# Photoelectric effect experimental data current vs. intensity vs frequency

1. Mar 6, 2009

### Kosta Dean

The photoelectric current is known to be directly proportional to the intensity of incident light with fixed frequency. Questions:
1) What are the experimental values of this proportionality constant for various fixed frequencies?
2) Is there a theoretical derivation that provides a formula for this proportionality constant?
3) What is the relationship of photoelectric current and frequency of incident light with fixed intensity?
4) What are the experimental and theoretical 'incongruencies' to the standard photoelectric effect explanation?

2. Mar 6, 2009

### map19

Since most experiments involve some sort of vacuum tube the results depend on the tube characteristics. This depends largely on the material of the target.

Stopping potential vs frequency is a straight-line graph.
If the electron is to escape the target it must pick up energy at least equal to the work function.

3. Mar 6, 2009

### Kosta Dean

Thanks for your response. I am looking for experimental data that provide the characteristics of the relationships between the photoelectric current, the intensity of the incident light, and the frequency of the incident light (all else being fixed, including the material and tube characteristics). More specifically, I want to know what is the slope of the Current vs. Intensity graph (for various fixed frequencies) and what are the characteristics of the graphs of Current vs. incident light Frequency (for fixed intensity of light). I assume this is not linear, but what is known about this relationship? And can this relationship be theoretically derived or is it just an experimental empirical fact?

4. Sep 4, 2009

### cyong

i have doubts on this whether
intensity is proportional to photo current and is proportional to frequency.
If the statement is true, when intensity is kept constant while the frequency is increased, will photo current decrease?

5. Sep 4, 2009

### map19

No. here's some info:
1.For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.
2.For a given metal, there exists a certain minimum frequency of incident radiation below which no photoelectrons can be emitted. This frequency is called the threshold frequency.
3.For a given metal of particular work function, increase in frequency of incident beam increases the intensity of the photoelectric current.
4.Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light but depends on the frequency of the incident light.
5.The time lag between the incidence of radiation and the emission of a photoelectron is very small, less than 10^−9 second.
6.The direction distribution of emitted electrons peaks in the direction of polarization (the direction of the electric field) of the incident light, if it is linearly polarized

6. Sep 5, 2009

### cyong

so when the intensity of light is kept constant while the frequency is increased, the rate of emission of photoelectron will keep constant?
can it be illustrated by a Current against Voltage graph with various frequency?

7. Jan 5, 2011

### antistrophy

I'm new here so sorry if this thread is dead, but I don't understand what happens to the photoelectric current if just the frequency of the incident photon is increased.

Wikipedia says this (which I think is wrong):
"For a given metal of particular work function, increase in frequency of incident beam increases the maximum kinetic energy with which the photoelectrons are emitted, but the photoelectric current remains the same, though stoppage voltage increases."

Clearly, unless I am missing something, increasing the frequency will increase the current, and http://phet.colorado.edu/en/simulation/photoelectric" [Broken] agrees with me.

Can someone clarify this for me? Thank you in advance.

Last edited by a moderator: May 5, 2017
8. Jan 6, 2011

### map19

See my previous post.
Or, think of it in energy terms.
Each photon with energy (E = hf ) that is at or above the metal's work function will spring loose an electron.
More photons = more electrons.
Work function for most metals is around the same as the energy in visible light photons.
eg. zinc 4.31 eV , copper 4.7 eV, Sodium 2.28 eV.
If the photon energy exceeds the work function the extra energy goes into kinetic energy of the electron.
Increasing the frequency (energy) for the same number of photons does NOT increase the number of electrons (one electron per photon) it just gives them extra kinetic energy.

9. Jan 6, 2011

### gomunkul51

If you agree that when a stream of electron passes through a point, the faster the stream - the larger the number of electrons that will pass in time t0 through that point:

By definition, current is dq/dt - please explain how the current will stay the same if on one hand k electrons will pass at some point in time t0, and on the other hand p>k electrons will pass through the same point in the same time t0 ?
k/t0 < p/t0

10. Jan 6, 2011

### antistrophy

Yes, in other words since energy is dependent on the velocity of the electrons and the current is the rate of flow of the electrons, if you increase energy you increase the velocity and therefore you increase the current.

Do I have this right? Then Wiki is wrong.

P.S. Map19 is talking about the intensity, not the frequency.

11. Jan 6, 2011

### SammyS

Staff Emeritus
Emphatically, No! I don't agree. The average spacing between electrons simply may increase.

12. Jan 6, 2011

### antistrophy

Then you have decreased the number of electrons being ejected in that time. Again this is the intensity not the frequency. keeping the same intensity then the rate does increase.

13. Jan 6, 2011

### map19

The only frequency to talk about here is the photon electromagnetic frequency, which is proportional to its energy. The number of photons is the intensity.
One photon - one electron knocked loose.
A change in photon frequency is a change in energy, not a change in the rate at which photons appear (that's intensity).
Thus the number of photons is the intensity, which determines the number of electrons produced, which is the current..
The photon frequency, AKA photon energy, determines the kinetic energy of the electrons as they spring loose.

14. Jan 6, 2011

### gomunkul51

explain how the spacing can be greater if in our case there is the same number of photons but with increased energy? or give me a more correct model.

15. Jan 7, 2011

### antistrophy

Okay, here's the deal. 10 photons of low frequency (but over the threshold) eject 10 electrons at 1 mm a second (random non realistic numbers to illustrate a point). And 10 photons of high frequency eject 10 electrons at 2 mm a second. 10 electrons ejected in both cases (intensity). Now the second 10 will pass a point in less time than the first 10 because they are going faster. This means that the change in the charge (which is the same for both) over the change in time is greater with higher velocity electrons; which is the current.

This http://amasci.com/miscon/speed.html" makes it quite clear; higher velocity electrons equal higher current. Thank you everyone for your help.

Last edited by a moderator: Apr 25, 2017
16. Jan 7, 2011

### map19

If your electrons are emitted by photons that are spaced by, say, ten a second, then the emitted electrons are also spaced by ten a second.
regardless of their velocity the current is 10 electrons/sec.

17. Jan 20, 2012

### rgoettler

The increased velocity of the electrons just shortens the transit time between when the electrons are ejected and when they are counted at your current meter. Be they fast electrons or slow electrons, the rate of their ejection and the rate of their being counted will be the same (10 electrons per second in the example in the previous post).