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Chapter 2
Full file at http://TestbankCollege.eu/Solution-Manual-Semiconductor-Physics-and-Devices-4th-Edition-Neamen
2.1
Sketch
______
2.2
Sketch
______
2.3
Sketch
______
2.4
From Problem 2.2, phase
= constant
Then
From Problem 2.3, phase
= constant
Then
______
2.5
Gold: eV J
So,
cm
or
m
Cesium: eV J
So,
cm
or
m
______
2.6
(a)
kg-m/s
m/s
or cm/s
(b)
kg-m/s
m/s
or cm/s
(c) Yes
______
2.7
(a) (i)
kg-m/s
m
or
(ii)
kg-m/s
m
or
(iii)
kg-m/s
m
or
(b)
kg-m/s
m
or
______
2.8
eV
Now
or
kg-m/s
Now
m
or
______
2.9
Now
and
Set and
Then
which yields
J keV
______
2.10
(a)
kg-m/s
m/s
or cm/s
J
or eV
(b)
J
or eV
kg-m/s
m
or
______
2.11
(a)
J
Now
VkV
(b)
kg-m/s
Then
m
or
______
2.12
kg-m/s
______
2.13
(a) (i)
kg-m/s
(ii)
Now
kg-m/s
so
J
or eV
(b) (i) kg-m/s
(ii)
kg-m/s
J
or eV
______
2.14
kg-m/s
m/s
______
2.15
(a)
s
(b)
kg-m/s
______
2.16
(a) If and are solutions to
Schrodinger's wave equation, then
and
Adding the two equations, we obtain
which is Schrodinger's wave equation. So
is also a solution.
(b) If were a solution to
Schrodinger's wave equation, then we could write
which can be written as
Dividing by , we find
Since is a solution, then
Subtracting these last two equations, we have
Since is also a solution, we have
Subtracting these last two equations, we obtain
This equation is not necessarily valid, which
means that is, in general, not a solution
to Schrodinger's wave equation.
______
2.17
so
or
______
2.18
or
______
2.19
Note that
Function has been normalized.
(a) Now
or
which yields
(b)
or
which yields
(c)
which yields
______
2.20
(a)
or
(b)
or
(c)
or
______
2.21
(a)
or
(b)
or
(c)
or
______
2.22
(a) (i) m/s
or cm/s
m
or
(ii)
kg-m/s
J
or eV
(b) (i) m/s
or cm/s
m
or
(ii) kg-m/s
eV
______
2.23
(a)
(b)
so m/scm/s
For electron traveling in direction,
cm/s
kg-m/s
m
m
or rad/s
______
2.24
(a)
kg-m/s
m
m
rad/s
(b)
kg-m/s
m
m
rad/s
______
2.25
J
or
or eV
Then
eV
eV
eV
______
2.26
(a) J
or eV
Then
eV
eV
eV
(b)
J
m
or nm
______
2.27
(a)
or
(b) mJ
(c) No
______
2.28
For a neutron and :
J
or
eV
For an electron in the same potential well:
J
or
eV
______
2.29
Schrodinger's time-independent wave
equation
We know that
for and
We have
for
so in this region
The solution is of the form
where
Boundary conditions:
at
First mode solution:
where
Second mode solution:
where
Third mode solution:
where
Fourth mode solution:
where
______
2.30
The 3-D time-independent wave equation in
cartesian coordinates for is:
Use separation of variables, so let
Substituting into the wave equation, we
obtain
Dividing by and letting , we
find
(1)
We may set
Solution is of the form
Boundary conditions:
and
where
Similarly, let
and
Applying the boundary conditions, we find
,
,
From Equation (1) above, we have
or
so that
______
2.31
(a)
Solution is of the form:
We find
Substituting into the original equation, we find:
(1)
From the boundary conditions,
, where
So ,
Also , where
So ,
Substituting into Eq. (1) above
(b)Energy is quantized - similar to 1-D result.
There can be more than one quantum state
per given energy - different than 1-D result.
______
2.32
(a) Derivation of energy levels exactly the same as in the text
(b)
For
Then
(i) For
J
or eV
(ii) For cm
J
or
eV
______
2.33
(a) For region II,
General form of the solution is
where
Term with represents incident wave and
term with represents reflected wave.
Region I,
General form of the solution is
where
Term involving represents the transmitted wave and the term involving represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that .
Then
(b)
Boundary conditions:
(1)
(2)
Applying the boundary conditions to the solutions, we find
Combining these two equations, we find
The reflection coefficient is
The transmission coefficient is
______
2.34
where
m
(a) For m
(b) For m
(c) For m
______
2.35
where
or m
(a) For m
(b) For m
(c) , where is the density of
transmitted electrons.
eVJ
m/scm/s
electrons/cm
Density of incident electrons,
cm
______
2.36
(a) For
or
m
Then
or
(b) For
=
or
m
Then
or
______
2.37
where
m
(a)
(b)
or m
______
2.38
Region I , ;
Region II ,
Region III ,
(a) Region I:
(incident) (reflected)
where
Region II:
where
Region III:
(b)
In Region III, the term represents a
reflected wave. However, once a particle
is transmitted into Region III, there will
not be a reflected wave so that .
(c) Boundary conditions:
At :
At :
The transmission coefficient is defined as
so from the boundary conditions, we want
to solve for in terms of . Solving
for in terms of , we find
We then find
We have
If we assume that , then will
be large so that
We can then write
which becomes
Substituting the expressions for and
, we find
and
Then
Finally,
______
2.39
Region I:
incident reflected
where
Region II:
transmitted reflected
where
Region III:
transmitted
where
There is no reflected wave in Region III.
The transmission coefficient is defined as:
From the boundary conditions, solve for
in terms of . The boundary conditions are:
At :
At :
But
Then, eliminating , , from the
boundary condition equations, we find
______
2.40
(a) Region I: Since , we can write
Region II: , so
Region III:
The general solutions can be written,
keeping in mind that must remain
finite for , as
where
and
(b) Boundary conditions
At :
At :
or
(c)
and since , then
From , we can write
or
This equation can be written as
or
This last equation is valid only for specific values of the total energy . The energy levels are quantized.
______
2.41
(J)
(eV)
or
(eV)
eV
eV
eV
eV
______
2.42
We have
and
or
To find the maximum probability
which gives
or is the radius that gives the greatest
probability.
______
2.43
is independent of and , so the wave
equation in spherical coordinates reduces to
where
For
Then
so
We then obtain
Substituting into the wave equation, we have
where
Then the above equation becomes
or
which gives 0 = 0 and shows that is
indeed a solution to the wave equation.
______
2.44
All elements are from the Group I column of
the periodic table. All have one valence
electron in the outer shell.
______