Question about the photoelectric effect

[Dima Petrukhin]

1. The question asks why the photodiode is less than 100% efficient.2. hf=work function+KEmax


3.- I reckon this is as some light will not cause e to be emitted if the wavelength of the light is less the needed.
-Or some photons will not hit any electron.

Is this correct?
Thinking about this question got me wondering if the work function may vary in a metal depending on how 'deep' the electron is in the metal?

And is it possible for a single photon to go through the metal without hitting an electron and not causing emission?

Thanks

 

[Dima Petrukhin]

I should say how close the electrons are to +ve metal ions

 

[kuruman]

You say "photodiode" but you are describing the photoelectric effect. In a photodiode incoming photons create electron-hole pairs which are then detected as a current. This is not described by the equation hf=work function+KEmax.

And is it possible for a single photon to go through the metal without hitting an electron and not causing emission?

A single photon that does not cause emission does not have to go "through" the material. The photon's energy may be absorbed by the lattice and be dissipated as heat. Read about quantum efficiency here https://en.wikipedia.org/wiki/Quantum_efficiency.