Energy levels and line spectraAQA A-Level Physics: Revision notes

Electrons in an atom can have only certain, discrete energy levels. The energies are negative, because energy must be supplied to remove an electron: $E = 0$ means the electron is free from the atom.

A line spectrum is a spectrum of separate bright lines at particular wavelengths, seen against a dark background when light from a gas discharge is split by a prism or grating. A continuous spectrum contains all wavelengths.

Line spectra are evidence for discrete energy levels: if electrons could have any energy, the spectrum would be continuous.

In a discharge lamp, collisions excite atoms to higher levels. When an electron falls from a higher level $E_1$ to a lower level $E_2$ it emits one photon:

$hf = E_1 - E_2$

Since $f = c/\lambda$, the wavelength is $\lambda = hc/(E_1 - E_2)$. A bigger energy difference gives a higher frequency and a shorter wavelength.

Each line in the spectrum matches one transition. Different gases have different levels, so each has its own pattern of lines.

In questions, energy levels may be quoted in joules or electronvolts. Convert to joules (× 1.60 × 10⁻¹⁹) before using $h$ and $c$.

For hydrogen the levels are $-13.6$ eV ($n = 1$), $-3.40$ eV ($n = 2$), $-1.51$ eV ($n = 3$) and $-0.85$ eV ($n = 4$).

The energy needed to move an electron from $n = 1$ to the free state is the ionisation energy, 13.6 eV for hydrogen.

An electron in hydrogen falls from $n = 3$ to $n = 2$.

• Energy difference = $-1.51 - (-3.40) = 1.89$ eV
• In joules: $1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19}$ J
• Frequency: $f = E/h = 4.56 \times 10^{14}$ Hz
• Wavelength: $\lambda = hc/E = 6.6 \times 10^{-7}$ m, which is red light

The transition $n = 2$ to $n = 1$ has an energy of 10.2 eV, which is ultraviolet.