Kittel Chapter 9 Solutions
Kittel Chapter 9 Solutions
Kittel Chapter 9 Solutions
Daniel Wolpert
9.1 Brillouin zones of rectangular lattice. Make a plot of the first two Brillouin zones of a primitive rectangular two-dimensional lattice with axes a, b=3a 2/a 2/3a
9.1 Brillouin zones of rectangular lattice. Make a plot of the first two Brillouin zones of a primitive rectangular two-dimensional lattice with axes a, b=3a
9.1 Brillouin zones of rectangular lattice. Make a plot of the first two Brillouin zones of a primitive rectangular two-dimensional lattice with axes a, b=3a 2/a 2/3a
2nd BZ First BZ
A two-dimensional metal has one atom of valency one in a simple rectangular primitive cell a = 2 A0 ; b = 4 A0.
a) Draw the first Brillouin zone. Give its dimensions in cm-1. b) Calculate the radius of the free electron fermi sphere. c) Draw this sphere to scale ona drawing of the first Brillouin zone.
2 * kF 2* 2 = N 2
4 * 0 2 4*2*(A )
2
=1
kF = 2 * A0 =
* 1012 cm 1
*1012 cm 1
* 1012 cm 1
Make another sketch to show the first few periods of the free electron band in the periodic zone scheme, for both the first and second energy bands. Assume there is a small energy gap at the zone boundary.
BZ periodic scheme
John Anzaldo
An open orbit in a monovalent tetragonal metal connects opposite faces of the boundary of a Brillouin zone. The faces are separated by
G = 2 108 cm 1.
A magnetic field B = 10 1T
is normal to the place of the open orbit. (a) What is the order of magnitude of the period of thek motion in
8 Take v = 10 cm / s
space?
9.5
From Eq. 25a we have , where I have decided to use SI units. v v v G d r q = e dt = h = ev B Letting we get = v , setting dk = G dt because v B since B is normal to the Fermi surface. Solving for gives Gh = . Plugging in the givens we evB get
v v dk dr v h = q B dt dt
Part b)
The electron will travel along the Fermi surface as shown. The velocity will change as the electron moves along the Fermi surface.
U(x)
-a/2
a/2 x
-U0
Chapter 9 Problem 7
Adam Gray
potassium on the free electron model. (b) What is the area in real space of the extremal orbit, for B = 10kG = 1T ?
Where
S = K
2 f
Plugging in:
1 2e ( ) = B hc(K 2 f )
1 2e ( ) = B hcK2 f
Note: The equation 34 was for cgs units, so all values used with this equation must be in this form. c=3x1010 cm/s h=1.05459x10-27 erg s e=4.803x10-10 erg1/2 cm1/2
This results in
1 ( ) = 5.55109 G 1 B
(b) To solve this part of the problem, go back to the equations we used for the cyclotron.
Be c = mc
Solve for r
r =
P = mv = hk
vf v f mc hk f c r = = = = c Be Be Be mc vf
Plugging in values from before and B=10kG r = 4.94x10-4 cm The orbit is circular, so the area is
r = 7.67 10 cm