Wave and particle confusion

1. Feb 13, 2008

i_island0

This is regarding photoelectric effect.

Suppose a power source (having power P) is at distance r from the metal plate and photons are assumed to strike the plate normally.
Intensity of light reaching the plate is: I = P/4.PI.r^2
If plate area is A, we say power received is: P(R) = I.A = nhv/t ----(1)

Now, my question is how suddenly we merged wave theory and quantum theory in equation (1).

2. Feb 13, 2008

peter0302

Because treating light like a single wave with an intensity is just a macroscopic representation of a _very_ large number of photons randomly distributed around the source according to their wave functions.

3. Feb 13, 2008

i_island0

can u explain in more detail please

4. Feb 14, 2008

i_island0

i never read anywhere that wave distribution represents photons

5. Feb 14, 2008

peter0302

The probabilistic distribution of the photons (which are particles) around a point light source is going to represent a wave eminating from a source. Photons are governed by wave functions and the Schrodinger Equation just like all particles. Therefore, like a wave, intensity will decrease with the square of the distance.

6. Feb 14, 2008

joyer2

'Quantum theory' is in hv----the energy of a photon is proportional to the frequency. That is completely in conflict with classical eletromagnetism, or electrodynamics, where the energy of the electromagnetic wave does not depend on frequency.

7. Feb 14, 2008

peter0302

That's right, but the question, I thought, was how could a wave concept like intensity be reconciled with the particle concept, and the answer is because a gazillion photons distributed randomly from a point-source look like a wave with intensity proportional to the number of photons divided by r^2.

8. Feb 14, 2008

marlon

But look at the experimental results of the photo-electric effect :

1) The kinetic energy of the electrons is proportional to he frequency of the EM radiation
2) The kinetic energy of the electrons is independent of the total energy of the EM radiation (ie the intensity).

marlon

9. Feb 14, 2008

joyer2

1) tells you the light is quantized---Ek~hv, the kinetic energy of an electron is transferred from one 'light particle'.
2) tells you the light is quantized---the number of photoelectrons N_pe is proportional to the intensity which is the number of photons N_p, i.e, N_pe = N_p, one photon knocks one electron out.

10. Feb 14, 2008

peter0302

What exactly is the question here?

11. Feb 14, 2008

Staff: Mentor

At what level are you studying? In the USA, most university physics students see this in their second-year "introductory modern physics" course and textbook.

12. Feb 14, 2008

i_island0

I am a mentor myself. Hopefully will become a good one soon.

13. Feb 15, 2008

marlon

OOPS, my post was adressed to I_Island, i should have quoted him.

LOL,

I_Island, the above two questions illustrate how the transistion from "wave to particle" is made and explained.

marlon

14. Feb 19, 2008

twinsen

Does one photon have energy=hv so n photons have energy nhv?
Isn't this another way of explaining the discrete energy levels planck used to explain the BB radiation. I take it this means that a particle can only act on one other particle and that one photon cant influence two electrons.

This would mean that if we increase n the quantity of photons and intensity then it wont result in an increase in the photoelectrons emitted as long as there are enough photons n to satisfy all the electrons availiable for release?

Im not sure I like the idea of having duality :(

Alex