Calculating Photoelectric Current in Amps

[Salvador]

Hi, how could i calculate the current I would get from the photoelectric effect, so that the end result would be in amps?
If I have a certain lightsource or source of powerful enough em radiation to conduct the photoelectric effect , how could I calculate the intensity needed for given current , also is photocurrent equivalent to ordinary electric current written in amps ?

Does the same power rule applies for photcurrent as for conventional current, for example 200 volts and 1 amp transmitted would equal 200 watts of power , versus 10 volts and 20 amps to achieve the same total power delivered.
So in photocurrent lower wavelength photons would require higher intensity while higher wavelength radiation source would require less intensity to get the same current , correct?

 

[sophiecentaur]

Current would always be measured and stated in Amps (in the case of photo current. it would be milliAmps).

To predict the photo current or Power from just the intensity of the incident light would be very hard / impossible. The Efficiency of energy transfer from light to electrical power would normally be pretty low. There are so many factors involved - the spectrum of the incident light (only frequencies higher than the threshold frequency would produce photoelectrons. The state of the surface of the photo emitting metal will have a massive effect on the efficiency of the transfer (Einstein's Equation for the kinetic energy of the photo electrons will only give the highest energy; many electrons will leave at slower velocities.
If you were looking for a way of generating serious power (Many Watts) from light, it would be better to think in terms of Photodiodes - which can be usefully efficient. They use a different technology.

Bottom line for relating light intensity to output current, you are looking at an absolute maximum of one electron per incident photon - but that would be ignoring many practical factors.

 

[ZapperZ]

Hi, how could i calculate the current I would get from the photoelectric effect, so that the end result would be in amps?
If I have a certain lightsource or source of powerful enough em radiation to conduct the photoelectric effect , how could I calculate the intensity needed for given current , also is photocurrent equivalent to ordinary electric current written in amps ?

Does the same power rule applies for photcurrent as for conventional current, for example 200 volts and 1 amp transmitted would equal 200 watts of power , versus 10 volts and 20 amps to achieve the same total power delivered.
So in photocurrent lower wavelength photons would require higher intensity while higher wavelength radiation source would require less intensity to get the same current , correct?

You will need to know the quantum efficiency(QE) of the photocathode at the particular wavelength of light that you are using. There is no material with 100% QE.

Most of us who have to deal with this simply MEASURE the photocurrent, rather than just calculate it.

Zz.

 

[Salvador]

First I would like to ask sophie , Why you say the current wuld be so low as miliamps? Well if I have a light source whose wavelength is higher than the threshold wavelength or energy needed to knock out photoelectrons , such as a laser or other source of radiation , I mean how much photons are emitted from a typical cd laser head for example in a second of time ? Shouldn't there be a large number of photons that go out a typical laboratory type laser and if shined upon a metal plate shouldn't they emitt a stream of electrons that are equivivalent to the number of incoming photons (assuming each of the incoming photons are above the threshold energy needed to knock out the electrons? )

So if I had a LED lamp like the ones you can now buy at hardware stores for outdoor lightning (mains voltage) and for the sake of argument if we assume the light from such led lamp is above threshold wavelength then is there any approximation to how high of a current one could get versus the wattage of the LED lamp or something along these lines?

It is kinda complicated to set up such an experiment at home due to lack of materials and devices , but I would like to experiment with two charged metal plates ( like a vacuum capacitor) with a PD across them and then " shine" a light on one the plates periodically to form a sort of switch for very high voltages.
But then I would need these plates with a proper distance and enclosed in a vacuum chamber or a special vacuum enclosure from glass etc and a special light source.

 

[sophiecentaur]

To work out the maximum possible current: Calculate how many photons per second will arrive from a 1W laser beam of a frequency of your choice. Multiply that by the charge on an electron and that will give you the Coulombs per second - i.e. the current. It's an easy calculation to do, as long as you are careful with the 'powers of ten'. And, btw, then you need to knock off a big chunk off that number because of the efficiency of the process.

You have a problem with your thought experiment. The threshold frequency for photoemission will depend upon the metal. It's only the alkali metals (afaik) that have a threshold frequency in the optical region. Their Work Function is low enough. An experiment with a Zinc plate is possible but it needs UV light to produce photoemission. The plates in a vacuum capacitor will not be made of Potassium or Sodium. They will probably be gold plated. Look up the Work Function of gold and see what transition frequency it would correspond to.

I am leaving all these the sums to you because it will be a good exercise in discovering the hard facts of practical Physics. Life can be tough for experimenters. :)

 

[ZapperZ]

Shouldn't there be a large number of photons that go out a typical laboratory type laser and if shined upon a metal plate shouldn't they emitt a stream of electrons that are equivivalent to the number of incoming photons (assuming each of the incoming photons are above the threshold energy needed to knock out the electrons? )

But you were already told, and I said this in my post, that there there are no material with 100% quantum efficiency! For copper, for example, the QE is of the other of 0.001%! This means that out of 1000 photons, you get, on average, ONE photoelectron!

