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1137
Chapter 35
Applications of the Schrödinger Equation
Conceptual Problems
1
•
True
2
•
Determine the Concept
Looking at the graphs in the text for the
n
= 1, 2, and 3 states,
we note that the
n
= 4 state graph of the wave function must have four extrema in the
region 0 <
x
<
L
and decay in toward zero in the regions
x
< 0 and
x
>
L
.
(
a
)
(
b
)
3
•
Determine the Concept
Looking at the graphs in the text for the
n
= 1, 2, and 3 states,
we note that the
n
= 5 state graph of the wave function must have five extrema in the
region 0 <
x
<
L
and decay in toward zero in the regions
x
< 0 and
x
>
L
.
(
a
)
(
b
)
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1138
Estimation and Approximation
*4
•
Picture the Problem
Assume a mass of 150 g for the baseball, 30 cm for the width of the
locker, and 1 cm/s for the speed of the ball, and equate the kinetic energy of the ball and
the quantummechanical energy and solve for the quantum number
n
.
The allowed energy states of a
particle of mass
m
in a
1dimensional infinite potential well
of width
L
are given by:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
2
2
2
8
mL
h
n
E
n
The kinetic energy of the ball is:
2
2
1
mv
K
=
For
E
n
=
K
:
2
2
2
2
2
1
8
mv
mL
h
n
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Solve for the quantum number
n
:
h
mvL
n
2
=
Substitute numerical values and
evaluate
n
:
( )( )( )
30
30
34
10
10
36
.
1
s
J
10
63
.
6
m
3
.
0
m/s
01
.
0
kg
15
.
0
2
≈
×
=
⋅
×
=
−
n
The Schrödinger Equation
5
••
Picture the Problem
We can show that
ψ
3
(
x
) is a solution to the timeindependent
Schrödinger equation by differentiating it twice and substituting in Equation 354.
Equation 354 is:
( )
() () ()
x
E
x
x
U
dx
x
d
m
=
+
−
2
2
2
2
h
Because
1
(
x
) and
2
(
x
) are
solutions of Equation 354:
( )
x
E
x
x
U
dx
x
d
m
1
1
2
1
2
2
2
=
+
−
h
and
( )
x
E
x
x
U
dx
x
d
m
2
2
2
2
2
2
2
=
+
−
h
Applications of the Schrödinger Equation
1139
Add these equations to obtain:
()
[]
)
(
)
(
)
(
)
(
)
(
)
(
2
2
1
2
1
2
2
2
2
1
2
2
x
x
E
x
x
x
U
dx
x
d
dx
x
d
m
ψ
+
=
+
+
⎥
⎦
⎤
⎢
⎣
⎡
+
−
h
(1)
Differentiate
)
(
)
(
)
(
2
1
3
x
x
x
+
=
twice
with respect to
x
to obtain:
dx
x
d
dx
x
d
dx
x
d
)
(
)
(
)
(
2
1
3
+
=
and
2
2
2
2
1
2
2
3
2
)
(
)
(
)
(
dx
x
d
dx
x
d
dx
x
d
+
=
Substitute in equation (1) to obtain:
)
(
)
(
)
(
2
3
3
2
3
2
2
x
E
x
x
U
dx
x
d
m
=
+
−
h
which shows that
)
(
)
(
)
(
2
1
3
x
x
x
+
=
satisfies Equation 354.
The Harmonic Oscillator
6
••
Picture the Problem
We can relate the spring constant to the mass of the hydrogen atom
and its angular frequency and then use the relationship between the allowed energy levels
and the angular frequency
ω
to derive an expression for the spring constant
k.
The spring constant
k
is related to
the mass
m
of the hydrogen
molecule and its angular frequency
:
2
m
k
=
(1)
Relate the energy spacing
∆
E
to the
angular frequency
:
π
h
=
=
=
∆
2
h
hf
E
Solve for
:
h
E
∆
=
Substitute for
in equation (1) to
obtain:
2
⎟
⎠
⎞
⎜
⎝
⎛ ∆
=
h
E
m
k
Substitute numerical values and evaluate
k
:
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1140
kN/m
14
.
1
s
J
10
05
.
1
J
10
7
.
8
u
kg
10
1.66
u
1
2
34
20
27
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
×
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
=
−
−
−
k
Remarks: Our result is very
similar to the stiffness constant of typical macroscopic
springs.
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This homework help was uploaded on 02/26/2008 for the course PHYSICS 11 taught by Professor Licini during the Spring '07 term at Lehigh University .
 Spring '07
 Licini

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