Kittel Chapter 9 Solutions

23
Homework # 8 Chapter 9 Kittel Phys 175A Dr. Ray Kwok SJSU

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Solution Manual to Kittel's Solid State Physics introductory textbook!

Transcript of Kittel Chapter 9 Solutions

Page 1: Kittel Chapter 9 Solutions

Homework # 8

Chapter 9 KittelPhys 175A

Dr. Ray Kwok

SJSU

Page 2: Kittel Chapter 9 Solutions

2π/a

2π/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

Prob. 1 – Brillouin zones of

rectangular lattice Daniel Wolpert

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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

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2π/3a

2π/a

First BZ

2nd BZ

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

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This is a Wigner-Seitz cell.

Prob. 2 – Brillouin zone,rectangular latticeGregory Kaminsky

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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 it’s 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.

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Calculation of the radius of

the Fermi sphere

2

4

4 2

1

2

2

0 2

* *

*

* *( )

π

π

k

A

F

=

kA

cmF = =−

π π

2 2100

12 1

**

22

2

2**π

π

k

L

NF

=

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π

21012 1* cm−

π *1012 1

cm−

Radius of free electron fermi sphere = π

2

Brilloin zone

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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.

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This is the first energy band

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Second energy band

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Prob. 4 – Brillouin Zones of Two-Dimensional Divalent Metal

Victor Chikhani

A two dimensional metal in the form of a square lattice has two conduction electrons per atom. In

the almost free electron approximation, sketch carefully the electron and hole energy surfaces. For

the electrons choose a zone scheme such that the Fermi surface is shown as closed.

Hole Energy surface

Electron Energy Surface

BZ periodic schemeSecond Zone periodic scheme

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Prob. 5 – Open Orbits

An open orbit in a monovalent tetragonal metal connects

opposite faces of the boundary of a Brillouin zone. The

faces are separated by . A magnetic field

is normal to the place of the open orbit. (a) What is the

order of magnitude of the period of the motion in space?

Take (b) Describe in real space the motion

of an electron on this orbit in the presence of the magnetic

field.

18102 −×= cmG TB

110−=

k

scmv /108=

John Anzaldo

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9.5

� From Eq. 25a we have , where I have decided to use SI units.

� Letting , setting we get

because since is normal to the Fermi surface.

� Solving for gives . Plugging in the givens we get

� Part b)

Bdt

rdq

dt

kd vvv

h ×=

vdt

rd=

v

Gkd =v

τ=dteq −= BevG v

h −=τ

Bv ⊥ B

τ τ=evB

Gh

The electron will travel along the Fermi surface

as shown. The velocity will change as the

electron moves along the Fermi surface.

s

sC

kg

cm

m

s

cmC

s

mkg

m

cm

cm

evB

G 10

1819

234

8

10315.1

10100

11010602.12

1062.6100102

−−

⋅=

⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅

=

π

h

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Mike Tuffley

5/12/09

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-a/2 a/2

-U0

x

U(x)

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Chapter 9 Problem 7 Adam Gray

(a) Calculate the period expected for

potassium on the free electron model.

(b) What is the area in real space of the extremal

orbit, for B = 10kG = 1T ?

)1

(B

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Starting with equation 34:

Where

Using Table 6.1 on pg. 139, for potassium we find

kf=0.75x108cm-1 .

cS

e

B h

π2)

1( =∆

2

fKS π=

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Plugging in:

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

)(

2)

1(

2

fKc

e

B π

π

h=∆

2

2)

1(

fcK

e

B h=∆

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This results in

(b) To solve this part of the problem, go back to

the equations we used for the cyclotron.

Solve for r

191055.5)1

( −−×=∆ G

B

kmvP h==c

fvr

ω=

mc

Bec =ω

Be

ck

Be

mcv

mc

Be

vvr

fff

c

f h==

==

ω

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Plugging in values from before and B=10kG

r = 4.94x10-4 cm

The orbit is circular, so the area is

272 cm1067.7 −×=rπ