- #1

BRN

- 108

- 10

## Homework Statement

The conduction band of a hypothetical crystal of one-dimensional Cesium reticular with step a=300 pm (1 atom per cell) is characterized by the ε dispersion law

##\epsilon (k) = V_0 + \frac{\hbar^2}{m_e}(\frac{1}{2}k^2 - \frac{a}{3\pi}|k|^3##

where ##V_0 = -4 eV##, is set so that the energy of a stationary electron in a vacuum on the outside of the crystal appears ε = 0. At zero temperature, determine the position of the Fermi level of the electrons of this crystal (one electron per atom in the conduction band) and the value of the work function. A finite temperature (but small compared to the Fermi temperature) the chemical potential is greater or less than the Fermi level?

## The Attempt at a Solution

The Fermi energy level is the highest occupied level in an fermions system at T=0. The dispersion law describes the shape of the conduction band and the Fermi level corresponds to ##\epsilon (k_F)##.

Then, I determine density of states ##g(k)## differentiating with respect k the number of states with wave vector ≥ k (which corresponds to the volume of a sphere of radius

##|\vec{n}|=\frac{V^{1/3}}{2\pi}|\vec{k}|##

##g(k)=g_s\frac{d}{dk}(\frac{4}{3}\pi \frac{V}{8\pi^3}k^3=g_s\frac{V}{2\pi}k^2##

Total number of electrons is:

##N=\int_{0}^{k_F}g(k) dk=\int_{0}^{k_F}g_s\frac{V}{2\pi}k^2 dk=g_s\frac{V}{6\pi^2}k_F^3##

with ##g_s=2##, I have:

##k_F=(3\pi^2\frac{N}{V})^{1/3}##

Crystaline structure is an simple cubic cell, then

##\frac{N}{V}=\frac{1/8*8}{a^3}=\frac{1}{a^3}##

so, ##k_F=1.0312*10^{10}[m^{-1}]##

and ##\epsilon (k_F)=-4.1704*10^{-19}[J]=-2.6023[eV]##

Work function is the minimum energy for extract an electron from metal and is the opposite of Fermi energy:

##W=-\epsilon (k_F)=2.6023[eV]##

Solutions: ##\epsilon_F = -3.304 eV##; W = 3.304 eV; μ remains constant (neither increases nor decreases), since the band dispersion happens to be symmetric around the Fermi point.

What is wrong?