First excite electrons into a higher state for lasers

[VVS]

Hi,
Could somebody please explain or refer to a good page which explains why it is necessary to first excite electrons into a higher state than the metastable state in LASERS? I checked so many sources but did not find anything.
thank you
VVS

 

[lightarrow]

Hi,
Could somebody please explain or refer to a good page which explains why it is necessary to first excite electrons into a higher state than the metastable state in LASERS? I checked so many sources but did not find anything.
thank you
VVS

I'm not sure but I think it's because in that way you can have a greater population inversion exciting the system with a common source; otherwise you should use a source of em energy with exactly the same frequency of the laser beam; in practice, you should use another similar laser to excite that one; not very useful!
With a white lamp (for example), you can instead excite the laser system to various, different, frequencies, higher than the lasing one; those excited levels then goes down (very fast) to the metastable one, which last much longer and so in this way you can fill it much more efficiently.
Just an idea, however, better to wait for confirmation.

 

[Cthugha]

Hi,
Could somebody please explain or refer to a good page which explains why it is necessary to first excite electrons into a higher state than the metastable state in LASERS? I checked so many sources but did not find anything.
thank you
VVS

You need to create a population inversion in order to start the lasing process. If you wanted to create a two level lasing system, you would not reach such an inversion because of your excitation being resonant with the lasing transition. So you would not only pump the system, but also cause stimulated emission (as your excitation energy drives both transitions at once) before the system is inverted, which prevents you from reaching inversion.

 

[VVS]

Oh ok thanx, that was very helpful.
I just have one more doubt:
When we pump electrons into the high state won't most of them just drop down to the lowest energy rather than the metastable one??
cheers again

 

[jdg812]

Oh ok thanx, that was very helpful.
I just have one more doubt:
When we pump electrons into the high state won't most of them just drop down to the lowest energy rather than the metastable one??
cheers again

Every transition has "branching ratio", which shows % of decays from high state to lowest or metastable state.

 

[VVS]

ok cheers, thanks =)

 

[VVS]

oh sorry another doubt. Since we are forcing the electrons into exited states and some of the electrons fall into the metastable state. Won't that cause the emission of the light of a different wavelength besides the one we want? I mean we won't have stimulated Emission, but it is still there won't it?

 

[Claude Bile]

Electrons can lose energy by other means other than radiating a photon, such as by colliding with another atom, or through the excitation of phonons in a solid state medium.

Typically electrons undergo lots of small transitions from the pump level to the metastable level, rather than one big one - this is also conducive for non-radiative transitions to dominate.

Claude.

 

[jdg812]

Typically electrons undergo lots of small transitions from the pump level to the metastable level, rather than one big one

What do you mean? Assume we have N atoms excited to the upper level, a*N of them may decay to lower level due to optical transitions and (1 - a)*N of them decay due to non-radiative (collisions with another atoms) transitions. Now, consider exactly one atom in upper state. May you apply your quoted statement to ONE excited atom?