Pass a current through hydrogen gas and it glows — but not with every colour. A spectroscope splits the glow into a handful of razor-sharp lines, always the same ones. Those lines are the atom’s energy ledger, printed in light.
Electrons in an atom cannot hold just any energy: they sit on a ladder of discrete energy levels. An atom absorbs a photon only if the photon’s energy E = hf exactly matches a gap between two levels, lifting the electron up; when the electron falls back it emits a photon carrying exactly the gap, E = E₂ − E₁.
Because every element has its own ladder, every element has its own set of spectral lines — a fingerprint. Hydrogen’s ladder, Eₙ = −13.6/n² eV, produces the Lyman (ultraviolet), Balmer (visible), and Paschen (infrared) series, and the very existence of line spectra rather than a continuous rainbow is the direct evidence that atomic energy is discrete.
Simple definition
Atomic electrons occupy discrete energy levels; a photon is emitted or absorbed only when an electron jumps between two levels, with photon energy E = hf equal to the exact energy difference.
Example
Sodium street lamps glow that particular orange because sodium’s ladder has a gap of 2.1 eV — a 589 nm photon — and essentially nothing else in the visible range.
Photon energyE = hf = hc/λ
Light delivers its energy in lumps called photons. High frequency (short wavelength) means a big lump — an ultraviolet photon carries more energy than a red one.
h = 6.63 × 10⁻³⁴ J s
Transition energyhf = E₂ − E₁
The photon pays for the jump exactly. No change, no credit: a photon with almost the right energy is simply not absorbed at all.
Emission going down, absorption going up
Hydrogen’s ladderEₙ = −13.6/n² eV
The levels are negative because the electron is bound — you must add energy to free it. The rungs crowd together as n grows, which is why each series has a short-wavelength limit.
n = 1 is the ground state; E∞ = 0
01
Emission and absorption fingerprints
A hot low-pressure gas emits its signature wavelengths only: bright lines on black. Put a cool gas in front of a continuous white source instead and it steals exactly those same wavelengths: black lines on a rainbow. Same ladder, same fingerprint — helium was discovered in the Sun’s absorption spectrum before anyone found it on Earth.
02
Reading the hydrogen series
Every drop that lands on n = 1 emits ultraviolet (Lyman series). Drops landing on n = 2 emit visible light (Balmer — the red 656 nm line is n = 3 → 2). Drops landing on n = 3 are infrared (Paschen). The bigger the fall, the bigger the photon energy and the shorter the wavelength.
03
Excite, then de-excite
Collisions or photons kick the electron up to an excited state; it typically stays there only nanoseconds before tumbling back down, sometimes in several small hops — each hop its own photon. Give the atom 13.6 eV or more and the electron leaves entirely: the atom is ionised.
02
Change one variable at a time
Make the relationship visible.
Hydrogen’s ladder Eₙ = −13.6/n² eV. Drop the electron and a photon leaves with exactly the gap. Land on n = 1 for ultraviolet (Lyman), n = 2 for visible (Balmer), n = 3 for infrared (Paschen).
Photon energy E₂ − E₁1.89 eV = 3.02e-19 J
Frequency f = E/h4.56e+14 Hz
Wavelength λ = hc/E658 nm
SeriesBalmer — visible
03
Catch the common trap
Explain before calculating.
Which observation is direct evidence that atomic energy levels are discrete?
Choose an answer to test the model.
04
Worked examples
State the rule, substitute, then check units.
EasyFind the energy of a 500 nm photon of green light, in joules and electronvolts.
Visible light means the Balmer series — the electron lands on n = 2.
E₅ − E₂ = −0.544 − (−3.40) = 2.86 eV ✓. The jump is n = 5 → n = 2.
Answern = 5 → n = 2, a Balmer transition
ChallengingWhat wavelength of light can just ionise hydrogen from its ground state, and why does each spectral series pile up against a short-wavelength limit?
Ionisation from n = 1 needs 13.6 eV = 2.18 × 10⁻¹⁸ J.
The rungs Eₙ = −13.6/n² crowd toward E = 0, so transitions into a given level approach a maximum energy — the series limit. Beyond it the spectrum turns continuous, because a freed electron can carry any surplus as kinetic energy.
Answer≈ 91 nm; the limit marks the jump from bound (discrete) to free (continuous) energies
Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.
IGCSE
01Fig. 7.1The nuclear model of the atomIGCSEOpen diagramClose diagram
Figure comment
Fig. 7.1The scattering apparatus, drawn inside an evacuated container. A source of α-particles sits in a lead block with a narrow channel cut through it, so a fine beam travels horizontally to a very thin vertical sheet of gold foil. Three paths are drawn from the point where the beam meets the foil: one carrying straight on in the original direction and labelled as the path of most of the particles, one deflected upwards through a moderate angle, and one turned back towards the source side of the foil through more than 90°.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The three lines are the paths of three different α-particles, not one particle bouncing about; each angle is measured from the original beam direction set by the channel in the lead.
aState State the purpose of the narrow channel cut through the lead block in Fig. 7.1.
