Fluorescence and phosphorescence Fluorescence is where UV light is absorbed then emits visible light right? Is it that you can only see it in the dark whereas phosphorescence is where UV light is absorbed and visible light is released but more slowly. With phosphorescence, is it that you can see it in the dark as well as in normal light? I'm confused about whether you see these things in the dark or light?
 A: You're having trouble telling the difference between the two because, as for many natural language words, I don't think there is a clearly defined difference and the difference arises from natural usage (as with the difference between "fruits" and "vegetables" in English). The following is my understanding of the difference.


*

*There is a well defined difference between the collective phenomena of fluorescence / phosphorescence on the one hand and general light-matter interactions on the other and that is that fluorescence and phosphorescence are inelastic and have a positive Stokes frequency shift; that is, each event transfers energy to the interacting lattice and only some of the incident light's energy is returned to the electromagnetic field as longer wavelength light;

*The main difference I perceive from the usage of the two words is that in fluorescence, the light producing decay from the upper metastable state to the ground band is by way of allowed transitions, whereas phosphorescence happens by way of the oxymoronically named forbidden transitions (sometimes there are many low energy states the decay happens to, with further nonradiative transitions happenning after the fluorescence / phosphorescence event). More on these weird words below, but what this means practically is that, generally, that fluorescence lifetimes are of the order of nanoseconds, whereas phosphorescence lifetimes are orders of magnitude longer - anything from microseconds to hours.


See my answer here to the Physics SE question "How long does an electron stay on a given orbital?" for more details on the difference between "allowed" and the oxymoronically named "forbidden" transitions. In summary, a simple, Fermi-golden-rule-type calculation of the transition lifetime shows it to be proportional to the reciprocal of the entity:
$$\tau_{1\to2} \propto \frac{1}{ \left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2}$$
and, by symmetry, the transition endpoints $\psi_1,\,\psi_2$ must be of opposite parity for $\left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2$ to be nonzero. The simple calculation therefore predicts that transition between like-parity states is impossible, i.e. with infinite lifetime, hence the name "forbidden transitions". What's happenning in practice is that (1) the calculation is only a good approximation, in particular (2) the transition endpoints $\psi_1,\,\psi_2$ aren't truly eigenstates of the fluorophore - electromagnetic field system: they are calculated by the quantum mechanics of the fluorophore in isolation. When the system is coupled to the electromagnetic field, the strict parity symmetry is broken, hence the possibility of the "forbidden" transition. Notwithstanding the self-contradictory nature of the words, the simple calculation does explain the vast lifetime difference between the two phenomena.
A: Fluorescence can occur at any wavelength, it is the transition between states of the same spin, e.g. most always excited singlet ($S_1$) to ground state singlet ($S_0$) which both have spin = 0 (and spin multiplicity 1). This can occur over a wide range of wavelengths; bacteriochlorophyll emits in the near infra red at approx 850 to 900 nm as it absorbs at a slightly shorter wavelength. In contrast benzene absorbs and emits in the ultraviolet, 200-300 nm and ethylene at even shorter wavelengths. Not many molecules have excited state lower in energy than about 10000 cm$^{-1}, or 1000 nm, so fluorescence has to be at shorter wavelengths or higher energy. 
Phosphorescence is the transition from an excited triplet state (spin = 1, spin multiplicity =3) to a lower energy singlet state, almost invariably a singlet ground state. 
Luminescence is the general name when the states are not singlet of triplets, quartet, doublets etc and usually occurs when transition metals are involved in organo-metallic compounds.
Fluorescence can occur between two triplet states, although I have only ever come across one example, so it is very rare. 
In electric dipole transitions (absorption, fluorescence, phosphorescence, luminescence) the photon has effectively no influence on the spin state. Singlet transitions are spin allowed, but can be symmetry forbidden. The angular momentum of the photon has to be conserved in a transition, which means that the difference in the initial and final states corresponds to a unit of angular momentum. Whether the transition is symmetry allowed or forbidden is usually determined by using the point group of the molecule. 
In singlet-triplet or triplet-singlet transitions there has to be an  angular momentum operator to make the forbidden transition occur, albeit only weakly. In T-S transitions this operator is spin-orbit coupling, the coupling of electron with orbital angular momentum.
