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Fermi Energy's plausible, but why define Fermi Temperature?

  1. Mar 17, 2015 #1
    I get that in a single particle of a metal, Fermi energy is defined at T = 0 as the maximum energy that electrons can reach.

    I get that, but my book defines this concept called Fermi Temperature.
    Is Fermi Temperature the temperature where electrons can reach the next empty energy band in support of heat energy, not by the direct influence of photons?

    Also, I would like to know if reaching Fermi temperature means the electrons will act upon Fermi-Dirac distribution rule(which I think means at that temperature electrons they would act like a regular gas, which otherwise they wouldn't.)

    (I looked up 'Fermi Energy' in wiki, and couldn't find the proper explanation for Fermi Temperature.)

    Thank you in advance.
     
  2. jcsd
  3. Mar 17, 2015 #2

    Simon Bridge

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  4. Mar 17, 2015 #3

    Thanks for your response but a simple formula like that means nothing to me when there is a specific parameter with no explanation given on what it is or what it does.
    What happens to the metal once it reaches to Fermi Temperature, is my question.
     
  5. Mar 17, 2015 #4

    Physics Monkey

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    It is helpful to compare the thermal de Broglie wavelength to the interparticle spacing. At high temperatures the kinetic energy is $$ \frac{p^2}{2m} \sim T $$ (this is just a classical particle formula), so the de Broglie wavelength is $$ \lambda = \frac{h}{p} \sim \frac{h}{\sqrt{Tm}}. $$ The classical approximation should be valid as long as the wavefunctions of the particles don't significantly overlap. Significant overlap will occur once the size of the wavefunction (de Broglie wavelength) is comparable to the interparticle spacing (set by the density). This occurs when when $$ \lambda \sim a = n^{-1/d} $$ with n the particle density and d the spatial dimension. If you solve this equation for T you will find that [itex] T \sim T_F [/itex] is roughly the temperature at which the gas transitions from being classical to quantum.
     
  6. Mar 17, 2015 #5
    Have you tried that formula? For a typical Fermi temperature in metals I get wavelengths of the order of 10^(-20) m.

    Fermi temperature is simply the temperature at which the particles in an ideal gas will have an average thermal energy equal with the Fermi energy.
    You cannot expect to have a metal at that temperature, so nothing will actually "happen" at the Fermi temperature.
     
  7. Mar 17, 2015 #6

    Simon Bridge

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    The Fermi temperature is just the name for the Fermi energy divided by the Boltzman constant... it is used because it crops up a lot in equations. It can be convenient. Reread the second link. Did you try using the PF search function and google as suggested?
     
  8. Mar 17, 2015 #7

    Physics Monkey

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    Unless I made a typo which I don't see, I think your calculation has an error. One should get length scales in the nm to angstrom range.

    I'm also not sure why you say you can't have a metal at temperatures near the Fermi temperature.
     
  9. Mar 17, 2015 #8
    Yes, now I see that you forgot a Boltzmann constant there.
    If you put in the right place you get indeed wavelengths of the order of angstroms or tens of angstroms.

    Fermi temperatures are of the order of a few tens of thousand K (over 10,000 K).
    The boiling point of tungsten is less than 6000 K. And is the highest. The melting point is below 4000 K.
    For common metals melting points are below 2000 K. Sure, the atoms of the metal may exist at the Fermi temperature, maybe in an ionized state, but this won't be a solid metal.
     
  10. Mar 18, 2015 #9

    I was already aware of the fact that a metal will evaporate even below Fermi temperature. But a metal in a form of gas would still be a metal, and now we can treat the metal as a gas.
    I'm wondering why it matters to be at that specific temperature even tho the metal would already be a gas at the temperature a lot less than Fermi temperature.
     
  11. Mar 18, 2015 #10

    I don't understand the notation you're using to describe the formulas.
    Maybe my firefox messed up the formulas.
     
  12. Mar 18, 2015 #11

    DrClaude

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    You're not treating the metal as a gas. You are treating the free electrons in the metal as a gas.

    The point is that electrons will show a quantum behavior for ##T < T_f##, and since ##T_f > T_\mathrm{melt}##, any solid metal will necessarily be in conditions where its electrons have to be considered as a quantum gas (compared to a classical gas).
     
  13. Mar 18, 2015 #12
    That's right! It's about the electrons! I'm an idiot.
    Thanks a lot !
    I don't really know the difference between a quantum gas and a classical gas, but I guess I can google it !
     
  14. Mar 18, 2015 #13

    DrClaude

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