Understanding the Photoelectric Effect: Electrons and Energy

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SUMMARY

The discussion centers on the photoelectric effect, specifically how electrons are emitted from a metal when exposed to electromagnetic radiation, such as ultraviolet (UV) or visible light. Participants clarify that UV radiation can cause photoemission if its photon energy exceeds the work function of the material. The conversation also addresses the importance of using a vacuum in photoelectric cells to prevent electron scattering, ensuring efficient electron flow to the anode. Additionally, it is established that the intensity of radiation affects the number of emitted electrons but not their kinetic energy.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Knowledge of electromagnetic radiation and photon energy
  • Familiarity with vacuum photocells and their components (anode and cathode)
  • Basic principles of electron emission and work function
NEXT STEPS
  • Research the relationship between photon energy and work function in various materials
  • Study the design and operation of vacuum photocells
  • Explore the effects of radiation intensity on electron emission
  • Investigate applications of the photoelectric effect in modern technology
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in the principles of electron behavior in electromagnetic fields and their applications in technology.

Physicsissuef
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Is in the photoelectric effect, the electrons are oppositing the electromagnetic field or they are excited from the energy?
 
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I don't understand your question. However, the photoelectric effect involves an electron absorbing a photon and, being more energetic, leaving the atom.
 
mathman said:
I don't understand your question. However, the photoelectric effect involves an electron absorbing a photon and, being more energetic, leaving the atom.

And do ultraviolet radiation pass to the metal, or just visible light?
 
Physicsissuef said:
And do ultraviolet radiation pass to the metal, or just visible light?

You really need to put a bit more effort in asking your question here, expecially if you want others to put effort into responding. If not, you will continue to get responses from other people in the form of "HUH?", and you will have to keep on explaining yourself.

UV radiation can cause photoemission if the work function is below the photon energy. Now what is it exactly that you want to know here in this thread?

Zz.
 
ZapperZ said:
You really need to put a bit more effort in asking your question here, expecially if you want others to put effort into responding. If not, you will continue to get responses from other people in the form of "HUH?", and you will have to keep on explaining yourself.

UV radiation can cause photoemission if the work function is below the photon energy. Now what is it exactly that you want to know here in this thread?

Zz.
Ok, sorry. I just want to know, if UV radiation passes through the glass bulb, the glass will absorb that energy. So what is causing the electrons to flow? UV radiation or visible light, or maybe something else?
 
This is getting utterly confusing. You WERE asking about the photoelectric effect, weren't you? Why is it optical conductivity now?

Glass bulb? When did that come in? Did you mention about glass anywhere till now? There's no "photoemission" using UV source on glass bulb, is there?

Why don't you start from the very beginning and ask your question once again. But this time, please put in as much effort and information in your question so that we know (i) what you know (ii) what exact it is the picture that you have in your mind. If you see that you have to ADD new stuff as you go along, it means clearly that you omitted important information in your original question.

Zz.
 
ZapperZ said:
This is getting utterly confusing. You WERE asking about the photoelectric effect, weren't you? Why is it optical conductivity now?

Glass bulb? When did that come in? Did you mention about glass anywhere till now? There's no "photoemission" using UV source on glass bulb, is there?

Why don't you start from the very beginning and ask your question once again. But this time, please put in as much effort and information in your question so that we know (i) what you know (ii) what exact it is the picture that you have in your mind. If you see that you have to ADD new stuff as you go along, it means clearly that you omitted important information in your original question.

Zz.
Sorry for being rude, but have you ever heard about photoelectric tube or photoelectric cell? There is glass balloon or bulb or whatever... Inside there is vacuum, so it is called vacuum photocell. There are anode and cathode. So when I bring some kind of radiation to some metal, the electrons are excited and pulled off the cathode. What is that kind of radiation? UV or visible light? Also I want to know why the anode and cathode are put in vacuum photocell?
 
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The vacuum is used so the electrons can stream without interacting with anything. The radiation is always electromagnetic, as long as the frequency is high enough to excite the electrons of the material being used.
 
