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The number of valence electrons in solid based on free electron theory, $N$ is given by

$$N = \frac{Vk_F^3}{3\pi^2} \tag{1}$$

  1. The $V$ in the formula, is it the volume of the whole, bulk crystal or the volume of one unit cell forming the crystal? I followed the proof towards (1), the $V$ is just $L^3$, and the $L$ is the $L$ used for the periodic boundary condition:

$$\psi(x) = \psi(x+L) \tag{2}$$

Crystal is described by unit cells, so the L in (2) should be the side length of the cubic unit cell, which means the V should be the volume of the unit cell forming the solid. However, the substitution of such a small volume into (1) will give an absurdly low number of valence electrons (e.g. 0.5).

  1. Actually what is the point of (1)? Say we have the electron density, n of material, and $k_F$ and n are related by:

$$k_F = \sqrt[3]{3\pi^2n}\tag{3}$$

we can use (3) to find the Fermi wavevector then use (1) to find N. However, if we already know n, can't we just find the total number of valence electrons in the system by multiplying n with the bulk volume?

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1 Answer 1

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If you're interested in the total number of electrons, then $V$ is the bulk volume of the crystal. You could treat $V$ as the volume of the unit cell, but then $N$ is the average number of electrons in that unit cell. Note that you can also derive this formula assuming the electrons live in a infinite potential square box with sides $L$ and then take $L$ to be large (e.g. see Griffiths Chapter 5.3.1).

Equation (3) is derived from Equation (1) by rearranging the terms and noting that $n = N/V$.

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