Energy Levels and Photon Emission

Atoms do not absorb or emit energy continuously. Instead, the electrons in an atom can only occupy specific, discrete energy levels. Transitions between these energy levels produce photons of precise wavelengths, giving rise to the characteristic line spectra that allow us to identify elements and understand the structure of matter.

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Discrete Energy Levels

In the early 20th century, experiments revealed that electrons in atoms can only exist at certain specific energies. These allowed energies are called energy levels (or energy states).

The ground state is the lowest energy level — the most stable state of the atom. All higher energy levels are called excited states.

Energy levels are usually represented on an energy level diagram, with the ground state at the bottom and higher levels above. The energies are typically given in electronvolts (eV), and they are negative because energy must be supplied to remove the electron from the atom (the zero of energy is defined as the energy of a free electron at rest, far from the nucleus).

Hydrogen Energy Level Data

Level	Energy (eV)	Energy (× 10⁻¹⁹ J)
n = ∞ (ionisation)	0	0
n = 5	−0.54	−0.864
n = 4	−0.85	−1.36
n = 3	−1.51	−2.42
n = 2	−3.40	−5.44
n = 1 (ground state)	−13.6	−21.8

The levels are not evenly spaced — they get closer together as they approach the ionisation energy. The gap between n = 1 and n = 2 (10.2 eV) is far larger than the gap between n = 4 and n = 5 (0.31 eV).

The pattern follows E_n = −13.6/n² eV for hydrogen.

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Excitation

Excitation occurs when an electron absorbs energy and moves from a lower energy level to a higher one. The electron can be excited by:

1. Photon absorption: The electron absorbs a photon whose energy exactly matches the difference between two energy levels. If the photon's energy does not exactly match any transition, it is not absorbed. This is a key difference from particle collisions.

2. Collisions with other particles: A fast-moving electron (or other particle) can transfer kinetic energy to a bound electron, exciting it to a higher level. Unlike photon excitation, the incoming particle does not need to have exactly the right energy — it just needs at least enough, and any excess remains as kinetic energy of the colliding particle.

The key principle: only specific, quantised amounts of energy can be absorbed (corresponding to transitions between discrete levels).

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De-excitation and Photon Emission

An electron in an excited state is unstable and will quickly (typically within ~10⁻⁸ s) drop back to a lower energy level. When it does, the energy difference is emitted as a photon. The energy of the emitted photon exactly equals the energy difference between the two levels:

$\Delta E = E_{\text{upper}} - E_{\text{lower}} = hf = \frac{hc}{\lambda}$ΔE=Eupper​−Elower​=hf=λhc​

The electron may drop directly to the ground state in a single transition, or it may cascade down through intermediate levels, emitting a photon at each step.

Worked example 1: An electron in hydrogen drops from n = 3 (E = −1.51 eV) to n = 2 (E = −3.40 eV). Calculate the wavelength of the emitted photon.

ΔE = (−1.51) − (−3.40) = 1.89 eV

Convert to joules: ΔE = 1.89 × 1.60 × 10⁻¹⁹ = 3.024 × 10⁻¹⁹ J

λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.024 × 10⁻¹⁹)

λ = 1.989 × 10⁻²⁵ / 3.024 × 10⁻¹⁹ = 6.58 × 10⁻⁷ m = 658 nm

This is red light — the Hα line of hydrogen, easily visible in hydrogen emission spectra.

Worked example 2: An electron in hydrogen drops from n = 4 to n = 1. Calculate the wavelength of the emitted photon.

ΔE = (−0.85) − (−13.6) = 12.75 eV = 12.75 × 1.60 × 10⁻¹⁹ = 2.04 × 10⁻¹⁸ J

λ = hc/ΔE = 1.989 × 10⁻²⁵ / 2.04 × 10⁻¹⁸ = 9.75 × 10⁻⁸ m = 97.5 nm

This is deep ultraviolet — part of the Lyman series. Not visible to the naked eye.

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How Many Transitions from a Given Level?

