Quantum Mechanics
4011
sinh
2
L
L
e
e
L
κ
κ
κ
−
−
=
For
1, sinh
2
L
e
L
L
κ
κ
κ
→
W
and
1
2
2
0
0
2
2
0
0
0
16
(
)
1
16
(
)
16
(
)
L
L
U e
E U
E
T
E U
E
E U
E
U e
κ
κ
−
⎡
⎤
−
→
+
=
⎢
⎥
−
−
+
⎣
⎦
For
2
2
2
2
0
0
0
1, 16
(
)
L
L
L
E U
E
U e
U e
κ
κ
κ
−
+
→
W
2
0
2
2
0
0
0
16
(
)
16
1
,
L
L
E U
E
E
E
T
e
U e
U
U
κ
κ
−
⎛
⎞⎛
⎞
−
→
=
−
⎜
⎟⎜
⎟
⎝
⎠⎝
⎠
which is Eq.(40.21).
(b)
0
2
(
)
.
L
m U
E
L
κ
−
=
=
So
1
L
κ
W
when
L
is large (barrier is wide) or
0
U
E
−
is large. (
E
is small compared to
0
.
U
)
(c)
0
2
(
)
;
m U
E
κ
κ
−
=
=
becomes small as
E
approaches
0
.
U
For
κ
small, sinh
L
L
κ
κ
→
and
1
1
2
2
2
2
2
0
0
0
2
0
0
2
(
)
1
1
4
(
)
4
(
)
U
L
U
m U
E L
T
E U
E
E U
E
κ
−
−
⎡
⎤
⎡
⎤
−
→
+
=
+
⎢
⎥
⎢
⎥
−
−
⎣
⎦
⎣
⎦
=
(using the definition of
κ
)
Thus
1
2
2
0
2
2
1
4
U L m
T
E
−
⎡
⎤
→
+
⎢
⎥
⎣
⎦
=
0
U
E
→
so
2
0
U
E
E
→
and
1
2
2
2
1
4
EL m
T
−
⎡
⎤
→
+
⎢
⎥
⎣
⎦
=
But
2
2
2
,
mE
k
=
=
so
1
2
1
,
2
kL
T
−
⎡
⎤
⎛
⎞
→
+
⎢
⎥
⎜
⎟
⎝
⎠
⎢
⎥
⎣
⎦
as was to be shown.
E
VALUATE
:
When
L
κ
is large Eq.(40.20) applies and
T
is small. When
0
,
E
U
→
T
does not approach unity.
40.48.
(a)
2
1
(
(1 2))
(
(1 2))
,
2
E
mv
n
ω
n
hf
=
=
+
=
+
=
and solving for
n
,
2
2
30
34
1
1
(1/2)(0.020 kg)(0.360 m/s)
1
2
1.3
10 .
2
(6.63
10
J
s)(1.50 Hz)
2
mv
n
hf
−
=
−
=
−
=
×
×
⋅
(b)
The difference between energies is
34
34
(6.63
10
J s)(1.50 Hz)
9.95
10
J.
ω
hf
−
−
=
=
×
⋅
=
×
=
This energy is too
small to be detected with current technology
40.49.
I
DENTIFY
and
S
ET
U
P
:
Calculate the angular frequency
ω
of the pendulum and apply Eq.(40.26) for the energy levels.
E
XECUTE
:
1
2
2
4
s
0.500 s
T
π
π
ω
π
−
=
=
=
The groundstate energy is
34
1
34
0
1
1
(1.055
10
J
s)(4
s
)
6.63
10
J.
2
2
E
ω
π
−
−
−
=
=
×
⋅
=
×
=
34
19
15
0
6.63
10
J(1 eV/1.602
10
J)
4.14
10
eV
E
−
−
−
=
×
×
=
×
1
2
n
E
n
ω
⎛
⎞
=
+
⎜
⎟
⎝
⎠
=
1
1
1
2
n
E
n
ω
+
⎛
⎞
=
+
+
⎜
⎟
⎝
⎠
=
The energy difference between the adjacent energy levels is
1
0
2
n
n
E
E
E
E
ω
+
Δ
=
−
=
=
=
=
33
15
1.33
10
J
8.30
10
eV
−
−
×
=
×
E
VALUATE
:
These energies are much too small to detect. Quantum effects are not important for ordinary size objects.
40.50.
I
DENTIFY
:
We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the
energy difference between the two energy states in the box.
S
ET
U
P
:
The energy of an electron in the
n
th
level is
2
2
2
.
8
n
n h
E
mL
=
We do not know the initial or final levels, but
we do know they differ by 1. The energy of the photon,
/ ,
hc
λ
is equal to the energy difference between the two states.
E
XECUTE
:
The energy difference between the levels is
34
8
7
(6.63
10
J
s)(3.00
10
m/s)
1.649
10
m
hc
E
λ
−
−
×
⋅
×
Δ
=
=
=
×
18
1.206
10
J.
−
×
Using the formula for the energy levels in a box, this energy difference is equal to
2
2
2
2
2
2
(
1)
(2
1)
.
8
8
h
h
E
n
n
n
mL
mL
⎡
⎤
Δ
=
−
−
=
−
⎣
⎦
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4012
Chapter 40
Solving for
n
gives
2
18
31
9
2
2
34
2
8
1
(1.206
10
J)8(9.11
10
kg)(0.500
10
m)
1
1
3.
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 Spring '06
 Buchler
 Physics, mechanics, Trigraph, NZ, wave function, dx

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