I wanted to start a discussion regarding the question posed at the end of the last class regarding how many states would be filled in a system of N atoms and N free electrons.
My thoughts are that N states would be filled when disregarding spin. I think this is sort of analogous to the problem we solved for the normal modes of masses connected by springs (although maybe it isn't). In the normal mode situation we found that for N degrees of freedom there are N normal modes and I think in the case of states if we have N atoms and N free electrons we will have we can kind of think of it as N degrees of freedom so N number of states filled. (I don't really like this explanation so if you have any better ideas please post them!)
Now when we include spin I think we will find that there will be N/2 states filled. I think that each electron has two possible states that it can fill. One of those being a shared state with another electron, but with opposite spin to the electron whose state it's sharing. If the probability of the electron occupying the same state as another electron is 1/2 which I think it is in our current model in which we disregard electron electron interaction than the number of states filled should be N/2.
Does this sound reasonable to you guys?
I think you're really close George. I didn't really have a good understanding as to why there would be N states but I think you're on the right track. It seems reasonable for the number of values of q to correspond to the number of states and your expression for q seems to give N possible values of N.
ReplyDeleteIsn't q quantized. Let's say you have N atoms each a distance a apart. (L=Na). What are the allowed values of q? (If we understood that before taking the N--> infinity limit maybe that would help?
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