Physicsissuef said:
Sorry for being rude, but have you ever heard about photoelectric tube or photoelectric cell? There is glass balloon or bulb or whatever... Inside there is vacuum, so it is called vacuum photocell. There are anode and cathode. So when I bring some kind of radiation to some metal, the electrons are excited and pulled off the cathode. What is that kind of radiation? UV or visible light? Also I want to know why the anode and cathode are put in vacuum photocell?

Not only have I heard of them, I've used them!

What kind of radiation? Any radiation in which the photon energy is higher than the work function! I thought I mentioned this already.

Electrons do not travel very far in air, and even if they do, a lot of them get scattered off and would not reach the anode. This is not what you want when you are using a photocell to detect EM radiation. If you are trying to detect UV, the "glass" has to be either quartz or fused silica, because ordinary glass absorbs UV.

Zz.
 
  • #10
ZapperZ said:
Not only have I heard of them, I've used them!

What kind of radiation? Any radiation in which the photon energy is higher than the work function! I thought I mentioned this already.

Electrons do not travel very far in air, and even if they do, a lot of them get scattered off and would not reach the anode. This is not what you want when you are using a photocell to detect EM radiation. If you are trying to detect UV, the "glass" has to be either quartz or fused silica, because ordinary glass absorbs UV.

Zz.
Ok, thanks. I understand now. I have two more questions. Why the cathode and anode are in vacuum balloon instead connected them with wire? Is this "[URL correct?
[/URL]
 
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  • #11
If they are connected to a wire, they would be the SAME thing, not two different things. The anode has to be at a higher potential then the cathode, the latter is usually grounded.

Zz.
 
  • #12
And why in my textbook says, that the kinetic energy of the electrons doesn't depends from the intensity of the radiation?
 
  • #13
Physicsissuef said:
And why in my textbook says, that the kinetic energy of the electrons doesn't depends from the intensity of the radiation?

Because "intensity" only increases the number of photons, not the energy within each photon.

Your textbook doesn't explain this? Please read for example, this:

http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html

Zz.
 
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  • #14
ZapperZ said:
Because "intensity" only increases the number of photons, not the energy within each photon.

Your textbook doesn't explain this? Please read for example, this:

http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html

Zz.
And what makes the current stronger? The more electrons ejected, or the speed of the electrons?
 
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  • #15
Physicsissuef said:
And what makes the current stronger? The more electrons ejected, or the speed of the electrons?

<scratching head>

Zapper said:
Because "intensity" only increases the number of photons

This is not clear?

Zz.
 
  • #16
ZapperZ said:
<scratching head>



This is not clear?

Zz.

Yes, I understand that with the intensity. Thank you very much. There is one more thing in my textbook. It says that the current is much stronger when I increase the intensity. So that means that more electrons are ejected and the current is getting stronger, right?
 
  • #17
Physicsissuef said:
It says that the current is much stronger when I increase the intensity. So that means that more electrons are ejected and the current is getting stronger, right?
Correct.
 
  • #18
I want to ask you what happens with bar code readers, when the light is touching the bar code (but the white parts of the bar code), do they reflect in the photodiode of the bar code http://www.keyence.com.sg/topics/barcode/bar_code/img/main.gif"
 
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  • #19
Yes. exactly so.
 
  • #20
dst said:
Yes. exactly so.

Can you recognize http://img218.imageshack.us/img218/8903/picture004copyrb7.jpg" Do you know what happens, when the light will touch the white parts? Will be sound produced? Shouldn't be black color instead of white?
 
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  • #21
It is some kind of Motion picture sound. When the light touches the white parts, white happens? Where the light is reflected? What sound it is produced?
 
  • #22
I find it hard to see what's at the back. A cathode ray tube? Photodiode?

If it's a photodiode or some sort of light sensitive resistor at the back, you should be able to get sound. I did a similar test with a laser beam and an amplifier and it makes for a reasonable listening tool.
 
  • #23
dst said:
I find it hard to see what's at the back. A cathode ray tube? Photodiode?

If it's a photodiode or some sort of light sensitive resistor at the back, you should be able to get sound. I did a similar test with a laser beam and an amplifier and it makes for a reasonable listening tool.

It is like http://history.sandiego.edu/GEN/recording/images2/89138b.jpg" I don't know the correct name. I want to ask you what happens when the light touch the white parts? Can you give me the English name of this process?
 
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