Photoelectric Effect: Why does Current Change?

In summary: As with anything in physics, if you go beyond the superficial description, it gets very involved and complex.
  • #1
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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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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!)
 
  • #6
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.
 

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon where electrons are ejected from the surface of a material when it is exposed to light of a certain frequency or higher. This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905.

2. Why does current change in the photoelectric effect?

Current changes in the photoelectric effect because the electrons that are ejected from the material's surface create an electric current when they move towards the positively charged electrode. The intensity of the current depends on the number of electrons ejected and their velocity.

3. How does the frequency of light affect the photoelectric effect?

The frequency of light is directly related to the energy of the photons. In the photoelectric effect, the energy of the photons must be greater than the work function of the material in order to eject electrons. Therefore, the higher the frequency of light, the greater the energy of the photons and the more electrons will be ejected.

4. What is the work function in the photoelectric effect?

The work function is the minimum amount of energy required to eject an electron from the surface of a material. It is different for each material and is dependent on factors such as the type of material, its temperature, and the surface conditions.

5. How does the intensity of light affect the photoelectric effect?

The intensity of light does not directly affect the photoelectric effect. However, it does indirectly affect it by increasing the number of photons and therefore, the number of electrons ejected. This is because the intensity of light is proportional to the number of photons.

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