An electron excited to level n can return to the ground state via multiple possible routes. The total number of possible transitions (and therefore distinct photon wavelengths) when an electron is in level n is:

$\text{Number of transitions} = \frac{n(n-1)}{2}$Number of transitions=2n(n−1)​

For n = 4: transitions = 4 × 3 / 2 = 6

These are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1

Each transition produces a photon of a different wavelength.

Worked example 3: List all possible photon energies when hydrogen is excited to n = 3.

Possible transitions: 3→2, 3→1, 2→1

Transition	ΔE / eV	λ / nm	Region
3 → 2	1.89	658	Red (visible)
3 → 1	12.09	103	UV (Lyman)
2 → 1	10.20	122	UV (Lyman)

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Emission Line Spectra

When a gas at low pressure is excited (by heating or passing an electric discharge through it), the atoms emit light. When this light is passed through a diffraction grating or prism, it produces a line emission spectrum — a series of discrete bright lines on a dark background.

Each bright line corresponds to a specific transition between energy levels. The pattern of lines is unique to each element — a spectral "fingerprint."

Key features:

• Discrete lines (not a continuous spectrum) — because only certain energy transitions are allowed
• Specific wavelengths — each line corresponds to a particular ΔE
• Unique to each element — because each element has a different set of energy levels
• Brighter lines indicate more common transitions (more electrons undergoing that transition)

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Absorption Line Spectra

When white light (containing all wavelengths) passes through a cool gas, atoms in the gas absorb photons that match their energy level transitions. These wavelengths are removed from the spectrum, leaving dark lines at exactly the same positions where emission lines would appear.

An absorption spectrum consists of dark lines on a continuous bright background — the exact inverse of an emission spectrum for the same element.

Application: stellar spectroscopy. The Sun's spectrum shows dark absorption lines (Fraunhofer lines) caused by elements in the Sun's outer atmosphere absorbing specific wavelengths from the continuous spectrum emitted by the hot interior. By identifying these lines, astronomers can determine the chemical composition of stars. Helium was first discovered in the Sun's spectrum (hence its name, from Greek "helios" = sun) before it was found on Earth.

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Ionisation

Ionisation occurs when an electron is given enough energy to completely escape from the atom. The electron is removed to infinity (n = ∞, E = 0).

The ionisation energy of hydrogen from the ground state is 13.6 eV — the energy needed to move the electron from n = 1 (E = −13.6 eV) to n = ∞ (E = 0).

If a photon has energy greater than the ionisation energy, the atom is ionised and the excess energy becomes kinetic energy of the freed electron:

$KE = hf - E_{\text{ionisation}}$KE=hf−Eionisation​

Worked example 4: A photon of energy 15.0 eV strikes a hydrogen atom in the ground state. What happens?

The ionisation energy from n = 1 is 13.6 eV. Since 15.0 eV > 13.6 eV, the atom is ionised.

KE of freed electron = 15.0 − 13.6 = 1.4 eV

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Fluorescent Lamps

A fluorescent tube works by combining excitation, de-excitation, and photon emission:

1. An electric current accelerates electrons through mercury vapour at low pressure
2. The fast-moving electrons collide with mercury atoms, exciting their electrons to higher energy levels
3. The mercury atoms de-excite, emitting UV photons (mainly at 254 nm and 185 nm)
4. The UV photons strike a phosphor coating on the inside of the tube
5. The phosphor atoms absorb the UV photons (excitation) and re-emit the energy as visible light of various wavelengths (de-excitation through multiple smaller energy transitions)

The phosphor converts a single high-energy UV photon into several lower-energy visible photons. Different phosphor mixtures give different colour temperatures (warm white, cool white, daylight).

Common exam mistake: Students often say "the mercury atoms emit visible light." In fact, mercury in a fluorescent tube primarily emits UV light. It is the phosphor coating that converts this UV into the visible light you see.

