Photon Absorption vs. Photoelectric Effect USually, photons are absorbed by substances only if their energies coincide with the orbital transitions of that particle. However, in the photoelectric effect, it seems that you can take the energy of any photon and calculate the KE of the particle ejected, so long as it covers the work function. There’s no matching of frequency or anything there. Why is that?
 A: Frequency-matching conditions appear when both the initial and the final state have definite energies, such as when you're exciting electronic transitions between different bound states in an atom. For the photoelectric effect the final state is a free electron, which can have any positive energy. A free electron at, say, 1eV, is a perfectly valid state, so the electron can indeed make a transition there.
Note also that this does not require a metal - you get exactly the same behaviour in the single-photon ionization of gas-phase atoms.
A: In the photoelectric effect you impinge photons on a metal which consists of many atoms. In isolation these atoms have well defined discrete level energies. However, when we combine several of these atoms to form a metal, the energy levels interact and start forming a so-called band structure. This structure becomes quasi continuous for large numbers of atoms and the electrons can move freely in this band, and, can therefore absorb the photon's energy. 
Note that band theory is also used to explain the behaviour of conductors, insulators and semiconductors. 
