Photoelectric effect in a a frequency range I know how to calculate the kinetic energy of the emitted electrons 
$$KE=hf-\phi$$
And the number of the emitted electrons, but this is valid for a certain frequency
$$N=\frac{IA}{hf}=\frac{P}{hf}$$
What happens if I have a frequency range? for example between λ1 and λ2?
BTW: It's not homework, I'm an engineering student,  not a physics student
 A: It depends on the cross-section of radiation reacting with the electrons in the sample. The cross-section depends on the energy of the photon and the atomic number of the target. 
The photoelectric cross-section for a particular geometry and energy is roughly given by:
$$\sigma_E = n_1\dfrac{Z^{n_2}}{E^3} $$
where 
$n_1$ and $n_2$ are fitted constants
Z the atomic number of the target
E the energy of the photon
So to the the total intensity you'd have to integrate the product of the flux of the photons for a particular energy E times the cross-section for that energy. 
All of this would be predicated on the restriction that the photon is of low energy and would be adsorbed near the surface. If the electrons have to escape from within the sample rather than from the surface of the sample, the the electron's cross-section would have to be considered. 
As seen in the diagram below (from Lecture Note on Photon interactions and Cross Sections, H.Hirayama) even a limited diagram the cross-section for photons in matter is complicated. The diagram is for high energy photons which leaves out all the detail of the elements absorption edges due to the binding energy of the electrons which occur at lower energies than shown. 

A: If there is a frequency range of source of light used then you will get a whole electron beam with different speeds accordingly.