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LEDs and Energy Levels

Light-emitting diodes (LEDs) also use the principle of energy transitions. In a semiconductor, electrons drop across the band gap (the energy gap between the conduction band and the valence band), emitting photons with energy equal to the band gap. The colour of the LED depends on the band gap:

LED colour	Wavelength / nm	Band gap / eV
Infrared	> 700	< 1.77
Red	620 – 700	1.77 – 2.00
Orange	590 – 620	2.00 – 2.10
Yellow	570 – 590	2.10 – 2.18
Green	495 – 570	2.18 – 2.50
Blue	450 – 495	2.50 – 2.76
White	Broad spectrum	Blue LED + phosphor

Real-world application: The development of efficient blue LEDs (Nobel Prize in Physics 2014, awarded to Akasaki, Amano, and Nakamura) enabled white LEDs, which are now the dominant lighting technology worldwide. A blue LED combined with a yellow phosphor produces white light, achieving far greater efficiency than incandescent bulbs (which waste ~95% of energy as heat).

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A-Level Deep Dive: Energy Levels and Photon Emission

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in its sub-strands on atomic energy levels and line spectra, requires students to recognise that electrons in atoms occupy discrete energy levels; describe photon emission as the result of electron transitions; relate photon energy to the difference between two levels via $\Delta E = hf = hc/\lambda$ΔE=hf=hc/λ; explain how line spectra evidence quantisation; and apply these ideas to fluorescence and gas-discharge sources (refer to the official Pearson Edexcel specification document for exact wording). The ideas interlock with the photoelectric effect (same $E = hf$E=hf), wave-particle duality (de Broglie $\lambda = h/p$λ=h/p) and diffraction-grating analysis (lines dispersed by $d\sin\theta = n\lambda$dsinθ=nλ), and reach forward into Topic 12 (turning points and nuclear) for ionisation and atomic structure. The Edexcel data-and-formulae booklet lists $E = hf$E=hf and $c = f\lambda$c=fλ, but does not tabulate hydrogen energy levels — the relationship $E_n = -13.6/n^2\ \text{eV}$En​=−13.6/n2 eV must be recalled.

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Worked example with full mark scheme

Question (8 marks):

A hydrogen atom has its lowest five energy levels at $E_1 = -13.6\ \text{eV}$E1​=−13.6 eV, $E_2 = -3.40\ \text{eV}$E2​=−3.40 eV, $E_3 = -1.51\ \text{eV}$E3​=−1.51 eV, $E_4 = -0.85\ \text{eV}$E4​=−0.85 eV and $E_5 = -0.54\ \text{eV}$E5​=−0.54 eV. The Planck constant is $h = 6.63 \times 10^{-34}\ \text{J s}$h=6.63×10−34 J s, the speed of light is $c = 3.00 \times 10^8\ \text{m s}^{-1}$c=3.00×108 m s−1, and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$1 eV=1.60×10−19 J.

(a) Calculate the wavelength of the photon emitted when an electron makes the transition from $n = 3$n=3 to $n = 2$n=2 in hydrogen. (4)

(b) State, with a reason, the region of the electromagnetic spectrum in which this photon lies, and identify the named series to which it belongs. (2)

(c) An electron in the $n = 1$n=1 ground state absorbs a photon of energy $12.75\ \text{eV}$12.75 eV. State and justify the highest level it can reach. (2)

Solution with mark scheme:

(a) Step 1 — calculate the energy difference.

$\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \text{eV}$ΔE=E3​−E2​=(−1.51)−(−3.40)=1.89 eV

M1 — magnitude of the difference between upper and lower levels (subtraction in correct order so the photon energy is positive). The common slip is $E_2 - E_3 = -1.89\ \text{eV}$E2​−E3​=−1.89 eV, which loses M1 because emitted photons carry positive energy.

Step 2 — convert to joules.

$\Delta E = 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19}\ \text{J}$ΔE=1.89×1.60×10−19=3.02×10−19 J

M1 — correct eV-to-J conversion. Skipping this and substituting $1.89$1.89 directly into $\lambda = hc/\Delta E$λ=hc/ΔE gives a wavelength in nonsense units.

