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I googled it a bit and found that photoelectric current is independent of frequency(of incident light). Some further look revealed that actually "saturation current" is independent of frequency.I could not find about the instantaneous current(current other than saturation current).

Speculation 1: If saturation current is not reached, then radiation with higher frequency will give greater photoelectric current.

Reason 1: Greater frequency means greater velocity of electrons, which will help them to counteract "space charge" and more electrons can reach the anode.


Problem 1:Let's say the intensity (W/m²) of the radiation remains the same, and only the frequency increases. The intensity multiplied by the Area of the plate results in the total energy that arrives the plate in each second. So IA = E/Δt, where E = nhf (n photons of frequency f) Let's say each photon is able to pull out one electron from the plate, so the current i = ne/Δt, where e is the charge of the electron. That gives n/Δt = i/e, and IAe=hfi -> i = IAe/hf, so when we increase the frequency, if the intensity of the radiation remains the same, the current decreases. Is this right?

Are my speculation and reason correct? And please help me to resolve my problem ?

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  • $\begingroup$ Counterpoint to Reason 1: a photovoltaic cell has a non-negligible resistance, and an excess of energy will be re-radiated before charges can reach either terminal, making gains by increased photon energy diminish. $\endgroup$ – KidElephant Jul 31 '14 at 20:02
  • $\begingroup$ @KidElephant Can you please explain yourself a little bit ? $\endgroup$ – user2369284 Jul 31 '14 at 20:05
  • $\begingroup$ For a photon to knock loose a pair within the cell, it must have at least the bandgap energy. Any more energy than this will contribute to the electron's kinetic energy. If the cell were undoped, the pair will bounce around until they lose enough energy and recombine. It is the doping, I would posit, that gives the pair the largest part of their energy, and so more incident photons gives a greater current than fewer, more energetic ones. $\endgroup$ – KidElephant Jul 31 '14 at 20:48
  • $\begingroup$ possible duplicate of Can the photoelectric effect be explained without photons? $\endgroup$ – Ali Aug 1 '14 at 0:36
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    $\begingroup$ Your reasoning looks fine to me, but note that by your logic, if you increase the frequency while keeping the intensity the same, then n decreases inversely proportional to f. Thus, you have less photons and so, by your methods, less current. The way to unify your speculation with your problem is do not keep intensity constant. That said, your physics isn't quite right, I'm just commenting on the logical process you empolyed $\endgroup$ – Jim Aug 1 '14 at 14:09
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Well it may help to first simply consider the famous the photoelectric effect, and see how photon frequency $\nu$ is related to the photoelectron's kinetic energy, once you are sure you understand up to this point, then we can move on to your question on photoelectric current. enter image description here

The picture is a typical diagram used to elaborate on the photoelectric effect, where the photon's energy $h\nu$ has to be larger than the electron's binding energy to its atom usually denoted as work function $E_0$, only then a photoelectron is created. As for its kinetic energy, just make use of the Energy Conservation in photoelectric effect: $$h\nu = E_0 + \frac{\hbar^2 k^2}{2m} $$ where the second term in the right hand side is the kinetic energy, $h$ Planck's constant, $m$ mass of the electron and $k$ its wavenumber (momentum $p=\hbar k$)

Now back to our photoelectric current:

When a photosensitive surface is subject to incident light (x-ray for example), photoelectrons can be ejected from the surface (metallic surface) if the photons' frequency is high enough to reach the necessary work function of the metal, then electrons can be ejected, and now you should already know how to define their kinetic energy. It is important to note again that the maximum kinetic energy depends on the frequency of the photons and not the intensity of the ray.

Next step: if the electrons now reach a collecting plate, a current can be detected. Furthermore if an external retarding potential is placed between the metallic surface and the collecting electrode, the current can be reduced, because at high enough potentials, even the fastest electrons will be prevented from reaching the collector. With the potential $U=qV$ (charge $q$, the voltage $V$), the work-energy theorem is written simply $W = \Delta KE = -\Delta U$. The electron starting from rest, strikes the plate at zero potential relative to its first plate: $\Delta KE = \frac{1}{2}mv^2$ and $-\Delta U = qV$, so an electron failing to reach the plate, must have had a kinetic energy of $KE = eV$, where $eV$ is the work done on charge moving through the retarding potential V. The kinetic energy of the fastest electrons can then be obtained by finding the critical retarding potential necessary to reduce the current flow to zero: $$eV_{crit}=h\nu - W $$

Finally as for the intensity of the ray and current saturation, if the intensity is high enough (very high number of incident photons) then all the electrons get the chance to be ejected (assuming $\nu$ high enough) and contribute to the current, once all possible electrons are ejected from the plate, saturation current can be reached.

This overview should give you the necessary tools to mull over your next questions on your own, but feel free to ask if you face new "understanding" problems.

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Your Speculation 1 is not really true, whereas what you consider in Problem 1 is correct.

The probability of a photocathode to emit an electron upon incident photon is called quantum efficiency. Contrary to your intuition, it is relatively constant for the range of wavelengths to which the photocathode is sensitive and it drops quickly outside of this range, see the picture below (taken from Hamamatsu webpage). Quantum efficiency of various photocathodes

I don't know if this property is true for all other materials - it is possible that people specifically looked for materials with flat photoefficiency spectrum to be used as photocathodes.

When you consider photocurrent produced per unit of light power, indeed, you have to take into account that there are less photons per Watt for shorter wavelengths. See how photo sensitivity look like when plotted on mA/W scale and note black curves of constant quantum efficiency:

Photocurrent vs wavelength (source)

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I think the original poster asked a question specifically excluding the saturation region of the I-V graph as such the photocurrent has not reached the maximum value yet. In other words, he/she is asking about the "linear" region of the I-V graph (i.e. when the applied voltage is between the stopping voltage and the saturation voltage).

I-V Graph

From the I-V graph, at a given applied voltage (between the stopping voltage and the saturation voltage), a higher frequency corresponds to a higher photocurrent.

I think what happens is the space-charge acted as the supplier of electrons rather than the cathode as the supplier. Also, high freq means higher KE so the electrons from the space-charge have higher chance to reach the anode; especially when the applied voltage is negative. Sorry no reference. I tried to Google this for a while but couldn't find anything. This is just what I came up as I have been pondering on this topic.

So speculation 1 is correct, Reason 1 not quite correct as I explained above. The liberation rate=current per charge equation in Problem 1 only holds if all liberated electrons are used as charge carriers for the photocurrent, in other words the photocurrent has reached maximum.

Hope this helps and if my thinking is incorrect, please let me know. Thanks.

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If we increase the frequency..the current will not increase because frequency is the power by which the photon is hitting the electron...the electron will move fast towards the anode but..the number of electrons will not increase therefore the current is not increasing..

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