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

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Prob. 1 – Brillouin zones of rectangular lattice

Daniel Wolpert

9.1 Brillouin zones of rectangular lattice. Make a plot of the first two Brillouin zones of a primitive primitive rectangular rectangular two-dimensional two-dimensional lattice with with axes a, b=3a 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

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 2π /a 2π /3a 2nd BZ First BZ

Prob. 2 – Brillouin zone,rectangular lattice Gregory Kaminsky

This is a Wigner-Seitz cell.

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.

π

Calculation of the radius of the Fermi sphere

2*

2

2 * π * k F

4 * π 0 2 4 * 2 * ( A ) 2

π

k F

=

2 * A

=

1

π 0

=

2

12

* 10 cm

−1

2

* k F

2π L

2

=

N

Brilloin zone Radius of free electron fermi sphere =

π

π

2

12

* 10 cm

−1

π

2

12

*10 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.

This is the first energy band

Second energy band

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 scheme

Second Zone periodic scheme

Prob. 5 – Open Orbits

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?

(b) Describe in real space the motion

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

9.5 v

d k

v

d r

v × B

From Eq. 25a we have dt dt , where I have decided to use SI units. v v v G d r = − q e = h = − ev B dt τ Letting we get = v , setting d k = G dt because v ⊥ B since B is normal to the Fermi surface. Solving for gives Gh = . Plugging in the givens we evB get

h

=

q

τ

τ

2 ⋅108 100cm

Gh evB

⋅

=

cm

2 ⋅ π ⋅1.602 ⋅10

m −19

⋅ 6.62 ⋅10

C ⋅10

8

cm s

⋅

− 34

kg ⋅ m 2 s

1m 100cm

⋅10

−1

kg

= 1.315 ⋅10

C ⋅ s

Part b) The electron will travel along the Fermi surface as shown. The velocity will change as the electron moves along the Fermi surface.

−10

s

Mike Tuffley 5/12/09

U(x)

-a/2

a/2 x

-U0

Chapter 9 Problem 7

Adam Gray

1 ( ) expected ∆ B

(a) Calculate the period for potassium on the free electron model. (b) What is the area in real space of the extremal orbit, for B = 10kG = 1T ?

Starting with equation 34: ∆(

1

B

Where

S

=

2π e

)=

hcS

K

π

2 f

Using Table 6.1 on pg. 139, for potassium we find kf=0.75x108cm-1 .

Plugging in:

∆(

1

B

∆(

)=

1

2π e 2

hc(π K f )

2e

)= 2 B hcK 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

B

) = 5.55 ×10−9 G −1

(b) To solve this part of the problem, go back to the equations we used for the cyclotron. ω c

=

Be

r

mc

=

v f

P = mv = hk

ω c

Solve for r r =

v f ω c

=

v f

Be mc

=

v f mc Be

=

h k f c

Be

Plugging in values from before and B=10kG r = 4.94x10-4 cm The orbit is circular, so the area is

r 2

π

=

7.67 × 10 −7 cm 2

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