Calculating Max Electric Current in Photoelectric Cell

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SUMMARY

The discussion focuses on calculating the maximum electric current generated by a photoelectric cell when subjected to a specific photon incidence and energy. With 2.5×1015 photons per second, each having an energy of 2.5 eV, and a work function of 2.2 eV, the effective current is determined to be 8×10-5 A, considering a photoelectric conversion efficiency of 20%. The role of the applied potential difference (V) is clarified, indicating that a stopping voltage of 0.3 V is necessary to halt the photocurrent, while a positive external voltage enhances electron emission by overcoming the work function barrier.

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Gabriel Maia
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Hi. The problem is the following:

I have 2.5×10^{15} 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^{14} photons. It means that 5×10^{14} photoelectrons will be taken away from the cell evey second. It is a current of

i=1.6×10^{-19}*5×10^{14}=8×10^{-5} 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.
 
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If your external voltage is negative you can adjust it until the photocurrent stops - this "stopping voltage" is the difference between your photon energy (2.5 eV) and the work function (2.2 eV) ... so here it would be 0.3 V.

There is a more complex effect based upon the accumulation of low-energy electrons just above the emitting surface: it serves to blockade the continued electron emission, effectively increasing the value of the work function. So in a working system you want your external plate to be positive so that it draws the electrons away ... and with the highest field strength that does not arc so that the electrons are swept away more rapidly.

In my ultrafast photo-electron gun there are 1,000 pulses per second, each with ~1,000 to 10,000,000 electrons (adjustable and calibrated). The photon pulses are about 150 femtoseconds in duration, and the extraction voltage is -30,000 V on the cathode (grounded grid anode) placed 6 mm distant; this gives a field strength of 5,000,000 V/m, or 5 V/um. This setup is "fast" enough to avoid any electron buildup, and produces a train of electron pulses each of about 250 fs for the smallest bunches, and 400 fs for 10,000 electrons, etc.
 

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