Simple frequency quantum question

[hvthvt]

Homework Statement

An electron (mass m) is contained in a cubical box of widths Lx = Ly = Lz. (a) How many different
frequencies of light could the electron emit or absorb if it makes a transition between a pair of the lowest five energy levels? What multiple of h2/8mL2 gives the (b) lowest, (c) second lowest, (d) third lowest, (e) highest, (f) second highest, and (g) third highest frequency?

Homework Equations

h/2m
E0=π2ℏ2/2mL2

The Attempt at a Solution

First we have E(111)? Which is E0 * 3
Then we have E(112?) Which is E0 * 6?
How should I continue? I do now understand how I can decide how many different frequencies of light there could be. I know that (Lx^ 2 + Ly^ 2 + Lz^2) is important But how to deal with this??

 

[collinsmark]

Homework Statement

An electron (mass m) is contained in a cubical box of widths Lx = Ly = Lz. (a) How many different
frequencies of light could the electron emit or absorb if it makes a transition between a pair of the lowest five energy levels? What multiple of h2/8mL2 gives the (b) lowest, (c) second lowest, (d) third lowest, (e) highest, (f) second highest, and (g) third highest frequency?

Homework Equations

h/2m
E0=π2ℏ2/2mL2

You really should be more careful with how you write your equations. The way you wrote it it is ambiguous whether you are squaring a variable or multiplying it by 2. Also, in your energy equation, you left out the n_x, n_y, and n_z part.

I'm pretty sure you meant to write:

E = \frac{\pi^2 \hbar^2}{2 m L^2} \left( \mathrm{times \ something \ with} \ n_x, n_y \ \mathrm{and} \ n_z \right)

The Attempt at a Solution

First we have E(111)? Which is E0 * 3
Then we have E(112?) Which is E0 * 6?

So far so good! Keep going with that idea. You'll notice that there are many degenerate energy states (this is mainly the result of L_x = L_y = L_z). For example, E(121) and E(211) all have the same energy as E(112). (Degenerate refers to different states having the same energy level.)

Thus far, even though you've examined several states, you have only found two distinct energy levels, because some of the states are degenerate.

So keep on going until you find the five lowest, distinct energy states. Plug in different combinations of positive integers until you find new, low numbers that you haven't already found. Be careful though, sometimes it's not obvious which number combinations will be lower until you try them out.

Once you have the five lowest distinct energy states, you need to find the differences between the energy states. In other words, you need to subtract one energy level from another to determine the transition energy.

Keep in mind that some of these transitions might produce the same energy difference as other transitions. That's another form of degeneracy.

Also, before we finish, do this and it will help you out:

Get rid of your \hbar by replacing it with \hbar = \frac{h}{2 \pi}.

Then replace your energies with frequencies by noting that E = hf \ \ \rightarrow \ \ f = \frac{E}{h}.

Trust me on this. It will make the second part of the problem make a lot more sense.

How should I continue? I do now understand how I can decide how many different frequencies of light there could be. I know that (Lx^ 2 + Ly^ 2 + Lz^2) is important But how to deal with this??

You don't need to worry about the ([STRIKE]L_x² + L_y² + L_z²[/STRIKE] Oops -- fixed this in edits:) 1/L_x², 1/L_y², 1/L_z² because L_x = L_y = L_z = L. That's already in your formula for energy. (Had the problem had different lengths in different dimensions, it would make this a whole different problem due to differences in degeneracy.)

On the other hand, you still need to work with the n_x, n_y, and n_z numbers.

[Edit: fixed the 1/L_x², 1/L_y², 1/L_z² mistake]

 

[Simon Bridge]

See also: http://home.comcast.net/~szemengtan/ "Statistical Mechanics" set.
1st page "Quantum Statistical Mechanics" I think... you may need to refer back through the notes for the terms.
You seem to be on the right path - keep going. As collinsmark points out, the key to this one is to play around with the numbers.