Use energy-level differences to calculate photon energies and classify transitions.
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Method and worked examples
An atom absorbs a photon when its energy matches an allowed upward transition and emits a photon on an allowed downward transition. The photon energy is the positive level difference: hf=Ei− Ef for emission. Bound levels lie below the free-electron zero. Ionisation from a specified level requires enough energy to reach zero; ground-state ionisation is therefore the largest such minimum.
Hydrogen transitions ending at n=1 form the Lyman ultraviolet series, those ending at n=2 form the Balmer series, and those ending at n=3 form the Paschen infrared series. The visible Balmer lines require hydrogen initially at n=2 for absorption. At room temperature almost all hydrogen atoms are in n=1, so ordinary visible Balmer absorption is absent.
Example 1
Levels are−5.00 eV and −2.00 eV. Find the emitted photon energy for the downward transition.
The atom moves from −2.00to−5.00 eV, releasing 3.00 eV. The photon energy is positive.
Example 2
An atom’s ground level is−8.00 eV and the free-electron level is zero. Find ground-state ionisation energy.
Eion =0−(−8.00)=8.00 eV.
Watch out:The zero is a reference; photon energy is a positive difference between levels.
Atomic transitions, hydrogen and ionisation practice | Year 12 Physics | Ascent Arcade