Photoelectric Effect: Why does Current Change?

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Discussion Overview

The discussion revolves around the photoelectric effect, specifically addressing why the current changes when light of the same intensity but different frequencies is used. Participants explore the relationship between photon energy, stopping potential, and the number of emitted photoelectrons, delving into the complexities of quantum efficiency and the behavior of light sources.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that increasing the frequency of light leads to higher photon energy, which in turn affects the stopping potential and the current, suggesting that the number of photons must decrease to maintain constant intensity.
  • Another participant emphasizes that the quantum efficiency may vary with photon energy, affecting the number of electrons emitted per photon.
  • A later reply questions the concept of quantum efficiency, proposing that it might imply one photon can eject multiple photoelectrons, but is corrected to reflect that quantum efficiency is typically less than 100%.
  • One participant expresses confusion about whether a photon can fail to eject a photoelectron, indicating a lack of clarity on the topic.
  • Another participant shares their extensive experience in the field of photoemission and suggests that the subject is complex and still holds many unanswered questions.
  • A participant references a paper on photoemission for further reading, indicating the depth of the topic.

Areas of Agreement / Disagreement

Participants express various viewpoints on the relationship between photon energy, current, and quantum efficiency, with no consensus reached on the implications of these factors. The discussion remains unresolved regarding the complexities of the photoelectric effect.

Contextual Notes

Participants highlight the dependence of quantum efficiency on photon energy and the implications of constant power in light sources, but these aspects remain underexplored and unresolved in the discussion.

chewchun
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Using the photoelectric experiment,it is known that if light of same intensity but different frequency is used, stopping potential is changed and current changed.

For instance,same intensity but frequency increased.
E=hf, energy of photon increased which leads to a higher K.E of photoelectrons,hence stopping potential increased.

But why is the current affected,in this case decreased.
I can deduce it from the formula Intensity= (Number of photons)(Energy of each photon)/( Time times area).
For I to be constant, number of photons must decrease since energy of each photon increases.
But i do not understand the concept behind it.

My argument is that K.E of photoelectrons increases,but the distance between each photoelectrons is still the same,which meant that number of photoelectrons arriving at the other end per area is still the same,meaning that intensity is constant even if frequency is increased?
 
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There are several issues involved here, and I'll just highlight the simplest ones:

1. It has to do with your light source. Even if you have a constant power for each frequency, as you increase the energy per photon, the number of photos emitted per second HAS to drop. This is because, each photon now carries more energy. Since the total energy per second is constant (power is a constant), there's less photons as you increase the photon energy.

2. The quantum efficiency may not be the same. This might be a minor effect, but the quantum efficiency (number of electrons emitted per incoming photon) is dependent on the photon energy.

Zz.
 
ZapperZ said:
There are several issues involved here, and I'll just highlight the simplest ones:

1. It has to do with your light source. Even if you have a constant power for each frequency, as you increase the energy per photon, the number of photos emitted per second HAS to drop. This is because, each photon now carries more energy. Since the total energy per second is constant (power is a constant), there's less photons as you increase the photon energy.

2. The quantum efficiency may not be the same. This might be a minor effect, but the quantum efficiency (number of electrons emitted per incoming photon) is dependent on the photon energy.

Zz.

Quantum efficiency,i presume to be the fact that 1 photon can eject 2 photoelectrons?
And WHAT IF my power vary with frequency? Or is power proportional to intensity?
What are the other issues?
 
chewchun said:
Quantum efficiency,i presume to be the fact that 1 photon can eject 2 photoelectrons?

Incorrect. QE is always (so far) less than 100%. In fact, for metals, the typical QE is ~0.01-0.001%!

Zz.
 
ZapperZ said:
Incorrect. QE is always (so far) less than 100%. In fact, for metals, the typical QE is ~0.01-0.001%!

Zz.

Oh,so its that 1 photon may not eject a photoelectron at all??!
I guess I am asking too much,even my teacher does not want to discuss this with me...!(Or he doesn't know!)
 
chewchun said:
Oh,so its that 1 photon may not eject a photoelectron at all??!
I guess I am asking too much,even my teacher does not want to discuss this with me...!(Or he doesn't know!)

As with anything in physics, if you go beyond the superficial description, it gets very involved and complex.

I've been working in the field of photoemission, photocathodes, etc. for more years than I want to count. There are still things we want to know, even though we already know a lot as it is.

If you want to read more about the photoemission process, you can read the Spicer paper that almost everyone in this field has read:

http://www.osti.gov/energycitations/purl.cover.jsp?purl=/10186434-Ekjh1W/10186434.PDF

Zz.
 

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