Mercury vapour will glow if placed in a tube in which electrons are accelerated down it by applying an accelerating potential difference across the tube.

This happens because the mercury atoms are excited to higher energy states by colliding electrons.

When mercury de-excites it emits photons giving rise to the glow in the discharge tube.

Mercury will absorb energy at 4.87 eV when a ground state electron rises to the first available excited state.

This energy is absorbed from the kinetic energy of the colliding accelerated electron.

The accelerated electron will slow down as its kinetic has been reduced.

If a colliding electron with kinetic energy 7.20 eV impacts a mercury atom it will leave with

$$ KE_{after} = KE_{before} - 4.87 \text{eV} $$

$$ KE_{after} = 2.33 \text{eV} $$

The experiment Franck and Hertz conducted passd current through a mercury vapour at high temperature and pressure.

The drops in current indicated the accelerated electrons had lost energy as they passed through the gas.

This experiment carried out in 1914 is credited as the first direct evidence for the quantisation of atoms energy states.

Notice the regular repeating pattern in voltage. Why does the pattern repeat?

The mercury atoms de-excite quickly after the collision.

To get back down to the ground state the atoms release a photon equal to the energy between the ernegy levels.

$$ hf = E_2 - E_1 $$

$$ hf = 4.87 \text{ eV} $$

$$ f = \frac{4.87 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} $$

$$ f = 1.81 \times 10^{15} \text{ Hz} $$

$$ \lambda = \frac{c}{f} $$

$$ \lambda = 254 \text{ nm} $$

This is an ultra violet photon.

In a mercury vapour lamp the outside of the lamp is coated in a phosphor powder. The powder will fluoresce in the visible if ultra violet is incident on it.

This allows the ultra violet photons to be converted into visible light so that the lamp can be a source of white light.