Photoelectric effect and kinetic energy According to my textbook, electrons which are emitted due to the photoelectric effect have kinetic energy given by:
$$KE_{electron}=h\nu-h\nu_0$$
Where $\nu_0$ is the activation frequency of the metal. 

My question is: when electrons are emitted in this way, is all of the photon's energy really converted to kinetic energy in the electron? Is there no loss of energy in this process? 

If there is energy lost (which I suspect must be true), is it a sufficiently small amount so that we can say it is negligible for most purposes?
$h\,\nu$ is photon's energy; $h\,\nu _0$ is the potential energy of the electron. All energy excess is converted in kinetic energy of the eletron - that's the left side of the equation.