So no, you simply cannot calculate the photon flux and then assume that you'll get the same number of photoelectrons out.

Zz.

 

[sophiecentaur]

But you were already told, and I said this in my post, that there there are no material with 100% quantum efficiency! For copper, for example, the QE is of the other of 0.001%! This means that out of 1000 photons, you get, on average, ONE photoelectron!

So no, you simply cannot calculate the photon flux and then assume that you'll get the same number of photoelectrons out.

Zz.

And what about the Work Function related to frequency of photons? The experiments, described in the textbooks all use Potassium. And that has to be a 'freshly cut' surface.
Your quoted value of quantum efficiency is a good reason for using solid state technology in practice.

 

[ZapperZ]

And what about the Work Function related to frequency of photons? The experiments, described in the textbooks all use Potassium. And that has to be a 'freshly cut' surface.

Not sure what the issue here in here. I never said anything about such issues, and certainly the work function is a factor here. In fact, QE is dependent of work function and photon frequency.

Your quoted value of quantum efficiency is a good reason for using solid state technology in practice.

Not all the time. Photomultiplier tubes are still necessary for many detectors, especially in high energy physics that require huge coverage. Read this already-funded research proposal and why solid-state photodetector is not suitable:

http://psec.uchicago.edu/other/Project_description_nobudgets.pdf

Furthermore, you can't use such a thing as an electron source for accelerator photoinjectors.

There is seldom a single technique suitable for all applications.

Zz.

 

[sophiecentaur]

There is seldom a single technique suitable for all applications.

Zz.

The OP was talking in terms of Amps so I inferred he was talking Solar Power.

 

[ZapperZ]

The OP was talking in terms of Amps so I inferred he was talking Solar Power.

That is not the only means to talk in terms of "amps". Bare electrons emitted from photocathodes (i.e. photocurrent) are measured in Amps as well. In fact, when I measure the QE of the photocathode that I make, what do you think the picoammeter measure?

Zz.

 

[sophiecentaur]

That is not the only means to talk in terms of "amps". Bare electrons emitted from photocathodes (i.e. photocurrent) are measured in Amps as well. In fact, when I measure the QE of the photocathode that I make, what do you think the picoammeter measure?

Zz.

That goes without saying.My point was that, earlier on, I was questioned for implying that the photocurrent would only be in milliAmps. Perhaps there was a Langauge problem there. There are situations when sub and multiples of base units are used in preference to base units. That's pretty general practice and used as a way of indicating ball park magnitudes, isn't it? Would you describe the length of your journey to work in Metres? (Unless you live over the shop.)

 

[ZapperZ]

That goes without saying.My point was that, earlier on, I was questioned for implying that the photocurrent would only be in milliAmps. Perhaps there was a Langauge problem there. There are situations when sub and multiples of base units are used in preference to base units. That's pretty general practice and used as a way of indicating ball park magnitudes, isn't it? Would you describe the length of your journey to work in Metres? (Unless you live over the shop.)

Oh, I get it now. You're talking about the magnitude of the current, instead of the unit itself.

One very seldom gets charge of the order of milliamps out of a photocathode. In fact, if one uses a higher powered light source, depending on the spot size, anything that might produce larger charge will tend to also cause an explosive emission.

In a typical photoelectric effect experiment, the currents that are often measured are usually in nano or micro amps.

Zz.

 

[sophiecentaur]

Right ZZ. We are now singing from the same hymn sheet. :)

 

[Salvador]

you said the QM efficiency of the photoelectric effect is very low , but if my photon source produces a stream of photons and all of them are above the energy threshold and say all strike the plate , then if the outgoing electron current doesn't match the incoming photon one, where do the extra photons go , do they get absorbed or what how come they don't knock out electrons?

 

[ZapperZ]

They get converted into HEAT.

Please note that it is Quantum Efficiency, not QM efficiency, which has no meaning.

As further education, look up Spicer 3-step model for photoemission.

Zz.

 

[sophiecentaur]

you said the QM efficiency of the photoelectric effect is very low , but if my photon source produces a stream of photons and all of them are above the energy threshold and say all strike the plate , then if the outgoing electron current doesn't match the incoming photon one, where do the extra photons go , do they get absorbed or what how come they don't knock out electrons?

'knock out" implies producing electrons on a path, normal to the surface. Well, for a start, half will end up going into the metal. The 'surface' has a finite depth and photoelectrons will be produced in all directionson the way towards the surface. That will involve energy loss, even for those photoelectrons with a 'forward' velocity component. This is a very classical picture of what goes on, of course but most photoelectrons, produced just below the surface will behave just like any other of the thermal electrons - just share its energy with the others and the lattice. Only a few are likely to escape.