Atomic Physics · 24.1
Discrete Energy & Spectra
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.
Build the model
Connect the measurement to the mechanism.
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.
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
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
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
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.
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.
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.
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
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.
Worked examples
State the rule, substitute, then check units.
EasyFind the energy of a 500 nm photon of green light, in joules and electronvolts.
- E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (500 × 10⁻⁹).
- E = 3.98 × 10⁻¹⁹ J.
- In eV: 3.98 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ ≈ 2.5 eV — atomic gaps are a few eV, which is why atoms trade visible light.
Answer≈ 4.0 × 10⁻¹⁹ J ≈ 2.5 eV
MediumA hydrogen electron drops from n = 3 (−1.51 eV) to n = 2 (−3.40 eV). Find the photon’s energy, frequency, and wavelength.
- ΔE = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV = 3.02 × 10⁻¹⁹ J.
- f = E/h = 3.02 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 4.56 × 10¹⁴ Hz.
- λ = c/f = 3.00 × 10⁸ ÷ 4.56 × 10¹⁴ ≈ 6.6 × 10⁻⁷ m — the red Balmer line.
Answer1.89 eV; 4.56 × 10¹⁴ Hz; ≈ 660 nm (red)
HardGlowing hydrogen shows a blue 434 nm line. Identify the transition, given levels −13.6, −3.40, −1.51, −0.850, −0.544 eV for n = 1…5.
- E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (434 × 10⁻⁹) = 4.58 × 10⁻¹⁹ J = 2.86 eV.
- 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.
- λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 2.18 × 10⁻¹⁸ ≈ 9.1 × 10⁻⁸ m = 91 nm, deep ultraviolet.
- 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