Atomic line spectra and energy levelsEdexcel A-Level Physics: Revision notes

The electrons in an atom can only have certain, fixed energies. These are the atom's energy levels. Energies are quoted as negative numbers because energy must be supplied to remove an electron: an electron that is just free of the atom has zero energy, so electrons bound to the atom have energy below zero.

The lowest level is the ground state. A level above the ground state is an excited state. An atom can change level only by gaining or losing exactly the right amount of energy, because there are no allowed energies in between.

For hydrogen the levels (n = 1, 2, 3, 4) are −13.6 eV, −3.40 eV, −1.51 eV and −0.85 eV, so the levels get closer together as n increases.

An atom can be excited by collisions or by absorbing energy. When an electron falls from a higher level $E_1$ to a lower level $E_2$, the atom emits a photon whose energy is the difference between the levels:

$hf = E_1 - E_2$

Because only certain energy differences are possible, the emitted radiation has only certain frequencies. Seen through a spectrometer, it appears as a set of bright lines on a dark background: an emission line spectrum. Each element has its own set of energy levels, so each has a characteristic pattern of lines.

A continuous range of frequencies would require continuous energy levels, so line spectra are direct evidence that levels are discrete.

When light with a continuous range of frequencies passes through a cool gas, an atom absorbs a photon only if the photon energy exactly equals the difference between two of its energy levels. The electron is excited to the higher level, and the atom soon re-emits a photon, but in a random direction.

So the frequencies that match transitions are largely missing from the transmitted beam, giving dark lines in the spectrum: an absorption line spectrum. The dark lines are at the same frequencies as the bright lines in the emission spectrum of the same element.

A photon whose energy matches no energy gap passes through, because an atom cannot absorb just part of a photon's energy.

To find the frequency for a transition:

1. Find the energy difference in eV, $\Delta E = E_1 - E_2$ (always positive).
2. Convert to joules: multiply by $1.60 \times 10^{-19}$.
3. Use $f = \Delta E / h$, with $h = 6.63 \times 10^{-34}$ J s. The wavelength follows from $\lambda = c/f$.

Worked example. An electron falls from −1.51 eV to −3.40 eV in hydrogen. $\Delta E = 1.89$ eV $= 3.02 \times 10^{-19}$ J, so $f = 3.02 \times 10^{-19} \div 6.63 \times 10^{-34} = 4.56 \times 10^{14}$ Hz, which is $\lambda = 658$ nm (red light).