406
Chapter 40
40.26.
0
2
00
2(
)
with
16
1
and
,
L
mU
E
EE
TG
e
G
UU
κ
−
⎛⎞
−
==
−
=
⎜⎟
⎝⎠
=
so
0
22(
)
16
1
.
E
L
Te
−−
=−
=
(a)
If
61
5
2
7
0
30.0 10 eV,
2.0 10
m,
6.64 10
kg
UL
m
=×
=
×
and
66
0
1.0 10 eV (
29.0 10 eV),
0.090.
UE
E
T
−= ×
=
×
=
(b)
If
0
10.0 10 eV (
20.0 10 eV),
0.014.
E
T
−=
×
=
×
=
40.27.
IDENTIFY
and
SET UP:
The energy levels are given by Eq.(40.26), where
.
k
m
ω
′
=
EXECUTE:
110 N/m
21.0 rad/s
0.250 kg
k
m
′
=
The ground state energy is given by Eq.(40.26):
34
33
19
15
0
11
(1.055 10
J s)(21.0 rad/s)
1.11 10
J(1 eV/1.602 10
J)
6.93 10
eV
22
E
−
−
×
⋅
=
×
×
=
×
=
1
;
2
n
En
=+
=
(1
)
1
1
2
n
+
+
=
The energy separation between these adjacent levels is
33
33
14
10
2
2(1.11 10
J)
2.22 10
J
1.39 10
eV
nn
EE E
E
−
+
Δ=
−
=
=
=
×
=
×
=
×
=
EVALUATE:
These energies are extremely small; quantum effects are not important for this oscillator.
40.28.
Let
2,
mk
δ
′
=
=
and so
2
d
ψ
x
ψ
dx
δ
and
2
2
(4
2
,
d
ψ
x
δδ
)
ψ
dx
and
ψ
is a solution of Eq.(40.21) if
2
E
δ
k/m
ω
.
m
′
=
=
40.29.
IDENTIFY:
We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the energy
difference between the two levels of the oscillator.
SET UP:
The energy of a photon is
/ ,
Eh
fh
c
γ
λ
and the energy levels of a harmonic oscillator are given by
.
n
k
n
m
′
EXECUTE: (a)
The photon’s energy is
34
8
6
(6.63 10
J s)(3.00 10 m/s)
0.21 eV
5.8 10 m
hc
E
−
−
×⋅ ×
=
×
(b)
The transition energy is
1
,
k
m
+
′
− =
=
which gives
2
.
π
ck
m
′
=
=
=
Solving for
,
k
′
we get
2
8
2
2
6
26
2
4
4
(3.00 10 m s) (5.6 10
kg)
5,900 N/m.
(5.8 10 m)
π
cm
π
k
−
−
××
′
=
×
EVALUATE:
This would be a rather strong spring in the physics lab.
40.30.
According to Eq.(40.26), the energy released during the transition between two adjacent levels is twice the ground
state energy
32
0
2
11.2 eV.
ω
E
−= =
=
=
For a photon of energy
E
34
8
19
(6.63 10
J s)(3.00 10 m s)
111 nm.
(11.2 eV)(1.60 10
J/eV)
ch
c
f
fE
−
−
×⋅
×
=
⇒
== =
=
×
40.31.
IDENTIFY
and
SET UP:
Use the energies given in Eq.(40.26) to solve for the amplitude
A
and maximum speed
max
v
of the oscillator. Use these to estimate
x
Δ
and
x
p
Δ
and compute the uncertainty product
.
x
xp
ΔΔ
EXECUTE:
The total energy of a Newtonian oscillator is given by
2
1
2
Ek
A
′
=
where
k
′
is the force constant and
A
is the amplitude of the oscillator. Set this equal to the energy
1
2
()
=
of an excited level that has quantum
number
n
, where
,
k
m
′
=
and solve for
A
:
2
kA
n
′
=
(2
1)
n
A
k
+
=
′
=
The total energy of the Newtonian oscillator can also be written as
2
1
max
2
.
Em
v
=
Set this equal to
1
2
=
and
solve for
max
:
v
2
max
mv
n
=
max
(2
n
v
m
+
=
=
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View Full DocumentQuantum Mechanics
407
Thus the maximum linear momentum of the oscillator is
max
max
(2
1)
.
pm
v
nm
ω
==
+
=
Assume that
A
represents
the uncertainty
x
Δ
in position and that
max
p
is the corresponding uncertainty
x
p
Δ
in momentum. Then the
uncertainty product is
(2
1)
1
(2
(2
1) .
x
xp
n
m
n
n
n
kk
ωω
+
⎛⎞
Δ
Δ
=
+
=+
⎜⎟
′′
⎝⎠
=
=
=
EVALUATE:
For
1
n
=
this gives
3 ,
x
ΔΔ
=
=
in agreement with the result derived in Section 40.4. The
uncertainty product
x
increases with
n
.
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 Spring '06
 Buchler
 Physics, dx

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