Photoelectric effect: Why does monochromatic radiation produces photoelectron with a spread of velocities? Recall that in photoelectric effect, $V = \frac{hf}{e} - \frac{Wo}{e}$. The incident photon with frequency f produces an electron with energy eV. This should result in a single velocity, why is there a velocity spread?
 A: The photoelectric effect is observed  as follows:

For a given metal, there exists a certain minimum frequency of incident radiation below which no photoelectrons are emitted. This frequency is called the threshold frequency. 
  Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron depends on the frequency of the incident light, but is independent of the intensity of the incident light so long as the latter is not too high.

You ask:

Recall that in photoelectric effect, V = hf/e - Wo/e. The incident photon with frequency f produces an electron with energy eV. This should result in a single frequency, why is there a frequency spread?

You are describing the threshold frequency with the formula, (the frequency describes the incoming photons with energy E=h*nu). Above that frequency electrons will be kicked out with a higher energy.
The effect happens because electrons in metals are in bands all over the lattice, bound loosely with small differences in the energy needed to be freed from the potential bindings. This is the minimum energy needed to kick them out. Any excess energy ( frequency of photons), becomes kinetic energy of the electron.
Thus the minimum frequency needed. It is not a transition between bound energy levels, but a transition from an energy level to free from potential bindings.
