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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 $E=\hbar^2k^2/2m$, where m is the electron mass. Instead, it is possible to use an effective mass $m^*$ given which of the following?

  1. $m^* = \frac{1}{2}\hbar^2 k \left(\frac{dk}{dE}\right)$
  2. $m^* = \frac{\hbar^2 k }{\left(\frac{dk}{dE}\right)}$
  3. $m^* = \hbar^2 k \left(\frac{d^2E}{dk^2}\right)^{1/3}$
  4. $m^* = \frac{\hbar^2}{\left(\frac{d^2E}{dk^2}\right)}$
  5. $m^* = \frac{1}{2} \hbar^2m^2 \left(\frac{d^2E}{dk^2}\right)$

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 \hbar v_g=dE/dk. Then, dv_g/dt = \hbar^{-1}(d^2E/dt^2dk/dt)= \hbar^{-1}(d^2E/dt^2F/\hbar). Thus, the effective mass is defined by F=\hbar^2/(d^2E/dk^2) dv_g/dt=mdv_g/dt.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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, dE/dK or d^2E/dk^2 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:
(d^2/dk^2)E = \hbar^2/m
Therefore:
m* = \hbar^2/(d^2E/dk^2)
You could also use units to eliminate E.
Alternate Solution - Unverified
jonestr
2005-11-12 00:50:59
a quick dimnesional analysis wors well here

Alternate Solution - Unverified
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)NEC
sina2
2013-10-08 04:01:01
I will chose D. I'm always caring to not forget in quantum \frac { dE }{ dk } isn't same is \frac {1}{\frac { dk }{ dE }}. This is so important. They maybe don't commute.NEC
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, dE/dK or d^2E/dk^2 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:
(d^2/dk^2)E = \hbar^2/m
Therefore:
m* = \hbar^2/(d^2E/dk^2)
You could also use units to eliminate E.
Alternate Solution - Unverified
nitin
2006-11-21 00:45:56
Let v_g=\frac{d\omega(k)}{dk} be the group velocity of the electron. Then

v_g=\frac{1}{\hbar}\left(\frac{dE(k)}{dk}\right), and

\begin{eqnarray*} \frac{dv_g}{dt}&=&\frac{1}{\hbar}\left(\frac{d^2E(k)}{dk^2}\frac{dk}{dt}\right)\\&=&\frac{1}{\hbar}\left(\frac{d^2E(k)}{dk^2}\right)\left(\frac{1}{\hbar}\frac{dp}{dt}\right)\\&=&\frac{1}{\hbar}\left(\frac{d^2E(k)}{dk^2}\right)\left(\frac{m^*}{\hbar}\frac{dv_g}{dt}\right),\end{eqnarray*}

where m^* is the effective mass. The answer (D) follows.
NEC
comorado
2006-10-27 13:07:45
You wrote \frac{\hbar^2}{m}d k=\frac{dE}{dk} \Rightarrow=\hbar^2(\frac{dE}{dk})

Must be: \frac{\hbar^2}{m}d k=\frac{dE}{dk}  \Rightarrow=\hbar^2(\frac{d^2E}{d^2k})
Typo Alert!
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).
Poop Loops
2008-10-25 20:39:17
Yeah, but then there's a 1/2 out front, which doesn't make sense.
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.
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
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...
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..
eighthlock
2013-09-15 12:39:01
Dimensional analysis does not distinguish between options (A) and (D)
Alternate Solution - Unverified
jonestr
2005-11-12 00:50:06
NEC

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