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GR9277 #97
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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?

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 .  Alternate Solutions
 drdoctor2013-09-17 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. Reply to this comment jonestr2005-11-12 00:50:59 a quick dimnesional analysis wors well here Reply to this comment 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. 2006-10-27 13:07:45
You wrote

Must be: jonestr
2005-11-12 00:50:59
a quick dimnesional analysis wors well here

 Jeremy2007-11-03 15:33:07 Dimensional analysis will only narrow it down to choices (A) and (D).
 Poop Loops2008-10-25 20:39:17 Yeah, but then there's a 1/2 out front, which doesn't make sense.
 ramparts2009-08-06 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.
 alemsalem2010-09-26 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
 flyboy6212010-10-23 05:26:41 Sort of, but that only narrows it down to A or D. Of course it's worth guessing at that point...
 asdfuogh2011-10-05 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..
 eighthlock2013-09-15 12:39:01 Dimensional analysis does not distinguish between options (A) and (D) jonestr
2005-11-12 00:50:06      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$