# Photon of intermediary energy - Spontaneous Emission

1. Apr 29, 2013

### jaumzaum

I'm studying spontaneous emission and I'm in trouble to understand the following: In a hydrogen atom we have distinct energy levels with different energy values associated to them.

E1 = -13,6 eV
E2 = -3,4 eV
E3 = -1,51 eV

To go from the first to the second energy level we need a photon of energy 10,2 eV, and to go from the first to the third we need a photon of energy 12,09 eV. What would happen if we inside a monochromatic beam of photons of energy 11 eV in a hydrogen atomic gas. Will the atoms be excited to the second energy level and release the remaining 0.8 eV? Will the atoms not be excited at all? Or will they use 10,2 eV to excite the electron to the second energy level and 0,8 eV to the kinetic/vibration energy of the atom?

Thanks
John

2. Apr 30, 2013

### jaumzaum

Noone?

3. Apr 30, 2013

### Staff: Mentor

To be clear, you are now talking about absorption, not spontaneous emission.

There is no vibrational degree of freedom for atoms and not much of the photon energy can transform into kinetic energy for the atom because of conservation of momentum.

Most photons will go through the gas without interacting. But if your light is bright enough, you can get Raman scattering, where you have simultaneous absorption and emission of a photon, with the emitted photon carrying away the "extra" energy.

4. Apr 30, 2013

### jaumzaum

You mean that for the atom absorb the photon "entirely", it should have the exact amount of energy needed to go up one level? I mean, would we need a photon of exactly 10,2eV? What if it have 10,20001eV? The question may seem somewhat stupid but I can't imagine a monochromatic light of exactly 10,2eV. Actually I don't think it's possible to produce a monochromatic light of an exact frequency or energy, but it is possible to produce beam of light whose photon frequencies oscillate between F ± Δf, being Δf very small. The question is, how small should Δf be so that absorption/spontaneous emission occurs? Another question: What's the difference between spontaneous emission and rayleigh scattering?

Thanks
John

5. May 1, 2013

### Staff: Mentor

You have to factor in the uncertainty principle. Since any excited state will eventually decay to the ground state, it has a finite lifetime. Therefore, its energy is "uncertain". This is often called the width of the energy level. This results in a Laurentzian lineshape (the reference mentions pressure broadning, but the same result is obtained because of the natural lifetime).

In a real spectroscopy experiment, you have to consider many other things, like pressure broadning (due to collisions) and the Doppler effect. Indeed, the light may also not be purely monochromatic, but with lasers you can come very close.

It usually is a continuum. The closer you are to resonance, the stronger the scattering. And as you detune, the signal decreases to the point you can't detect it anymore.

I'll try to answer that question later.

6. May 1, 2013