Why is the Fermi wavelength in a semiconductor larger than in a metal? I am a high school student, trying to better understand quantum point contacts. Would appreciate a simple explanation if possible.
Source: https://arxiv.org/abs/cond-mat/0512609

Since the conductance quantum $e^2/h$ contains only constants of nature, the conductance quantization might be expected to occur in metals as well as in semiconductors.A quantum point contact in a semiconductor is a mesoscopic object, on a scale intermediate between the macroscopic world of classical mechanics and the microscopic world of atoms and molecules.This separation of length scales exists because of the large Fermi wave length in a semiconductor.In a metal, on the contrary, the Fermi wave length is of the same order of magnitude as the atomic separation.A quantum point contact in a metal is therefore necessarily of atomic dimensions.

 A: Fermi wave length is $$\lambda_F=\frac{2\pi}{k_F}, \text{ where } k_F=\frac{1}{\hbar}\sqrt{2m^*\epsilon_F},$$
where the Fermi energy, $\epsilon_F$, is measured in respect to the bottom of the conduction band.
The Fermi energy of a (n-doped) semiconductor is typically rather close to the band edge. In fact, by controlling the doping, it can be made it as close to the band edge as we want. On the other hand, in metal, where the conduction band is partially filled with electrons, the Fermi energy is necessarily quite big.
Remarks:

*

*Note the difference between the Fermi level and Fermi energy, see here, which are often confounded.

*We are talking here about an n-dope semiconductor, which is essentially a metal.

A: The Fermi wavelength of a free electron gas is $$\lambda_F = \left( \frac{8\pi V}{3N} \right)^{\frac{1}{3}} \,,$$
so it is proportional to the separation of the electrons. In a semiconductor this distance is much larger than in a metal, for which the electron density is much higher.
