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  GR9277 #23
Problem
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\prob{23}
The Fermi temperature of Cu is about 80,000 K. Which of the following is most nearly equal to the average speed of a conduction electron in Cu?

  1. 2E-2 m/s
  2. 2m/s
  3. 2E2 m/s
  4. 2E4 m/s
  5. 2E6 m/s

Statistical Mechanics}Fermi Temperature

(Much of the stuff I classified as Stat Mech might also be considered Condensed Matter or Solid State Physics. They are classified as thus because the Stat Mech book I mentioned in the booklist on the site http://grephysics.yosunism.com is perhaps the best intro to all this.)

The Fermi velocity is related by \epsilon_F=k T_F =\frac{1}{2}m v^2\Rightarrow v=\sqrt{\frac{2kT_F}{m}}, where \epsilon_F is the fermi energy, and T_F is the Fermi temperature.

One should know by heart the following quantities, k=1.381E-23 and m=9.11E-31 (but then again, they are also given in the table of constants included with the exam). Plug these numbers into the expression above to find v,

v=\sqrt{\frac{2kT_F}{m}}\approx \sqrt{\frac{2\times 1.4 E-23 \times 8E5}{9E-31}}=\sqrt{\frac{2.8 E-23 \times 8E5}{9E-31}}\approx\sqrt{0.3 E8 \times 8E5}=\sqrt{24/10 E13}\approx \sqrt{10^{12}}=10^6,
the choice that comes closest to this order is choice (E).

Note that the hardest part of this problem is the approximation bit. No calculators allowed. Sadness.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
claire
2009-09-12 11:31:59
If you don't want to have to bother with the big numbers for the approximation, you can use your knowledge that for room temperature (~300K) you have kT=\frac{1}{40}eV (this is a useful thing to know). And you know that an electron mass is .5MeV/c^2.

The problem uses 80,000K, but do it for 90,000, since thats easier with the numbers you know.
Then kT~\frac{30}{4}, so v_{f}=\sqrt{\frac{30eV\times9\times10^{16}\frac{m^2}{s^2}}{10^6eV}} ~ \sqrt{2.7\times 10^{12}} ~ 2 \times 10^6 m/s, option (E).
Alternate Solution - Unverified
Comments
rwmurphy
2012-11-08 12:58:40
I have a slight issue with the wording of this problem: it only states that the Fermi temperature of copper is at 80000K, not that the copper is actually at the fermi temperature. I mistakenly thought the ETS was just trying to be tricky and asking for a reasonable drift velocity in Cu, which then (A) would be the answer.

*sigh*

When I look at it now, it easy to assume that they mean copper is at that temperature, but under a crunch for time, your brain does funny things! Either way, they should make it less ambiguous.
NSF Fellow
2013-09-18 11:13:11
It says conducting electrons! they are at fermi energy!

NoPhysicist3
2017-03-23 09:38:59
same sh*t man
NoPhysicist3
2017-03-23 09:42:00
but as i noticed GRE problems tend to use all information given in the task
NEC
Quark
2011-10-25 14:35:18
Why do we use kT for the energy and not \frac{3}{2}kT for the three translational degrees of freedom in finding the kinetic energy?
liliapunto
2013-07-14 03:15:52
This is exactly what I did and what I want to ask about.
Prufrock
2013-09-03 15:04:01
Because condensed matter physicists don't recognize factors as small as 3/2 - and in this problem it is certainly true that the 3/2 is insignificant.
NSF Fellow
2013-09-18 11:14:37
it's not a gas!
NSF Fellow
2013-09-18 13:08:25
and it says the Fermi temperature of Cu, not the temperature Cu is at.
NEC
neon37
2010-11-11 03:59:18
Here we can clearly get away with order of magnitude approximation, so dont bother about any numbers except 10^n ones. You will get the answer in less than a minute. NEC
nyuko
2009-10-30 10:10:08
I did this problem in seconds. I just remember the typical Fermi speed for electron in metals is of the order \10^{6}m/s
I think there are some typical values for physicists to remember. Fermi speed of electrons in metal is one.
NEC
claire
2009-09-12 11:31:59
If you don't want to have to bother with the big numbers for the approximation, you can use your knowledge that for room temperature (~300K) you have kT=\frac{1}{40}eV (this is a useful thing to know). And you know that an electron mass is .5MeV/c^2.

The problem uses 80,000K, but do it for 90,000, since thats easier with the numbers you know.
Then kT~\frac{30}{4}, so v_{f}=\sqrt{\frac{30eV\times9\times10^{16}\frac{m^2}{s^2}}{10^6eV}} ~ \sqrt{2.7\times 10^{12}} ~ 2 \times 10^6 m/s, option (E).
Alternate Solution - Unverified
ramparts
2009-08-02 17:36:13
I did this much more quickly by just looking at orders of magnitudes. Ignoring dimensionless constants (which will never change the answer by a factor of 2), temperature depends on the energy divided by the Boltzmann constant, and the energy is roughly kinetic energy, mv^2. So we have

v \sim \sqrt{\frac{T}{k m_e}}

Plug in the appropriate factors of 10, and you get 10^{6.5}. Excellent.

I have no idea if the physics is right, but the dimensions work out and the answer is right ;)
Albert
2009-11-03 05:05:17
Yes, your method is really bad. Not only your physics is abominable, your math is plain wrong too. You got the formula wrong and that's besides the fact that you did basically the same thing as Yosun, just didn't write the steps. And then got it all mixed up. I wonder how you ever reached the answer (or did you?). Sorry, but it's important that truth be spoken.
NEC
none
2008-10-19 23:01:52
If you write things a bit differently you get \sqrt{x.y \times 10^{12}} = \sqrt{x.y} \times 10^6. The only answer with the right order of magnitude is (E)NEC
student2008
2008-10-14 09:58:44
Actually, ETS means the average velocity modulus, \langle v\rangle=\sqrt{\frac{8kT}{\pi m_e}}\approx 1.75\cdot 10^{6}\ \frac{m}{s}, which can be obtained using the Maxwell distribution. Though such rigorous formula isn't relevant for the real exam, of course.
student2008
2008-10-14 10:08:31
Even more precise would be T=\langle \epsilon_{kin}\rangle =3/5 \epsilon_F and \langle v\rangle \approx 1.36\cdot 10^6 \ \frac ms, but apparantely it's not for the exam :)
student2008
2008-10-15 00:14:53
Stupid me, all I wrote above is wrong. In fact, the concept of Fermi-gas (and Fermi-energy) of electrons requires T \ll T_F, not T = \frac35T_F. What is relevant here is the low-temperature Fermi distribution of electrons f(p)\simeq\theta(p_F - p) (which is true up to melting temperatures: T_{melt}\sim 10^3 K \ll 80 000 K), not the Maxwell distribution. However, the problem asks about the mean speed of the \bf conduction electrons. And these are the electrons near the Fermi surface, since only their energy may change (under the not-extremely-high voltage). So, Yosun's solution is right, with this reservation.
NEC
prismofmoonlight
2007-10-31 11:22:40
I think T_F is 8E4, not 8E5 as given in your equation (though the change is not enough to affect the approximation too terribly much).

Thanks for the great site!
NEC

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