# A question about Photoelectric Effect

1. Apr 14, 2008

### Hevonen

[SOLVED] A question about Photoelectric Effect

According to my teacher "photoelectric current is independent of frequency but dependent on intensity". This statement seems to conflict the following exercise:

"In an experiment to investigate the photoelectric effect, monochromatic light is incident on a metal surface. The photoelectric current and the maximum kinetic energy of the photoelectrons are measured.

Which one of the following correctly shows the change, if any, in the photoelectric current and in the maximum kinetic energy of the photoelectrons when light of the same intensity but higher frequency is incident on the same metal surface?"

A.
Photoelectric current is the same
Maximum kinetic energy increases

B.
Photoelectric current decreases
Maximum kinetic energy increases

The solution is B. Hence, my teacher's statement is ambiguous. Clearly, photoelectric current is NOT independent of the frequency as an increase in frequency has decreased the current. Hence, i think I don't understand the statement of independence. Please, help me!

2. Apr 14, 2008

### ZapperZ

Staff Emeritus
You are correct. It appears that you have a better understanding of the photoelectric effect phenomenon than your teacher.

Zz.

3. Apr 14, 2008

### Hevonen

Yet, your response conflicts with other exercise in which the independence of photoelectric current about frequency is clearly stated:

State three pieces of evidence provided by the photoelectric effect that support the particle nature of electromagnetic radiation.

1. photoelectric current / rate of emission independent of frequency;
2. photoelectric current / rate of emission depends on intensity of radiation;
3. (max) kinetic energy of electron dependent on frequency;
etc.

It is the solution 1. that confirms my teacher's statement that "photoelectric current is independent of frequency". Sorry but i still do not get the meaning of independence. What does it mean here then? Please, help me.

4. Apr 14, 2008

### ZapperZ

Staff Emeritus
I had to read this several times because these things are worded in a rather convoluted way.

"Intensity", it appears in this case, is the # of photons per unit time. So if intensity remains the same, then the amount of photoelectrons being emitted per unit time should remains the same.

However, if the the frequency is higher, then on average, the photoelectrons should have higher average kinetic energy as well. It also means that the max kinetic energy should be higher. Off the top of my head, I would say that the photocurrent remains the same if the intensity remains the same because you still have the same number of photoelectrons per unit time being emitted. So (1) is correct. This means that in your original post, the solution should have been (A).

Zz.

5. Apr 14, 2008

### Shooting Star

Last edited: Apr 14, 2008
6. Apr 14, 2008

### Shooting Star

If intensity is the number of photons, then if you increase the frequency, the current must increase. Without going into any calculations, the number of electrons with higher energies will be passing through the positive electrode, per unit time.

7. Apr 14, 2008

### ZapperZ

Staff Emeritus
Not necessarily, because the number of photon passing a unit area per unit time will still be the same. It means that the photoelectron density in a unit volume is different.

Zz.

8. Apr 14, 2008

### Shooting Star

In this case, do you mean higher or lower by different?

If you mean lower, then I will have to read the "detailed" details of photoelectric effect all over again.

If the the photo electron density in a unit volume is indeed higher, would it not knock out more electrons, thus increasing the current?

9. Apr 14, 2008

### ZapperZ

Staff Emeritus
If the KE is higher, than the density of photoelectron per unit forum is lower. This is because they are all moving faster, but they are, on average, further apart. This is the only way to get the same number of photoelectrons passing a unit cross-sectional area per unit time with the case of photoelectrons having lower KE.

One can approximate this by assuming the photoelectrons in a cylindrical beam. The ones with a higher KE but with the same number of photoelectrons in 1 second will have a longer cylinder than the ones with a smaller KE but also having the same number of photoelectrons in 1 second.

Zz.

10. Apr 14, 2008

### Shooting Star

I think I am bit tired, because all this while I meant photon density when I was writing photoelectron density. I will clarify this whole thing afterward. The last post of mine was meaningless. Sorry.

However, my question in post #6 has not been answered properly, I believe. I'll frame it properly and come back to here.

11. Apr 14, 2008

### ZapperZ

Staff Emeritus
All you need to keep in mind is that the # of photons per unit time hitting the cathode is the same. This means that the number of photoelectrons emitted per unit time is the same, no matter how fast they are emitted. So if I collect the same number of electrons in 1 second, no matter what their KE are, then I have the same current.

