22.4 Energy levels in atoms and line spectra
- Syllabus
- 9702–2028–2029
- Topic
- 22.4
- Level
- A2
An isolated atom allows electrons only in particular energy states, not every energy between them.
An electron changes level by absorbing or emitting a photon whose energy equals the gap: ΔE=hf. Larger gaps correspond to higher-frequency, shorter-wavelength lines.
A hydrogen emission spectrum contains separate lines because only certain transitions are allowed between its levels.
The electron does not radiate continuously while remaining in a stationary level; emission occurs during a transition between allowed states.
Emission spectrum: excited atoms in a hot, low-pressure gas make downward electron transitions and emit photons. Against a dark background, only the allowed photon wavelengths appear as separate bright lines.
Absorption spectrum: continuous light passes through a cooler, low-pressure gas. Electrons absorb only photons whose energies match upward level gaps, so those wavelengths appear as dark lines in the transmitted continuous spectrum.
Excited electrons later fall and re-emit photons, but in random directions. Along the original viewing direction, fewer photons remain at the absorbed wavelengths, so the dark lines persist.
For the same element, absorption and emission lines occur at the same characteristic wavelengths because both use the same energy-level gaps.
Each line has one frequency f and photon energy hf. Discrete line frequencies therefore imply discrete photon energies, discrete energy gaps and hence discrete atomic electron levels.
A line spectrum is not a continuous rainbow with uneven brightness: only specific transitions are possible, and absorption does not permanently destroy the photon energy.
Ephoton=hf=hc/λ=∣Eupper−Elower∣
Downward transition: one photon is emitted. Upward transition: one matching photon is absorbed. In both cases photon energy is the positive magnitude of the level difference.
For hydrogen E3=−1.51 eV and E2=−3.40 eV, the n=3→2 gap is (−1.51)−(−3.40)=1.89 eV=3.02×10⁻¹⁹ J.
Then f=ΔE/h=(3.02×10⁻¹⁹)/(6.63×10⁻³⁴)=4.56×10¹⁴ Hz and λ=c/f=6.58×10⁻⁷ m=658 nm.
To find an unknown upper level from an emitted photon, use Eupper=Elower+hf. Keep signed level energies until after the subtraction.
Do not use the energy of either level alone as the photon energy. Convert 1 eV=1.60×10⁻¹⁹ J before using h in J s, and choose the transition direction from the wording or arrow.