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
+
=
=
This preview has intentionally blurred sections. Sign up to view the full version.
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
.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics

Click to edit the document details