recall2 marks
Check answer · 2 marks
the lead absorbs alpha particles emitted in all other directions (1)
so a narrow beam of known direction reaches the foil, against which deflection angles can be measured (1)
bDescribe Describe the force that acted on the α-particle following the middle path in Fig. 7.1, the one deflected upwards through a moderate angle, and describe where in the gold atom that force acted.
routine3 marks
Check answer · 3 marks
an electrostatic force of repulsion (1)
between the positively charged α-particle and the positively charged nucleus (1)
acting as the particle passed close to a nucleus, but not directly at it (1)
cExplain Explain how Fig. 7.1 would have to be redrawn if the positive charge of each gold atom were spread evenly throughout the whole atom, as in the earlier model of the atom.
demanding4 marks
Check answer · 4 marks
only the straight-through path, with at most very small deflections, would be drawn (1)
no path turned back through more than 90° would appear (1)
charge spread through the whole atom produces a much weaker repulsion at any point in it (1)
this force is far too small to reverse the motion of a fast, massive α-particle (1)
dSuggest The gold foil in Fig. 7.1 is replaced by an aluminium foil of the same thickness. An aluminium nucleus holds 13 protons and a gold nucleus holds 79. Suggest how the three paths drawn would change.
top of the paper4 marks
Check answer · 4 marks
most α-particles would still pass straight through, so that path is unchanged (1)
the aluminium nucleus carries a much smaller positive charge than the gold nucleus (1)
so the repulsive force at the same distance of approach is much smaller (1)
fewer particles are deflected through large angles, and far fewer are turned back through more than 90° (1)
Transfer challenge
An α-particle is fired straight at a gold nucleus, along a line through its centre. Explain, in terms of energy, what happens as it approaches, and explain how its closest approach would differ if it were fired faster.
Check answer · 5 marks
as it approaches, the repulsion does work against its motion and its kinetic energy falls (1)
the kinetic energy is transferred to electrostatic potential energy in the field of the nucleus (1)
it stops momentarily at its closest approach, then is pushed back along its original path (1)
a faster particle starts with more kinetic energy (1)
so it travels closer to the nucleus before stopping (1)
IB
02Fig. 1.1Structure of the atomIBOpen diagramClose diagram
Figure comment
Fig. 1.1An energy level diagram for a hydrogen atom, drawn not to scale, with energy increasing up the page. Five horizontal levels are shown: the ground state n = 1 at −13.6 eV, then n = 2 at −3.40 eV, n = 3 at −1.51 eV, n = 4 at −0.85 eV, and n = ∞ at 0 eV. A vertical arrow drawn between the −1.51 eV level and the −3.40 eV level points downwards, marking the electron transition the question describes.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The arrow length means nothing here, the diagram is not to scale: take every energy from the printed labels, and note every level is negative because 0 eV is the just-free electron.
aState State the energy that must be supplied to a hydrogen atom to remove its electron completely when that electron occupies the level marked n = 2, and state why every level below n = ∞ carries a negative value.
recall2 marks
Check answer · 2 marks
3.40 eV, the difference between the −3.40 eV level and the 0 eV level marked n = ∞
Zero is defined as the electron free of the atom and at rest, so a bound electron has less energy than this and its value is negative
bCalculate Calculate the wavelength of the photon emitted when the electron falls from the level marked n = 4 to the level marked n = 2. Use hc = 1240 eV nm.
routine3 marks
Check answer · 3 marks
ΔE = −0.85 − (−3.40) = 2.55 eV
λ = 1240/2.55
λ = 486 nm
cDetermine An electron begins in the level marked n = 4. Using only the levels drawn, determine how many different photon wavelengths could be emitted as it returns to the ground state, and determine the shortest of those wavelengths.
demanding4 marks
Check answer · 4 marks
Every downward transition between the four bound levels is available: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1
Six different wavelengths
Shortest wavelength comes from the largest energy drop, 4→1: ΔE = −0.85 − (−13.6) = 12.75 eV
λ = 1240/12.75 = 97.3 nm
dExplain Hydrogen atoms in the ground state are bombarded first with free electrons of kinetic energy 12.5 eV, and then with photons of energy 12.5 eV. Explain, using the levels drawn, why the electrons excite the atoms but the photons do not.
top of the paper4 marks
Check answer · 4 marks
Excitation from the ground state requires exactly 10.20 eV, 12.09 eV or 12.75 eV to reach n = 2, n = 3 or n = 4
A photon is absorbed whole or not at all, and 12.5 eV matches none of these gaps, so the photons pass through unabsorbed
A free electron transfers energy by collision and need not give up all of it, so it can supply exactly 12.09 eV and raise the atom to n = 3
The bombarding electron then moves on with the remaining 12.5 − 12.09 = 0.41 eV of kinetic energy
Transfer challenge
A sodium street lamp emits strongly at 589 nm. Determine the energy gap in the sodium atom responsible for this emission, in eV, and explain why cool sodium vapour placed in front of a white-light source produces a dark line at exactly the same wavelength.
Check answer · 4 marks
E = 1240/589
E = 2.11 eV
Atoms in the vapour absorb only photons whose energy matches one of their own level differences, so 589 nm photons are removed from the beam
The absorbed energy is re-emitted in all directions, and often as a cascade at other wavelengths, so the transmitted beam is depleted at 589 nm and a dark line appears