GR9277 #97
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Problem
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\prob{97}
Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and the wave number k is not the classical equation , where m is the electron mass. Instead, it is possible to use an effective mass given which of the following?
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Advanced Topics}Solid State Physics
This is a result one remembers by heart from a decent solid state physics course. It has to do with band gaps, which is basically the core of such a course.
Then again, one can easily derive it from scratch upon recalling some basic principles: , , where k is the wave vector, E is the energy, m is the mass, and p is the momentum.
From the above, one has .
Set the two 's equal to get . Cancel out the 's to get , after differentiating with respect to k on both sides.
Alternatively, one can try it Kittel's way:
Start with . Then, . Thus, the effective mass is defined by .
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Alternate Solutions |
drdoctor 2013-09-17 08:48:06 | Here's how I solved the problem:
Presumably, you want m* to be written in terms of some sort of constants--ie. m* shouldn't depend on k or E explicitly. However, or could potentially be constants, so they could be included in m*. So, that eliminates A, B, and C. It's fairly easy to see from here that answer D is simply the second derivative of E with respect to k for the equation given in the problem statement, except that you need to rearrange to solve for m after taking the double derivative:
Therefore:
You could also use units to eliminate E. | | jonestr 2005-11-12 00:50:59 | a quick dimnesional analysis wors well here
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Comments |
honeybunches 2016-09-12 20:04:35 | Here is a quick approach: If the dispersion relationship was the classical equation, we would expect the effective mass to be equal to the electron mass. This very quickly gets you (D) | | sina2 2013-10-08 04:01:01 | I will chose D. I'm always caring to not forget in quantum isn't same is . This is so important. They maybe don't commute. | | drdoctor 2013-09-17 08:48:06 | Here's how I solved the problem:
Presumably, you want m* to be written in terms of some sort of constants--ie. m* shouldn't depend on k or E explicitly. However, or could potentially be constants, so they could be included in m*. So, that eliminates A, B, and C. It's fairly easy to see from here that answer D is simply the second derivative of E with respect to k for the equation given in the problem statement, except that you need to rearrange to solve for m after taking the double derivative:
Therefore:
You could also use units to eliminate E. | | nitin 2006-11-21 00:45:56 | Let be the group velocity of the electron. Then
, and
where is the effective mass. The answer (D) follows. | | comorado 2006-10-27 13:07:45 | You wrote
Must be: | | jonestr 2005-11-12 00:50:59 | a quick dimnesional analysis wors well here
Jeremy 2007-11-03 15:33:07 |
Dimensional analysis will only narrow it down to choices (A) and (D).
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Poop Loops 2008-10-25 20:39:17 |
Yeah, but then there's a 1/2 out front, which doesn't make sense.
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ramparts 2009-08-06 23:30:04 |
Yep - it's possible I screwed something up (and I didn't bother looking at E after I looked at A through D :P) but I'm pretty sure the first three were not units of mass.
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alemsalem 2010-09-26 08:28:20 |
you might reason that the mass shouldn't be dependent on momentum (k) otherwise it wouldn't be useful to use an "effective mass" so it cannot be A.
but i admit when i did the exam i solved this one based on units then just picked A as a guess
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flyboy621 2010-10-23 05:26:41 |
Sort of, but that only narrows it down to A or D. Of course it's worth guessing at that point...
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asdfuogh 2011-10-05 19:04:12 |
I would pick D) if I had to pick from A) and D)... A) looks like the original mass equation, except with differentials. Not a great reason, but still..
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eighthlock 2013-09-15 12:39:01 |
Dimensional analysis does not distinguish between options (A) and (D)
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| | jonestr 2005-11-12 00:50:06 | | |
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