1. Watch the two videos on Boch states and post comments to that post about something related to the videos.
2. Make a wave function graph for \(k=3\pi\)/(4a). Take a picture of your graph and email it to me.
3. This is a "comment question". You can post your thoughts here in the comments. You can say: " I have no idea..." or whatever reflects your thoughts on this problem. We can all work on it together here.
a) Estimate, guess or calculate the energy of the ground state of a square well 0.4 nm wide and 10 eV deep. In units of eV, post your guess, estimate or calculated result in here as a comment. We can perhaps discuss and refine this. (Try using \(\hbar c= 197.3\) eV-nm and mc^2=.511 x 10^6 eV.)
Participate! Have fun.
b) how about the 1st excited state?
I am going to add a comment here to model what I am looking for from you. You might say, for example:
(Spoiler Alert)
For a 0.4 nm wide well the infinite square well ground state energy would be at about 2.35 eV (from \(\hbar^2 c^2 \pi^2)/[2mc^2 L^2]\)). I guess for a finite well the ground state would be a bit lower. Maybe about …?
And the first excited state for an infinite square well would be at 4 times that, the next state 9 times, etc…
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ReplyDeleteI really don't know, I keep getting a small number of order 10^-34. For both a and b.
ReplyDeleteI think what you got seems pretty reasonable. Maybe I did it wrong. I'll put a picture of my calculation in the bottom of the post.
ReplyDeleteAlso, I think maybe you got E1 and E3, cause they differ by a little less than 9. The 9 suggests 1, 3 correspondence. And that it is less than 9 suggests hat maybe the excited state, at an energy near the barrier energy, has a energy shifted lower than the corresponding infinite square well state even more than the ground state. Does that make sense at all?
PS. If you would like to send me a screenshot of your wolfram alpha calculation and graph, I can add it to the post.
ReplyDeletePPS. This site supports MathJax, which is essentially Latex.
I think you are are correct that it is negative. The eqn in the spoiler alert is referenced to the bottom of the well, like how much above that is it. So really when you get 1.28 eV that would mean -10 eV + 1.28 eV = -8.72 eV. I was definitely too careless about that.
ReplyDelete. Well, first, it is really good that you are looking at units. That is really helpful in quantum!
ReplyDeleteThey don't all cancel out. in the numerator you have eV^2 (from the (hbar c)^2