The Fermi temperature of a solid is related to Fermi energy by relation $$ { E }_{ F } ={ k }_{ B }\times{ T }_{ F } $$ where $ { k }_{ B } $ is Boltzmann constant. But what is the significance of Fermi temperature?

  • 4
    $\begingroup$ "The Fermi temperature can be thought of as the temperature at which thermal effects are comparable to quantum effects associated with Fermi statistics". Source: wikipedia article on the Fermi Energy. Does this answer your question? $\endgroup$ – AccidentalFourierTransform Nov 2 '15 at 18:24
  • 2
    $\begingroup$ Hi, I am guessing you have already read this: en.wikipedia.org/wiki/Fermi_energy $\endgroup$ – user81619 Nov 2 '15 at 18:25
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient research effort. $\endgroup$ – Danu Mar 18 '16 at 10:33

If you want to decide whether a gas of fermions is degenerate, then you would compare the temperature of the gas with its Fermi temperature. If $T \ll T_F$ then the gas can be considered completely degenerate. If $T \sim T_F$ then the gas is partially degenerate. If $T > T_F$ then the gas is not degenerate.


When we measure the temperature of a material, we do not typically measure the temperature of a single atom or electron. What we measure is the average temperature of the material. There is invariably going to be a distribution of energy within the material. In this distribution, an extremely small thermal mass, consisting of a very small fraction of the nearly free electrons (which is itself a very small fraction of the total electrons in the system), is at the Fermi energy, and the temperature corresponding to that energy is the relatively high Fermi temperature. Therefore the high Fermi temperature is not inconsistent with the low temperature or the solid as a whole.

Reference: http://nptel.ac.in/courses/113106040/Lecture25.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.