# Photoelectric effect

## Main Question or Discussion Point

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 arguement 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?

Related Other Physics Topics News on Phys.org
ZapperZ
Staff Emeritus
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.

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?

ZapperZ
Staff Emeritus
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.

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 im asking too much,even my teacher does not want to discuss this with me....!(Or he doesn't know!)

ZapperZ
Staff Emeritus
Oh,so its that 1 photon may not eject a photoelectron at all??!!
I guess im 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.