Acceleration in photoelectric effect In all the books I read, only the initial speed with which an electron is removed from a plate is mentioned, what happens to the acceleration?
Consider a plate (charge neutral and not connected to anything) and light of appropriate wavelength hits the plate
Suppose that an electron is removed from the plate, it leaves a hole (a positive charge) in the plate, therefore the removed electron will feel a force of attraction with the plate and therefore there is an acceleration
I am right?
If n electrons are removed, then there are n holes, therefore the acceleration remains constant for n electrons?
 A: Suppose that you have a neutral plate whose potential relative to the "ground" is zero.  
As photo-electrons are emitted from the plate, the potential of the plate increases, and so subsequent photo-electrons are attracted to the plate but if they have sufficient kinetic energy they will escape from the plate.  
However, there will come a time when the potential of the plate, $V_{\rm plate}$, is such that the photo-electrons have a maximum kinetic energy, $K_{\rm max}$, which is insufficient for them to escape from the plate, $eV_{\rm plate} > K_{\rm max}$.  
So after escaping the photo-electrons will be "dragged" back to the plate. 
So the photoelectrons will suffer negative acceleration of magnitude $eE$ where $E$ id the magnitude of the electric field and equal to the potential gradient.
A: After being ejected from the material, the electron is treated as a free particle, therefore acceleration is zero.
The metallic plate is macroscopic, namely it has $10^{23}$ and more electrons. Then, for all purposes, it remains neutral. If you want, you can think that it is connected to ground. Then, the hole which is left by the ejected electron is immediately filled.
