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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 $E=\hbar^2k^2/2m$, 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$\Rightarrow$}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
 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, $dE/dK$ 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 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 $\frac { dE }{ dk }$ isn't same is $\frac {1}{\frac { dk }{ dE }}$. 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, $dE/dK$ 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.
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.
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: $\frac{\hbar^2}{m}d k=\frac{dE}{dk} \Rightarrow=\hbar^2(\frac{d^2E}{d^2k})$
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 $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$