半导体物理与器件第四版课后习题答案.doc

Chapter 3

If a o were to increase, the

bandgap energy

would decrease and the material

would begin

to behave less like a

semiconductor and more

like a metal. If a o were to

decrease, the

bandgap energy would increase and

the

material would begin to behave

more like an

insulator.

_______________________________________

Schrodinger's wave equation is:

2 2 x, t

V x x, t

2m x2

j x, t t

Assume the solution is of the form:

x, t u x exp j kx E t

Region I: V x 0 . Substituting the

assumed solution into the wave equation, we

obtain:

2

E

2m x

jku x exp j kx t

u x

exp j kx E t

x

j jE u x exp j kx E t which becomes

2

2 u E

2m

jk x exp j kx t

2 jk u x

exp j kx E t

x

2 u x

exp j

E

t

x2

kx

Eu x exp j kx E t

This equation may be written as

k 2 u x

u x 2 u x 2mE

2 jk

x

2 2 u x 0

x

Setting u x u1 x for region I,

the equation

becomes:

d 2 u1 x 2 jk du1 x k 2 2 u1 x 0

dx2 dx

where

2 2mE

2

In Region II, V x V O. Assume the

same

form of the solution:

x, t u x exp j kx E t

Substituting into Schrodinger's

wave

equation, we find:

2

2 u E

jk x exp j kx t

2m

2 jk

u x

exp j kx E t

x

2

2

x

exp j

u kx E t

x

V O u x exp j kx E t

Eu x exp j kx

E

t

This equation can be written as:

k 2u x 2 jk u x 2u x

x x2

2mV O

u x

2mE

u x 0

2 2

Setting u x u 2 x for region II, this

equation becomes

d 2

u 2 x du 2 x dx 2 2 jk

dx

k 2

2 2mV O

u 2 x 0

2

where again

2

2mE

2

We have

2

2 jk du 1 x

d u 1 x k 2

2

u 1 x 0

dx 2

dx

Assume the solution is of the form:

u 1 x

A exp j

k x

B exp j

k x

The first derivative is

du 1 x j

k A exp j

k x

dx

j

k B exp

j

k x

and the second derivative becomes

d 2 u 1 x j

k

2

k x

dx 2

A exp j

2 B exp

j k j

k x

Substituting these equations into the

differential equation, we find

k 2 A exp j

k x

2

B exp j

k x

k

2 jk j k A exp j

k x j

k B exp j

k x

k 2

2

A exp j

k x

B exp j

k x

Combining terms, we obtain

2

2 k k

2

2k

k

k

2

2

A exp j

k x

2

2 k k 2

2k

k

k 2

2

B exp j

k x

We find that

0 0

For the differential equation in u 2 x

and the

proposed solution, the procedure is exactly

the same as above.

_______________________________________

We have the solutions

u 1 x

A exp j k x

B exp j

k x

for

0 x a and

u 2 x C exp j

k x

D exp j

k x

for b x 0 .

The first boundary condition is

u 1 0 u 2 0

which yields

A B C D

The second boundary condition is

du 1 du 2

dx

x 0

dx

x 0

which yields

k A

k B

k C

k D 0

The third boundary condition is

u 1 a u 2 b

which yields

A exp j

k a

B exp j

k a C exp j

k

b

D exp j

k

b

and can be written as

A exp j

k a

B exp j

k a

C exp j

k b

D exp j

k b

The fourth boundary condition is du 1 du 2

dx

x a

dx

x

b

which yields

j

k A exp j

k a

j

k B exp j k a

j

k C exp j

k

b

j

k D exp j k

b

and can be written as

k A exp j

k a

k B exp j

k a

k C exp j

k b

k D exp j k b 0

_______________________________________