Step 3 — apply $\Delta E = hc/\lambda$ΔE=hc/λ.

$\lambda = \frac{hc}{\Delta E} = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{3.02 \times 10^{-19}}$λ=ΔEhc​=3.02×10−19(6.63×10−34)×(3.00×108)​

M1 — correct rearrangement and substitution into the Planck-Einstein relation written as $\lambda = hc/\Delta E$λ=hc/ΔE.

Step 4 — evaluate.

$\lambda = \frac{1.989 \times 10^{-25}}{3.02 \times 10^{-19}} = 6.58 \times 10^{-7}\ \text{m} \approx 658\ \text{nm}$λ=3.02×10−191.989×10−25​=6.58×10−7 m≈658 nm

A1 — final answer $\lambda \approx 6.6 \times 10^{-7}\ \text{m}$λ≈6.6×10−7 m or $660\ \text{nm}$660 nm (2–3 sf), with units. A unit error here ("$658\ \text{m}$658 m" or "$658\ \text{nm}^{-1}$658 nm−1") loses the A1 even with the numerical value correct.

(b) Step 1 — identify the region.

A wavelength of $\sim 660\ \text{nm}$∼660 nm falls in the visible region of the EM spectrum, in the red part. B1 — visible (red) accepted.

Step 2 — name the series.

All transitions ending at $n = 2$n=2 in hydrogen form the Balmer series. The $n = 3 \to n = 2$n=3→n=2 line is the H-alpha line. B1 — Balmer series identified, with reasoning that lower level is $n = 2$n=2.

(c) Step 1 — find the level whose energy from the ground state matches.

From $n = 1$n=1, the energy required to reach $n = 4$n=4 is $E_4 - E_1 = -0.85 - (-13.6) = 12.75\ \text{eV}$E4​−E1​=−0.85−(−13.6)=12.75 eV. M1 — selecting the correct upper level by computing each gap from $n = 1$n=1 and matching $12.75\ \text{eV}$12.75 eV.

A1 — the electron reaches $n = 4$n=4. (A photon of exactly this energy is absorbed only if its energy matches a level difference; intermediate energies are not absorbed at all because no level lies between $n = 1$n=1 and the target. This quantisation is the central physical idea.)

Total: 8 marks (M3 A1 B2 M1 A1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A hot tube of low-pressure gas is observed through a diffraction grating, and a series of bright lines on a dark background is recorded.

(a) Explain what the appearance of a line spectrum tells you about the energy levels of the atoms in the gas. (2)

(b) The shortest-wavelength line in the visible part of the spectrum corresponds to a photon of energy $3.03\ \text{eV}$3.03 eV. Calculate this wavelength, in nm. (3)

(c) State and explain how the spectrum would change if the gas were replaced by a different element. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1) — stating that each line corresponds to a transition between two specific energy levels.
• A1 (AO2) — concluding that energy levels are discrete (quantised) because only certain photon energies, and therefore only certain wavelengths, appear.

(b)

• M1 (AO1) — converting $3.03\ \text{eV}$3.03 eV to joules: $3.03 \times 1.60 \times 10^{-19} = 4.85 \times 10^{-19}\ \text{J}$3.03×1.60×10−19=4.85×10−19 J.
• M1 (AO1) — substituting into $\lambda = hc/E$λ=hc/E.
• A1 (AO1) — final answer $\lambda \approx 4.10 \times 10^{-7}\ \text{m} = 410\ \text{nm}$λ≈4.10×10−7 m=410 nm (violet end of visible).

(c)

• B1 (AO2) — different element, different set of energy levels, hence a different set of wavelengths (a different "fingerprint" line spectrum), with the underlying principle named.

Total: 6 marks split AO1 = 4, AO2 = 2. This question is typical of Edexcel's preference for questions that mix a short procedural calculation (AO1) with two short reasoning items (AO2) testing whether the candidate has internalised quantisation rather than just the formula.

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