# Photons transfer energy to the atoms in the metal plate?

1. Aug 7, 2004

### wolram

When sunlight hits an object, for instance a metal plate, the plate will
get hotter, now as i understand, heat is caused by atoms vibrating, so
the photons hitting the metal plate must have caused the atoms in it
to vibrate, my question is, how do massless? photons transfer energy
to the atoms in the metal plate?

2. Aug 7, 2004

### Petrushka

The photons have no rest mass, but they do have energy (kinetic energy). Therefore when the photon collides with the atom, I presume it would transfer some of this kinetic energy to the atom which could heat it up.

3. Aug 7, 2004

### wolram

Petrushka
I must admit that i am stupid, i did not realise that if something is massless, that
it could have kinetic energy, is this related to relativistic mass?

4. Aug 7, 2004

### Gonzolo

No no no no. No relativity here. No stupidity either. That is pretty much what is special about photons, they have energy but no mass. If we speak of photons, then the energy of the IR beam is E = hNf (Planck's constant x number of photons x frequency of the photons). But we don't even need Plank. You know about electric fields? A beam of light is a wave of electric fields (and magnetic but not important for now). When the electric fields reach the metal, they make its atomes and electrons move, like any field does. Relations for the energy of a EM wave can be derived from Maxwell's work.

5. Aug 7, 2004

### pmb_phy

Even photons have mass. Many people call that mass relativistic mass. In such a collision of photon with atom the sum of the (relativistic) mass is conserved. Momentum is conserved too and photons have momentum. If the atom absorbes the photon then the atom will be in a higher energy state. That is not called an increase in heat. Heat pertains to a collection of particles like a solid made of atoms or a gas of atoms etc.

Pete

6. Aug 7, 2004

when a photon hits a metal plate, the photon hits an electron and sends it to a higher energy level and then back down to its original level emitting another photon. This is why metals are shiny. Is it this oscilation from one energy level and back that causes the heat to be produced?

7. Aug 7, 2004

### wolram

Thanks for the replies so far, but im still confused between the,
photon "hitting", an electron, as in a collision between two bodies,
and the atom "absorbing", the energy of the photon.

8. Aug 7, 2004

### Gonzolo

Heat is also photons. Suppose a visible photon makes an electron go from level 2 to level 8. The electron may well drop back to level 6, then 4, then 2 again, emmiting 3 lower energy (= IR = heat) photons instead. This is why the color black is hot. Metals are shiny because they reflect the photons. they bounce back without any electrons changing levels.

pmb_phy, photons have momentum, but no mass (Not in conventionnal theories anyway, and I would need a very good reference to be convinced if it did in even in TOE efforts). Relativistic mass is the mass of an object when it is moving very fast, which is a function of the object's mass at rest. In the case of a photon, they are both zero. The problem from wolram's first post doesn't seem to have anything to do with relativity and should be completely answerable without it.

9. Aug 7, 2004

### pmb_phy

I said they have relativistic mass. They don't have rest mass.
Inertial mass, (aka relativistic mass), m, is defined such that mv is a conserved quantity. Momentum is then defined as p = mv. Therefore anything that has momentum has relativistic mass - by definition. Any decent relativity text will tell you that.

Pete

Last edited: Aug 7, 2004
10. Aug 7, 2004

when the photon hits the electron, the electron is in a discreet energy level in which it stays in by the attractive elecromagnetic force towards the positively charged nucleus. Once the photon hits, it may give the electron enough energy to escape this energy level and to go into a higher energy level, (further from the nucleus). This photon, if it has enough momentum can even knock out the electron totally out of the metal (photoelecric effect) or it can have a low enough energy and just bounce off the electron causing no effect. Its not hard to understand. The energy of the photon is
$$E=nhf \\$$
$$E=energy\ n=integer \\$$
$$h=planck's\ constant\ = 6.63x10^{-34J/s}\\$$
$$f=frequency\\$$
or
$$E = \frac{hc}\lambda \\$$

This energy equation wan be used to determine what will happen when the photon hits the metal. For example, the photoelectric effect can be determined weather it will occur or not.

$$E = W_o + E_k$$
$$W_o = work\ function\ of\ the\ metal$$
$$E_k = kinetic\ energy\ left\ after\ collision\ with\ electron$$

if the energy of the photon is equal to or greater than the work function of the metal, then the photoelectric effect will occur, if it is a lot greater, then the effect will occur and the electron will have a lot of kinetic energy left over. If the photon's energy is less than the work function, then the electron will oscilate.

Last edited: Aug 7, 2004
11. Aug 8, 2004