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EnglandPhysicsSyllabus dot point

How do discrete electron energy levels in an atom explain the emission and absorption of photons of specific wavelengths?

Discrete energy levels in atoms, excitation and de-excitation, ionisation, the relationship between photon energy and energy level difference, line spectra, and the operation of the fluorescent tube.

A focused answer to AQA A-Level Physics 3.2.2.2 and 3.2.2.3, covering discrete energy levels, excitation and ionisation, the photon energy equation linking level differences to wavelength, line spectra, and how the fluorescent tube works.

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  1. What this dot point is asking
  2. Discrete energy levels
  3. Excitation, de-excitation and ionisation
  4. Photon energy and energy levels
  5. Line spectra
  6. The fluorescent tube
  7. Try this

What this dot point is asking

AQA specification points 3.2.2.2 and 3.2.2.3 want you to explain that electrons in atoms occupy discrete energy levels, describe excitation, de-excitation and ionisation, link the energy of an emitted photon to the difference between two levels, explain line spectra, and describe the operation of a fluorescent tube.

Discrete energy levels

The levels are negative because energy must be supplied to free a bound electron; the deeper (more negative) the level, the more tightly the electron is held. An electron cannot have an energy between two allowed levels.

Excitation, de-excitation and ionisation

Electrons can be excited by absorbing a photon of exactly the right energy, or by collision with a free electron, which need only have at least the energy gap (any surplus stays as the colliding electron's kinetic energy).

Photon energy and energy levels

Line spectra

Because only certain transitions are possible, only certain photon energies (and so wavelengths) appear. This produces an emission line spectrum of bright lines on a dark background, unique to each element, acting as a fingerprint that lets astronomers identify the elements in distant stars. An absorption spectrum shows dark lines where those same wavelengths have been absorbed by atoms in the path of white light.

The fluorescent tube

Try this

Q1. An electron falls from a level at −3.4 eV-3.4 \text{ eV} to a level at −13.6 eV-13.6 \text{ eV}. Find the energy of the emitted photon in joules. [2 marks]

  • Cue. Energy gap =10.2 eV= 10.2 \text{ eV}, then multiply by 1.60×10−191.60 \times 10^{-19} to get 1.63×10−18 J1.63 \times 10^{-18} \text{ J}.

Q2. Explain why each element produces a unique line spectrum. [2 marks]

  • Cue. Each element has its own set of discrete energy levels, so only its specific transition energies (wavelengths) appear.

Q3. State what is meant by ionisation of an atom. [1 mark]

  • Cue. An electron gains enough energy to leave the atom completely (reaching the zero level).

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksAn electron in an atom falls from an energy level of −1.5 eV-1.5 \text{ eV} to a level of −3.4 eV-3.4 \text{ eV}. Calculate the wavelength of the emitted photon. Take h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}, c=3.00×108 m s−1c = 3.00 \times 10^8 \text{ m s}^{-1} and 1 eV=1.60×10−19 J1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}.
Show worked answer →

The photon energy equals the difference between the levels: ΔE=(−1.5)−(−3.4)=1.9 eV\Delta E = (-1.5) - (-3.4) = 1.9 \text{ eV}.

Convert to joules: E=1.9×1.60×10−19=3.04×10−19 JE = 1.9 \times 1.60 \times 10^{-19} = 3.04 \times 10^{-19} \text{ J}.

Use E=hcλE = \dfrac{hc}{\lambda}, so λ=hcE=(6.63×10−34)(3.00×108)3.04×10−19=6.5×10−7 m\lambda = \dfrac{hc}{E} = \dfrac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{3.04 \times 10^{-19}} = 6.5 \times 10^{-7} \text{ m} (red light).

Markers reward the energy difference, conversion to joules, and the wavelength from λ=hcE\lambda = \dfrac{hc}{E}.

AQA 20215 marksExplain how a fluorescent tube produces visible light, referring to excitation, de-excitation and the role of the phosphor coating.
Show worked answer →

A high voltage accelerates free electrons along the tube. These collide with mercury atoms and excite their electrons to higher energy levels.

The excited mercury electrons de-excite, falling back to lower levels and emitting photons, mainly in the ultraviolet.

The ultraviolet photons strike the phosphor coating on the inside of the tube, exciting the phosphor's electrons. These de-excite in several smaller steps, emitting lower-energy photons in the visible range, which we see as light.

Markers reward electron collisions exciting mercury, de-excitation emitting ultraviolet, and the phosphor absorbing UV and re-emitting visible light through multiple smaller transitions.

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