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Quantum Mechanics
4011
sinh
2
LL
ee
L
κκ
κ
−
−
=
For
1, sinh
2
L
e
→
W
and
1
22
00
0
16 (
)
1
16 (
)
16 (
)
L
L
Ue
EU
E
T
E
E
U e
−
⎡⎤
−
→+
=
⎢⎥
−−
+
⎣⎦
For
00 0
1, 16 (
)
L
E Ue
−+
→
W
2
0
0
16 (
)
16
1
,
L
L
E
E
E
Te
U
U
−
⎛⎞
⎛
⎞
−
→=
−
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
which is Eq.(40.21).
(b)
0
2(
)
.
Lm
U E
L
−
=
=
So
1
L
W
when
L
is large (barrier is wide) or
0
UE
−
is large. (
E
is small compared to
0
.
U
)
(c)
0
)
;
mU
E
−
=
=
becomes small as
E
approaches
0
.
U
For
small, sinh
→
and
11
222
2
2
0
2
)
4(
)
)
UL
Um
L
T
E
E
⎡
⎤
−
=+
⎢
⎥
⎣
⎦
=
(using the definition of
)
Thus
1
0
2
2
1
4
ULm
T
E
−
=
0
→
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.
EVALUATE:
When
L
is large Eq.(40.20) applies and
T
is small. When
0
,
→
T
does not approach unity.
40.48.
(a)
2
1
((
1
2
)
) ((
1
2
)
)
,
2
Em
vn
ω
nh
f
==
+
=
+
=
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.
IDENTIFY
and
SET UP:
Calculate the angular frequency
ω
of the pendulum and apply Eq.(40.26) for the energy levels.
EXECUTE:
1
4
s
0.500 s
T
ππ
ωπ
−
=
The groundstate energy is
34
1
34
0
(1.055 10
J s)(4 s )
6.63 10
J.
E
−
×
⋅
=
×
=
34
19
15
0
6.63 10
J(1 eV/1.602 10
J)
4.14 10
eV
E
−
=×
×
1
2
n
En
=
1
1
1
2
n
+
+
=
The energy difference between the adjacent energy levels is
10
2
nn
EE E
E
+
Δ=
− =
=
=
=
33
15
1.33 10
J
8.30 10
eV
×=
×
EVALUATE:
These energies are much too small to detect. Quantum effects are not important for ordinary size objects.
40.50.
IDENTIFY:
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.
SET UP:
The energy of an electron in the
n
th
level is
2
.
8
n
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.
EXECUTE:
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
−
−
×⋅ ×
=
=
×
18
1.206 10
J.
−
×
Using the formula for the energy levels in a box, this energy difference is equal to
(1
)
(
21
)
.
88
hh
En n
n
mL
mL
− −
=
−
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View Full Document4012
Chapter 40
Solving for
n
gives
21
8
3
1
9
2
23
4
2
8
1 (1.206 10
J)8(9.11 10
kg)(0.500 10 m)
11
3
.
2(
6
.
6
2
6
1
0
J
s
)
EmL
n
h
−−
−
−
⎛⎞
⎛
⎞
Δ×
×
×
=+
=
+
=
⎜⎟
⎜
⎟
×⋅
⎝⎠
⎝
⎠
The transition is from
3
n
=
to
2.
n
=
EVALUATE:
We know the transition is not from the
4
n
=
to the
3
n
=
state because we let
n
be the higher state
and
1
n
−
the lower state.
40.51.
IDENTIFY:
If the given wave function is a solution to the Schrödinger equation, we will get an identity when we
substitute that wave function into the Schrödinger equation.
SET UP:
The given wave function is
22
/ 2
00
()
x
xA
e
α
ψ
−
=
and the Schrödinger equation is
2
.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, mechanics

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