When photons hit metals, does the electron created go in a random direction?

Jarfi

Some metals generate electrons when photons hit them, If so then what direction will the electron go into, will it be random and the uncertainty principle and quantum randomness. Or will it go in a direction relative to the point of impact from the photon, So if you shot a photon at an atom, Would the electron come out the oposite site? or in the same direction as the photon came from???

Is it random or not?

sophiecentaur

I think you are visualising the event as being a bit like a bullet hitting a wall. That would tend to throw shards of wall off in the direction that the bullet might 'bounce off' (i.e. laws of reflection). I don't think that the photoelectric effect is quite so mechanical. afaik, the photon arrives and adds its energy to the total energy around the surface. One (lucky) electron which happens to be near the surface and already, perhaps, 'travelling outwards', can have its kinetic energy increased enough to escape with some extra KE. It is hardly like a coconut shy because the ball (photon) has no well defined size and many electrons (coconuts) occupy one wavelength of the incident light.

I guess someone must have done photo electric experiments and looked at the angular distribution of photoelectrons. The result would be interesting but i would predict that there would be a maximum intensity normal to the surface, rather than at the 'reflected angle'. (Momentum conservation is not violated as long as one considers the whole mass of metal as the target and not the individual photoelectron.)

Jarfi

Well, the electron traveling near the surface is completely random in quantum mechanics so you are saying that the electrons created in solar sells go in a random direction, and also the opposite, when creating light by forcing an electron in an atom, does the photon go in a random direction? When photons are created do they go in a random direction?

sophiecentaur

The electrons in all solids have a range of energies, corresponding to KE in every direction. An electron in an isolated gas atom has no defined velocity - just an energy. An incident photon will merely raise its energy with respect to the nucleus (no direction is involved). At some later time, according to the probability situation, another photon will be emitted totally randomly. This is the basis of absorption spectroscopy, which shows dark bands where the light from one direction has been absorbed and re-radiated in all directions, leaving only a small amount of energy flowing in the original direction.
In a photo cell, the energy given by an incident photon will increase the total energy in part of the lattice, allowing electrons that happen to be in an appropriate state (a bit non-committal here, I'm afraid) to make it across the junction, against the potential difference. If an electron happens to gain energy to take it away from the junction, then I guess the photon energy is 'wasted'. This would imply a maximum efficiency of 50% and I wonder if this is actually the case. I must look into this. [edit: the pv description above is a bit over-simplified haha but again I don't think direction comes into it. The photon energy produces an electron-hole pair which causes a current to flow]

xts

The answer strongly depends on photon energy.

For photons in single eV range (visible light, near ultraviolet), the distribution is nearly uniform. (However it is not so easy to measure, as electrons of so low energies are vulnerable to even weakest electric fields, and we must use such fields to detect those electrons)
For energetic photons - keV-GeV (X-rays, gammas) - scattered electrons follow the direction of the hitting photon.

I would really like to know where the border between those two lies. I've never played with 100 eV or so photons...