What is the purpose of a 2 stage photon emission in Fluorescent lights? In fluorescent light tubes, mercury atoms are first ionised causing the release of UV photons and then the UV photons are absorbed by a phosphor coating which excites the electrons and causes the release of visible light photons. 
Surely one of these 2 stages could be cut? i.e. using an element other than mercury to directly emit visible light photons or using electrons to directly excite the electrons in phosphor
 A: Gaseous emission spectra have well defined frequencies and produce coloured light, when what people want is white light. Sodium is an excellent example of this. Sodium lights are a very efficient way of producing light if what you want is yellow light.
To get white light we take advantage of a phenomenon called fluorescence. In solids narrow bandwidth light can be absorbed then re-emitted with a much broader spectrum. This is because when a photon is absorbed a variable amount of its energy is lost as heat in the solid, and the re-emitted photons have a broad spread of (lower) energies.
And this is exactly what happens in fluorescent lights and indeed the new LED bulbs. Narrow wavelength uv light (blue light in LED bulbs) excites fluorescence over a broad spread of visible wavelengths. A mixture of phosphors is carefully chosen to produce as close to white light as possible.
As for using electrons directly: generating intense electron beams is hard while generating lots of electrons using an avalanche is very easy. Any light based on directly irradiating a phosphor with electrons would be a lot more complicated and expensive than just letting the electron avalanche excite the gas molecules. It can be done of course - cathode ray tubes, as in old style TVs, generate light this way - but it's hard to see how this would make a good replacement for fluorescent lights.
