半导体物理与器件第四版课后习题答案.doc

Chapter 3
3、1
If 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 were to decrease, the
bandgap energy would increase and the
material would begin to behave more like an
insulator、
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3、2
Schrodinger's wave equation is:

Assume the solution is of the form:

Region I: 、 Substituting the
assumed solution into the wave equation, we
obtain:

which bees

This equation may be written as

Setting for region I, the equation
bees:

where
Q、E、D、
In Region II, 、 Assume the same
form of the solution:

Substituting into Schrodinger's wave
equation, we find:

This equation can be written as:

Setting for region II, this
equation bees

where again
Q、E、D、
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3、3
We have

Assume the solution is of the form:

The first derivative is

and the second derivative bees

Substituting these equations into the
differential equation, we find

bining terms, we obtain

We find that
Q、E、D、
For the differential equation in and the
proposed solution, the procedure is exactly
the same as above、
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3、4
We have the solutions

for and

for 、
The first boundary condition is

which yields

The second boundary condition is

which yields

The third boundary condition is

which yields

and can be written as

The fourth boundary condition is

which yields

and can be written as

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3、5
(b) (i) First point:
Second point: By trial and error,

(ii) First point:
Second point: By trial and error,

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3、6
(b) (i) First point:
Second point: By trial and error,

(ii) First point:
Second point: By trial and error,

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3、7

Let ,
Then

Consider of this function、

We find

Then

For ,
So that, in general,

And

So

This implies that
for
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3、8
(a)

J
From Problem 3、5

J

J
or eV
(b)

J
From Problem 3、5,

J

J
or eV
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3、9
(a) At ,

J
At , By trial and error,

J

J
or eV
(b) At ,

J
At 、 From Problem 3、5,

J

J
or eV
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3、10
(a)

J
From Problem 3、6,

J

J
or eV
(b)

J
From Problem 3、6,

J

J
or eV
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3、11
(a) At ,

J
At , By trial and error,

J

J
or eV
(b) At ,

J
At , From Problem 3、6,

J

J
or eV
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3、12
For K,

eV
K, eV
K, eV
K, eV
K, eV
K, eV
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3、13
The effective mass is given by

We have

so that
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3、14
The effective mass for a hole is given by

We have that

so that
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3、15
Points A,B: velocity in -x direction
Points C,D: velocity in +x direction
Points A,D:
negative effective mass
Points B,C:
positive effective mass
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3、16
For A:
At m, eV
Or J
So

Now
kg
or

For B:
At m, eV
Or J
So

Now
kg
or

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3、17
For A:

kg
or

For B:

kg
or

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3、18
(a) (i)
or
Hz
(ii)
cmnm
(b) (i)
Hz
(ii)
cmnm
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3、19
(c) Curve A: Effective mass is a constant
Curve B: Effective mass is positive
around , and is negative
around 、
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3、20

Then

and

Then

or

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3、21
(a)

(b)

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3、22
(a)

(b)

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3、23
For the 3-dimensional infinite potential well,
when , , and
、 In this region, the wave equation
is:

Use separation of variables technique, so let

Substituting into the wave equation, we have

Dividing by , we obtain

Let

The solution is of the form:

Since at , then
so that 、
Also, at , so that
、 Then where

Similarly, we have
and
From the boundary conditions, we find
and
where
and
From the wave equation, we can write

The energy can be written as

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3、24
The total number of quantum states in the
3-dimensional potential well is given
(in k-space) by

where

We can then write

Taking the differential, we obtain

Substituting these expressions into the density
of states function, we have

Noting that

this density of states function can be
simplified and written as

Dividing by will yield the density of
states so that

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3、25
For a one-dimensional infinite potential well,

Distance between quantum states

Now

Now

Then

Divide by the "volume" a, so

So
mJ
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3、26
(a) Silicon,

(i) At K, eV

J
Then
m
or cm
(ii) At K,
eV

J
Then

m
or cm
(b) GaAs,

(i) At K, J

m
or cm
(ii) At K, J

m
cm
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3、27
(a) Silicon,

(i)At K, J

m
or cm
(ii)At K, J

m
or cm
(b) GaAs,

(i)At K, J

m
or cm
(ii)At K, J

m
or cm
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3、28
(a)

For ;
eV; mJ
eV; mJ
eV; mJ
eV; mJ
(b)

For ;
eV; mJ
eV; mJ
eV; mJ
eV; mJ
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3、29
(a)
(b)
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3、30
Plot
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3、31
(a)

(b) (i)

(ii)

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3、32

(a) ,

(b) ,

(c) ,

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3、33

or

(a) ,
(b) ,
(c) ,
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3、34
(a)
;
;

;

;

;

(b)

;
;

;

;

;

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3、35

and

So

Then
Or
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3、36

For , Filled state

J
or eV
For , Empty state

J
or eV
Therefore eV
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3、37
(a) For a 3-D infinite potential well

For 5 electrons, the 5th electron occupies the quantum state ; so

J
or eV
For the next quantum state, which is empty, the quantum state is 、 This quantum state is at the same energy, so
eV
(b) For 13 electrons, the 13th electron occupies the quantum state
; so

J
or eV
The 14th electron would occupy the quantum state 、 This state is at the same energy, so
eV
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3、38
The probability of a state at
being occupied is

The probability of a state at
being empty is

or

so Q、E、D、
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3、39
(a) At energy , we want

This expression can be written as

or

Then

or

(b)
At ,

which yields

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3、40
(a)

(b) eV

(c)

or

or

which yields K
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3、41
(a)

or 0、304%
(b) At K, eV
Then

or 14、96%
(c)
or 99、7%
(d)
At , for all temperatures
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3、42
(a) For
Then

For , eV
Then

or

(b) For eV,
eV
At ,

or

At ,

or

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3、43
(a) At

or

At , eV
So

or

(b) For ,
eV
At ,

or

At ,

or
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3、44

so

or

(a) At K, For

At
(b) At K, eV
For ,
For ,
At ,
(eV)
(c) At K, eV
For ,
For ,
At ,
(eV)
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3、45
(a) At ,

Si: eV,

or

Ge: eV

or

GaAs: eV

or

(b) Using the results of Problem 3、38, the answers to part (b) are exactly the same as those given in part (a)、
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3、46
(a)

or
eV

so K
(b)

or K
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3、47
(a) At K,
eV

eV
By symmetry, for ,
eV
Then eV
(b) K, eV
For , from part (a),

eV
Then eV
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