller steps — emitting photons of visible light. — This process is called fluorescence. The phosphor converts high-energy UV into lower-energy visible photons. Different phosphor compositions produce different colours of white light.

Answer: Electrons are accelerated → ionise and excite mercury atoms → excited mercury electrons emit UV photons → phosphor absorbs UV and emits visible light via fluorescence.

Notice that both ionisation and excitation are essential to the tube's operation. Ionisation sustains the discharge by producing the free electrons needed to keep the current flowing. Excitation produces the UV photons that ultimately drive the phosphor to emit visible light.

5. The Photon Energy Equation and Energy Level Transitions

Every time an excited electron drops between energy levels, it releases a photon whose energy is precisely equal to the difference between those two levels. This is not approximate — it is exact, and it is why atomic spectra consist of sharp lines rather than a continuous spread of colours.

The energy of the transition is the difference between the two levels: $\Delta E = E_{\text{upper}} - E_{\text{lower}}$. This $\Delta E$ equals the photon energy $hf$. When energy levels are given in electronvolts (eV), convert to joules before substituting: $1 \text{ eV} = 1.6 \times 10^{-19}$ J.

Worked example: Calculating the Wavelength of an Emitted Photon

Note: the energy values used in this example are illustrative/hypothetical and are provided for calculation practice only — they do not represent mercury's actual measured energy levels. An electron falls from an excited state at $-1.56$ eV to a lower level at $-10.38$ eV. Calculate the wavelength of the photon emitted. ($h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^{8}$ m s$^{-1}$, $1$ eV $= 1.6 \times 10^{-19}$ J)

1. $\Delta E = E_{\text{upper}} - E_{\text{lower}} = -1.56 - (-10.38) = -1.56 + 10.38 = 8.82$ eV — The photon carries away the energy released by the transition. Always subtract the lower (more negative) level from the upper level to get a positive energy difference. [M1]
2. $\Delta E = 8.82 \times 1.6 \times 10^{-19} = 14.112 \times 10^{-19}$ J $= 1.41 \times 10^{-18}$ J — Multiply by the electronvolt conversion factor. Keep the full value for the next step to avoid rounding errors. [M1]
3. $E = \frac{hc}{\lambda}$ so $\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{1.41 \times 10^{-18}}$ — Rearrange $E = hc/\lambda$ to make $\lambda$ the subject. [M1]
4. $6.63 \times 10^{-34} \times 3.00 \times 10^{8} = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$ J m — Multiply $h$ by $c$: $6.63 \times 3.00 = 19.89$, and $10^{-34} \times 10^{8} = 10^{-26}$.
5. $\lambda = \frac{1.989 \times 10^{-25}}{1.41 \times 10^{-18}} = 1.41 \times 10^{-7}$ m $= 141$ nm — $1.989 \div 1.41 \approx 1.41$ and $10^{-25} \div 10^{-18} = 10^{-7}$. This wavelength falls in the ultraviolet region, illustrating how energy level gaps of this magnitude produce UV photons. [A1]

Answer: $\lambda \approx 1.41 \times 10^{-7}$ m (141 nm) — ultraviolet radiation.

6. Ionisation Energy and the Threshold Condition

Ionisation has a strict threshold. An electron colliding with a mercury atom will only ionise it if the colliding electron's kinetic energy is greater than or equal to the ionisation energy of mercury. Below that threshold, the collision can only cause excitation — or if the energy does not match any level gap, the collision may be elastic (no energy transferred to the atom at all).

This threshold behaviour is why the voltage applied to a fluorescent tube matters. At low voltages, electrons do not gain enough kinetic energy to ionise mercury atoms, and the tube will not sustain a discharge. The starter circuit provides an initial high-voltage pulse to ionise enough mercury atoms to get the discharge going; after that, the process is self-sustaining.

There is an important distinction between energy supplied by a photon and energy supplied by a colliding electron. A photon must have exactly the right energy to cause excitation (matching a specific level gap) — it cannot cause partial excitation. A colliding electron, however, can transfer any amount of energy up to its full kinetic energy, making collisional excitation and ionisation more flexible in practice.

7. Exam Technique and Mark-Winning Language

Questions on this topic tend to follow predictable patterns. You will be asked to 'explain' or 'describe' what happens in the fluorescent tube, or to 'compare' excitation and ionisation. Each command word demands a different response structure, and using the wrong one costs marks even if your physics is correct.

For 'explain' questions, always include a cause-and-effect chain: state what happens, then state why using a named physical principle. For 'compare' questions, address both similarities and differences explicitly — a response that only describes one process will not access the comparison marks.

This lesson covers the single specification requirement for this topic: ionisation and excitation, and their roles in the fluorescent tube. Every exam question on this topic is ultimately testing whether you can distinguish the two processes precisely and trace the complete sequence from electron acceleration to visible light emission.