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401
Q
UANTUM
M
ECHANICS
40.1.
IDENTIFY
and
SET UP:
The energy levels for a particle in a box are given by
22
2
.
8
n
nh
E
mL
=
EXECUTE: (a)
The lowest level is for
1,
n
=
and
34
2
67
1
2
(1)(6.626 10
J s)
1.2 10
J.
8(0.20 kg)(1.5 m)
E
−
−
×⋅
==
×
(b)
2
1
2
Em
v
=
so
67
33
2
2(1.2 10
J)
1.1 10
m/s.
0.20 kg
E
v
m
−
−
×
=
×
If the ball has this speed the time it would take it
to travel from one side of the table to the other is
33
33
1.5 m
1.4 10 s.
1.1 10
m/s
t
−
×
×
(c)
2
12
1
2
,
4 ,
8
h
EE
E
mL
so
67
67
21
1
3
3(1.2 10
J)
3.6 10
J
EE E E
−−
Δ=
− =
=
×
=
×
(d) EVALUATE:
No, quantum mechanical effects are not important for the game of billiards. The discrete,
quantized nature of the energy levels is completely unobservable.
40.2.
1
8
h
L
mE
=
34
15
27
6
19
(6.626 10
J s)
6.4 10
m.
8(1.673 10
kg)(5.0 10 eV)(1.602 10
J eV)
L
−
−
×
××
×
40.3.
IDENTIFY:
An electron in the lowest energy state in this box must have the same energy as it would in the ground
state of hydrogen.
SET UP:
The energy of the
n
th
level of an electron in a box is
2
2
.
8
n
nh
E
mL
=
EXECUTE:
An electron in the ground state of hydrogen has an energy of
13.6 eV,
−
so find the width
corresponding to an energy of
1
13.6 eV.
E
=
Solving for
L
gives
1
8
h
L
mE
=
34
10
31
19
(6.626 10
J s)
1.66 10
m.
8(9.11 10
kg)(13.6 eV)(1.602 10
J eV)
−
−
×
EVALUATE:
This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the
K shell.
40.4.
(a)
The energy of the given photon is
3
34
18
9
(3.00 10 m/s)
(6.63 10
J s)
1.63 10
J.
(122 10
m)
c
Eh
f h
λ
−
×
== =
×
⋅
= ×
×
The energy levels of a particle in a box are given by Eq.40.9
2
2
2
2
()
.
8
h
En
n
mL
−
3
4
22 2
10
31
20
(
6
.
6
3
1
0
J
s
)
(
2
1
)
3.33 10
m.
8
8(9.11 10
kg)(1.63 10
J)
hn n
L
mE
−
−
−×
⋅
−
=
×
Δ×
×
(b)
The ground state energy for an electron in a box of the calculated dimensions is
23
4
2
19
11
0
2
(6.63 10
J s)
5.43 10
J
3.40 eV
8
8(9.11 10
kg)(3.33 10
m)
h
E
mL
−
−
=
×
=
(onethird of the original photon energy),
which does not correspond to the
13.6 eV
−
ground state energy of the hydrogen atom. Note that the energy levels for
a particle in a box are proportional to
2
,
n
whereas the energy levels for the hydrogen atom are proportional to
2
1
.
n
−
40
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View Full Document402
Chapter 40
40.5.
IDENTIFY
and
SET UP:
Eq.(40.9) gives the energy levels. Use this to obtain an expression for
21
EE
−
and use the
value given for this energy difference to solve for
L.
EXECUTE:
Ground state energy is
2
1
2
;
8
h
E
mL
=
first excited state energy is
2
2
2
4
.
8
h
E
mL
=
The energy separation
between these two levels is
2
2
3
.
8
h
EE E
mL
Δ= − =
This gives
3
8
Lh
mE
==
Δ
34
10
31
19
3
6.626 10
J s
6.1 10
m
0.61 nm.
8(9.109 10
kg)(3.0 eV)(1.602 10
J/1 eV)
L
−
−
−−
=×
⋅
=
×
=
××
EVALUATE:
This energy difference is typical for an atom and
L
is comparable to the size of an atom.
40.6.
(a)
The wave function for
1
n
=
vanishes only at
0
x
=
and
xL
=
in the range 0
.
≤≤
(b)
In the range for
,
x
the sine term is a maximum only at the middle of the box,
/ 2.
=
(c)
The answers to parts (a) and (b) are consistent with the figure.
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 Spring '08
 Jabbour
 Materials Science And Engineering

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