Zz.

12. Apr 14, 2008

### Parlyne

This is not correct. The intensity of light is not equivalent to the number of photons per unit time.

Intensity is the power delivered to a surface per unit area. Power is rate at which energy is delivered. So, if each photon has an energy $h\nu$, and the rate at which photons hit a unit area of the cathode is $R$, the intensity of the light will be $I=Rh\nu$. This means that, if we assume constant intensity, the rate at which photons hit a unit area of the cathode is $R=\frac{I}{h\nu}$.

Since the number of photons in is assumed to be the same as the number of electrons out (and, even if it isn't the relationship should just involve a constant factor), the photocurrent should look like $J=eRA=\frac{IeA}{h\nu}$, where A is the area (or, really, effective area) of the cathode.

Clearly, if we increase the frequency of the light without changing the intensity, the photocurrent must decrease because there are fewer photons to knock out electrons.

13. Apr 15, 2008

### Hevonen

You are correct. This is the main essence of this problem. Your analysis shows clearly that photoelectric current is dependent on frequency. Yet, a problem arises. If we suppose that each electron has energy $h\nu$ and "intensity is the power delivered to a surface per unit area", then why intensity would not be defined as "the number of electrons per unit time delivered to a surface per unit area"? Clearly, the number of electrons is proportional to the energy. But then I think there would be no sense in having frequency as the definition would be the same. Hence, I reckon the best definition for intensity is yours and not to mix it with the number of electrons.

As for the independence, it is very hard to accept that there would be a mistake in IBO's final exams. Can the following sentence in post #3 mean then something else: "1. photoelectric current / rate of emission independent of frequency;"? In what sense, photoelectric current could be independent of frequency? Is there any other meaning for frequency in this context except "the number of photons striking the metal surface per unit time"?

Last edited: Apr 15, 2008
14. Apr 15, 2008

### ZapperZ

Staff Emeritus
Yup, you are correct. I was trying to simplify the question way too much by making intensity directly proportional to # of photons per unit time. Thanks for the correction.

Zz.

15. Apr 15, 2008

### Shooting Star

Well, that was the confusion I had yesterday from reading your post, but things seem to have been resolved in my absence.

"The number of photons striking the metal surface per unit time" has got nothing to do with the frequency of the photons. The frequency of a photon is an intrinsic property, equal to c/λ, where λ is the wavelength of the EM wave in the classical limit. (The energy of a photon is hf, where f is the frequency.)

Last edited: Apr 15, 2008
16. Apr 15, 2008

### Hevonen

Last edited: Apr 15, 2008
17. Apr 15, 2008

### peter0302

It's very simple. One photon liberates one electron, and current is equal to charge per unit time, i.e., the number of electrons per unit time. Therefore, the number of photons per unit time (i.e. intensity) determines the current. Increasing the frequency makes each electron liberated more energetic but it does not increase the number of electrons liberated per unit time, and thus, does not affect the current.

Last edited: Apr 15, 2008
18. Apr 15, 2008

### Hevonen

Thanks peter0302! I understand it now. Independence means here that one photon correspond to the liberation of one electron. In that sense, photoelectric current is independent of frequency. However, the mathematical relationship was shown in post #12 that frequency is inversely proportional to the photoelectric current. In that sense, one could argue frequency to be dependent on the photoelectric current. Clearly, photoelectric effect has many dimensions -- fascinating.

19. Apr 15, 2008

### ZapperZ

Staff Emeritus
You need to be a bit careful here in extending the simple description too far. The equation that you are dealing with here is the most naive description of the phenomenon. It assumes a quantum efficiency of 100%, i.e. one photon liberates one photoelectron. That's the assumption one has to make to talk about the amount of photocurrent. This assumption doesn't work, especially for metals, which has a quantum efficiency, at best, in the 0.01% range. When you peel off the simple picture, a number of things will pop out. For example, at higher frequency, one can get more photoelectrons even if one kept the photon flux (which I initially tied to "intensity") to be the same. This is because the higher energy photons can go deeper into the band structure of the material and thus, have a higher probability of producing photoelectrons.

So in reality, the change in photocurrent is actually quite complicated, more so than simply looking at the highest KE of the emitted photoelectrons.

Zz.

20. Apr 15, 2008

### peter0302

Also I think your teacher WAS treating intensity as simply the rate of photons striking the metal.