Photon of intermediary energy - Spontaneous Emission

In summary: I think now I understand it better. In summary, the conversation discusses the energy levels of a hydrogen atom and how photons of specific energies can excite the atom to different energy levels. It also touches on the uncertainty principle and how it affects the accuracy of monochromatic light. The difference between spontaneous emission and Rayleigh scattering is also brought up.
  • #1
jaumzaum
434
33
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
 
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  • #2
Noone?
 
  • #3
jaumzaum said:
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.
To be clear, you are now talking about absorption, not spontaneous emission.

jaumzaum said:
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?
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
DrClaude said:
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.

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
jaumzaum said:
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?
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).


jaumzaum said:
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.
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.

jaumzaum said:
The question is, how small should Δf be so that absorption/spontaneous emission occurs?
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.

jaumzaum said:
Another question: What's the difference between spontaneous emission and rayleigh scattering?
I'll try to answer that question later.
 
  • #6
Thanks, I've just read about Lorentzian Lineshape and pressure broadening, it was exactly the doubt I had.
 

What is a photon of intermediary energy?

A photon of intermediary energy is a type of electromagnetic radiation that falls within the middle range of the electromagnetic spectrum, between high-energy gamma rays and low-energy radio waves. It has a specific amount of energy and can travel at the speed of light.

What is spontaneous emission?

Spontaneous emission is a process in which an excited atom or molecule releases energy in the form of a photon without any external stimulation. This is a random and natural occurrence that happens when the atom or molecule returns to its ground state.

How does spontaneous emission relate to photons of intermediary energy?

Photons of intermediary energy can be emitted through spontaneous emission when an excited atom or molecule transitions to a lower energy state. The energy of the emitted photon will correspond to the energy difference between the two states.

What is the significance of photons of intermediary energy and spontaneous emission in quantum physics?

In quantum physics, photons of intermediary energy and spontaneous emission play a crucial role in understanding the behavior of atoms and molecules. These processes are essential for explaining how energy is transferred and how atoms and molecules interact with each other.

How are photons of intermediary energy and spontaneous emission used in practical applications?

Photons of intermediary energy and spontaneous emission have many practical applications, including in laser technology, medical imaging, and telecommunications. They are also used in spectroscopy techniques to study the energy levels and properties of atoms and molecules.

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