# B Relationship among Intensity, Frequency, and Current

1. Mar 5, 2016

### Aldnoahz

Hi all, I am really confused about the relationship among these three properties and I cannot find an answer satisfying enough..

If frequency of incident light increases but intensity remains the same, the photocurrent will decrease. Does this mean that the new curve on the current vs. voltage graph will have a stopping voltage more to the left and lower current? But I thought same intensity means same current so how is it possible for the current to decrease if intensity remains the same?

The relationship among these three variables is kind of confusing. Any explanation will be appreciated. Thanks.

2. Mar 5, 2016

### Qwertywerty

This is incorrect. Does this clarify your doubt?

3. Mar 5, 2016

### Aldnoahz

Thanks for replying. I do know that this is correct but what is the reasoning behind it? Why does photocurrent decrease when intensity remains the same?

4. Mar 5, 2016

### Aniruddha@94

I suppose you're taking about the photoelectric effect? What you've said is wrong. If the intensity of the light is the same then the photoelectric current doesn't change. This current is the number of electrons being emitted per second. That is directly proportional to the intensity of the light. So greater intensity means more photons striking the surface and hence an increase in the emitted electrons.
The frequency of the light tells is about the energy of the photon. Increased energy of the photon means greater kinetic energy of the electron and hence a larger potential ( stopping voltage) to stop it.
Intensity has nothing to do with voltage and the frequency of the light has no effect on the photocurrent.

5. Mar 5, 2016

### Drakkith

Staff Emeritus
I was under the impression that intensity in this context was measured in power, not photons. Is that incorrect?

6. Mar 5, 2016

### Qwertywerty

Doesn't a greater intensity imply a larger number of photons, as compared to a less intense light, for a particular wavelength of light?

7. Mar 5, 2016

### Drakkith

Staff Emeritus
Sure, but that's true no matter how you define intensity here pretty much.

8. Mar 5, 2016

### Qwertywerty

So a larger number of photons striking would mean more photoelectrons emitted, right (for appropriate frequencies) ? I believe that was @Aniruddha@94's point.

9. Mar 5, 2016

### Qwertywerty

Photocurrent depends on the incident light's intensity, and the collector potential.
The energy of the photoelectrons emitted depends on the frequency of incident light, and the collector potential.

10. Mar 5, 2016

### Drakkith

Staff Emeritus
That was my question. If the intensity is referring to power, then if you keep the intensity the same as you increase the frequency, the electrical current should decrease because you have fewer photons and thus fewer electrons being ejected from the metal.

11. Mar 6, 2016

### Aniruddha@94

I'm pretty sure it's the other way around. All of the books where I read this refer to intensity as the number of photons striking the surface (Resnick Halliday, Beiser, Quantum Physics by Robert Eisberg and Resnick and if it's reliable, Wikipedia)
That's not correct. It's an experimental fact that keeping the intensity same and changing the frequency doesn't change the saturation photocurrent.
Also, I think your point is a little flawed. Keeping the intensity same and increasing the energy (and hence power) means that you'd have to increase the area as well. So there shouldn't be a decrease in current.
In the pure particle picture of light, changing the energy of the individual photons (frequency) has no effect on the number of photons (intensity). I'm not sure if we're allowed to mix the wave and particle concepts.

12. Mar 6, 2016

### Qwertywerty

That is equivalent to reducing the number of photons in unit area.
Why?
Which concepts?

13. Mar 6, 2016

### Qwertywerty

I have apparently misunderstood the OP's question. Apologies.

14. Mar 6, 2016

### Drakkith

Staff Emeritus
Alright. I guess they are talking about intensity as the number of photons per second in this context.

15. Mar 6, 2016

### Qwertywerty

16. Mar 7, 2016

### Aniruddha@94

17. Mar 7, 2016

### sophiecentaur

That is not how intensity is defined - simple as that. Remember that quantities like Intensity were defined and used long before we were aware of photons. Intensity is a Classical quantity and you should really use the accepted definition - if you want people to understand what you're saying and if you want good answers.

18. Mar 7, 2016

### Aniruddha@94

You're right, I didn't use the terms correctly, so sorry about that. And like I said, I still have to spend some time to study about light

19. Mar 8, 2016

### sophiecentaur

To get this straight
The bottom line to this is that the Photo current will go down if the frequency is increased with the intensity held constant. There are fewer photons at the higher frequency. (One electron released per photon - at the simplest level of argument)

20. Mar 8, 2016

The question in the original post was a good one. Usually the response curves of the photodiodes, e.g. silicon photodiodes are given as a graph of response R vs. wavelength, and the response does increase nearly linearly with wavelength until the photodiode reaches the long wavelength cut-off where the photon energy is insufficient to generate a photoelectron. The intensity is measured in watts or watts/cm^2 and a theoretical response R for 100% quantum efficiency can be computed where R=photocurrent/incident power(intensity) which equivalently is R=electron charge/photon energy= $e/(hc/\lambda)=e*\lambda/(hc)$ Response computed (using this formula) is $R=.80*\lambda$ Coulombs/Joule(equivalent to Amp/Watt) where wavelength $\lambda$ is in microns . For a silicon photodiode which peaks (and cuts off) at wavelength $\lambda=1.0$ microns, peak response at $\lambda=$ 1.0 microns is R=.80 Coulombs/Joule=.80 Amps/Watt (which is really more practically .80 microamps/microwatt (or milliamps/milliwatt)). A good commercially available silicon photodiode typically responds for wavelength $\lambda$ from .4 microns (4000Angstroms)< $\lambda < 1.0$ microns and the response of a good one at its peak wavelength (1.0 micron) is typically R=.65 A/W. If you google silicon photodiodes, I think you should find a typical response curve included, and the response is nearly linear with wavelength(per the above formula) up to 1.0 microns,where it reaches its peak, and then an abrupt cut-off for wavelengths longer than 1.0 microns. (A quantum efficiency factor $\eta$ is often included in the response formula.)

Last edited: Mar 8, 2016
21. Mar 8, 2016