CH274 - Electrons in Solids and Materials Exercise 2.1
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CH274 - Electrons in Solids and Materials
Exercise 2.1 – Fermi level from free electron model
Use the following data to calculate the Fermi level EF for potassium, rubidium and
caesium.
(Hint: assume the validity of the free electron model, i.e. that EF = (32 n)2 / 3 )
Model answer
In the free electron model, the Fermi level is expressed as
EF = (32 n)2 / 3 ,
where n is the number density of free electrons.
Potassium has one electron in its outer 4s shell which can be considered to be nearly free in the solid state. Therefore, it can be supposed that each atom contributes with one electron to the density of free electrons, implying that
nK = = 1 . ,
where Nelectrons is the number of electrons contained in the volume V and Natoms is the number of atoms contained in the same volume. In this problem we are given the mass density pand must determine the atomic number density, Natoms/V, from it. In order to do so we can express
M
where M is the mass contained in the volume V, and write
M = matom . Natoms ,
where matom is the mass of a single atom. This latter quantity can be expressed as
m = ma ,
A
where NA is the Avogadro number and ma is the atomic mass.
So
p = = matom . = . , and = p. .
As a consequence,
nK = p = 862 3一3 = 1.331028 m-3,
and
E1(一)0(34))2一31 (3几2 )2 / 3 (1.33 10 28 )2 / 3 = 5.79 10 一38 (1.33 10 28 )2 / 3 =
= 3.25×10- 19 J = 2.03 eV.
Rubidium has one electron in its outer 5s shell which can be considered to be free in the solid state. Therefore,
nearly
nRb = p = 1532 3一3 = 1.081028 m-3 and
EF ,Rb = 5.7910一38 (1.061028 )2 / 3 = 2.83×10- 19 J = 1.77 eV.
Finally Cesium has one electron in its outer 6s shell which can be considered to be nearly free in the solid state. Therefore,
nCs = p = 1873 3一3 = 8.481027 m-3 and
EF ,Cs = 5.7910一38 (8.481027 )2 / 3 = 2.41×10- 19 J = 1.50 eV.
2022-09-17
Fermi level from free electron model