GR 8677927796770177 | # Login | Register

GR9677 #99
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Quantum Mechanics$\Rightarrow$}Two-State Systems

Recall the mnemonic for remembering what a LASER is---Light Amplified Stimulated Emission Radiation.

A laser consists of two states, with a metastable-state in between the top and bottom state. Initially, one has all the atoms in the ground-state. But, photons come in to excite the atoms (through absorption), and the atoms jump into the top state; this is called a population inversion, as the ground-state atoms are now mostly in the top inverted" state. More photons come in to excite these already excited atoms, but instead of absorption, emission occurs, and the atoms jump to a lower meta-stable state while emitting photons (in addition to the incident photons). The atoms stay in this metastable state due to selection rules, where a transition back to the ground-state is forbidden.

One doesn't need to know all that to solve this problem. Instead, merely the idea of a laser requiring two main states and a metastable state in between would suffice. Since the question gives the bottom state as $n=1$ and top state as $n=3$, one deduces that the metastable state must be $n=2$, as in choice (B).

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
hanin
2009-09-30 02:46:12
So if the transition is, for example, n=1 to n=4 then n=2 and n=3 have to be metastable?
casaubon
2008-10-09 00:02:30
So I still don't get links on this site. Shows up fine in the preview. Anyways,

http://idol.union.edu/malekis/ESC24/Seyffie%27s%20Pages/Quantum%20Applications/threelevelatom.gif
2007-10-30 17:59:51
I don't understand why the system must have an n=2 level, the selection rules do not forbid the electron from jumping straight from n=1, l=0, m=0 to say n=3, l=1, m=0rnwithout ever touching the n=2
 Ethan2008-10-08 19:50:28 I believe the explanation goes as follows: in order for a laser too work, ie amplify the light put in, there must be an excess of electrons in the high energy state. N(n=3)>N(n=1). If n=2 were not metastable then it would not be possible to maintain this excess concentration in n=3, as the electrons would fill up the n=1 level too quickly. In multiple gas lasers (I think the most common) the active gas need not have any metastable state.
 casaubon2008-10-09 00:01:58 Almost. The excess population exists at n=2 (the metastable state). The laser line is the n=2 to n=1 transition. See " target="_blank">http://idol.union.edu/malekis/ESC24/Seyffie%27s%20Pages/Quantum%20Applications/threelevelatom.gif">See this figure.
Richard
2007-09-28 16:45:33
Sorry to be punctilious, but "Light Amplification by Stimulated Emission" is hardly a mnemonic. It is an acronym. Do you have to remember that PHRASE to remember the word LASER?

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$