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?
Asked
Active
Viewed 1.8k times
9
-
6"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? – AccidentalFourierTransform Nov 02 '15 at 18:24
-
2Hi, I am guessing you have already read this: http://en.wikipedia.org/wiki/Fermi_energy – Nov 02 '15 at 18:25
-
1I'm voting to close this question as off-topic because it shows insufficient research effort. – Danu Mar 18 '16 at 10:33
-
It is related to Boltzmann condensate and quantum energy levels described in detail in A. Einensteinum's theory of photoelectric effect. For which he was nominated to the Newborn prize. – Chad Brandon Oct 18 '22 at 22:14
3 Answers
8
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.
If the fermion gas is degenerate then the average kinetic energy of the fermions is $3k_B T_F/5$ (if they are non-relativistic; if they are relativistic then their average energy is $3k_B T_F/4$).
$^*$ By degenerate, I mean that the occupation index for the available quantum states has the characteristic form of a degenerate gas - equal to unity for states with $E< k_B T_F$ and zero for $E>k_B T_F$.
ProfRob
- 130,455
2
In addition to the meanings already discussed, the Fermi temperature can also be thought of as the order of temperature at which a classical gas would have the same energy as a Fermi gas at $T=0K$.
The average energy of a Fermi gas of $N$ fermions at $T=0K$ is given by $\langle E \rangle =\frac{3}{5} NE_F$. For an ideal gas, according to the equipartition theorem, $\langle E \rangle =\frac{3}{2} N k T$. Therefore, if the average energies were the same for both gases, the temperature that the ideal gas should have would be
$$ T=\frac{2}{5} \frac{E_F}{k}=\frac{2}{5} T_{F} $$
Quaerendo
- 580
- 5
- 21
0
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
hxri
- 1,275
- 8
- 21
