- #1

Gabriel Maia

- 72

- 1

I have 2.5×10[itex]^{15}[/itex] photons inciding every second on a photoelectric cell. Each photon has 2.5eV of energy and the work function of the cell is 2.2eV. I know that the photoelectric conversion efficiency is 20% and I'm asked to find the maximum electric current through the cell when a potential difference V is aplied to the system.

So... I know that only 20% of the incoming photons will contribute to the current generation. That's 5×10[itex]^{14}[/itex] photons. It means that 5×10[itex]^{14}[/itex] photoelectrons will be taken away from the cell evey second. It is a current of

i=1.6×10[itex]^{-19}[/itex]*5×10[itex]^{14}[/itex]=8×10[itex]^{-5}[/itex] A

What is the role of V here? If the light has energy enough to take the photoelectrons, unless V=0.3eV (or anything >0.3eV) the potential will have no effect on the current, right?

Thank you.