GREPhysics.NET: GR9277 #97

GR9277 #97

Problem

GREPhysics.NET Official Solution

\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?

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

Advanced Topics $\Rightarrow$ }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: $E=p^2/(2m)=\hbar^2 k^2/(2m)$ , $p=\hbar k=mv$ , where k is the wave vector, E is the energy, m is the mass, and p is the momentum.

From the above, one has $dv/dt = \frac{1}{m}\frac{dp}{dt}=\frac{\hbar}{m}\frac{dk}{dt}$ .

$dE/dk = \hbar^2 k/m = \hbar p/m = \hbar v \Rightarrow v = \frac{1}{\hbar}\frac{dE}{dk} \Rightarrow dv/dt = \frac{1}{\hbar}\frac{d^2E}{dtdk}$

Set the two $dv/dt$ 's equal to get $\frac{\hbar}{m}\frac{dk}{dt}=\frac{1}{\hbar}\frac{d^2E}{dtdk}$ . Cancel out the $dt$ 's to get $\frac{\hbar^2}{m}dk=\frac{dE}{dk} \Rightarrow m = \hbar^2/(\frac{dE}{dk})$ , 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 $F=\hbar^2/(d^2E/dk^2) dv_g/dt=mdv_g/dt$ .

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. Reply to this comment
jonestr 2005-11-12 00:50:59	a quick dimnesional analysis wors well here Reply to this comment

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)

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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.

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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.

Reply to this comment

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.

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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: $\frac{\hbar^2}{m}d k=\frac{dE}{dk} \Rightarrow=\hbar^2(\frac{d^2E}{d^2k})$

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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)

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jonestr
2005-11-12 00:50:06

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