Atomic line spectra and energy levelsEdexcel International A Level Physics: Revision notes

Section 1

Discrete energy levels

Electrons in an atom can only have certain discrete energies, called energy levels. They are shown as negative values (for example −13.6 eV for hydrogen) because energy must be supplied to free an electron; a free electron at rest has energy 0.

The lowest level is the ground state.
Higher levels are excited states.
An electron cannot have an energy between the levels.

Key termsenergy levelground stateexcited state

Section 2

Emission line spectra

When an electron falls from a higher level $E_2$ to a lower level $E_1$ , the atom emits a photon:

$hf = E_2 - E_1$

Only certain energy differences exist, so only certain frequencies are emitted. Seen through a diffraction grating or prism, they form separate bright lines on a dark background: a line emission spectrum. Each element has its own pattern, because its energy levels are different.

Electrons are excited by collisions in a discharge tube, or by absorbing energy.

Key termsline emission spectrum

Section 3

Absorption line spectra

If light with a continuous spectrum passes through a cool gas, an atom can absorb a photon only if its energy exactly equals the difference between two levels. These frequencies are removed, and dark lines appear in the spectrum at the same frequencies as the element's emission lines. The absorbed energy is re-emitted in random directions, so the transmitted intensity is reduced.

Key termsabsorption spectrum

Exam tip

An atom absorbs a photon only when hf exactly matches an energy gap. A photon with slightly more or less energy is not absorbed.

Section 4

Calculating frequency and wavelength

Find the energy difference $\Delta E$ between the levels, in eV.
Convert to joules: multiply by $1.60 \times 10^{-19}$ .
Use $f = \Delta E / h$ , and then $\lambda = c/f$ if needed.

Worked example. Hydrogen n = 2 to n = 1: ΔE = 13.6 − 3.40 = 10.2 eV = 1.63 × 10⁻¹⁸ J, so f = 1.63 × 10⁻¹⁸ / 6.63 × 10⁻³⁴ = 2.5 × 10¹⁵ Hz.

Hydrogen n = 3 to n = 2: ΔE = 1.89 eV = 3.02 × 10⁻¹⁹ J, so f = 4.6 × 10¹⁴ Hz and λ = 6.6 × 10⁻⁷ m, which is visible (red).

Common mistake

Subtract the level energies; do not add them. Using the sign correctly, the gap from −3.40 eV to −1.51 eV is 1.89 eV.

Section 5

Counting possible lines

If an atom is excited to a level with several levels beneath it, it can return in more than one way. From n = 3 there are three transitions (3 to 2, 2 to 1, 3 to 1), giving three lines. Larger energy gaps give higher frequencies, so the line from n = 3 to n = 1 has the highest frequency.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Atomic line spectra and energy levels

A gas discharge tube containing hydrogen at low pressure emits light when a high voltage is applied across it. When the light is viewed through a diffraction grating, a small number of separate coloured lines is seen against a dark background.
Explain why dark lines appear in an absorption spectrum at the same frequencies as the bright lines in the emission spectrum of the same gas.2 marks
The lowest three energy levels of a hydrogen atom are n = 1 at −13.6 eV, n = 2 at −3.40 eV and n = 3 at −1.51 eV. Use the Planck constant h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.60 × 10⁻¹⁹ J.
Calculate the frequency of the photon emitted when an electron moves from n = 2 to n = 1.2 marks
An atom of an element has energy levels of −5.4 eV (the ground state), −3.1 eV and −1.5 eV. A sample of the element is excited so that some of its atoms are in the −1.5 eV level. Use the Planck constant h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.60 × 10⁻¹⁹ J.
Calculate the frequency of the photon emitted when an atom moves from the −1.5 eV level to the ground state.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).