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29
3
The Quantum Theory of Light
31
(a)
ππ
=
2
2
dB
E
rr
dt
( 29
=
2
r
dB
E
dt
(b)
If
r
remains constant, then:
==
2
r
dB
E
E
qe
dt
so that
2
e
r
dB
Fdt
dt
m dv
dt
, or
+∆
=
=
∆=
∫∫
0
2
2
2
e
vv
B
e
v
e
re
dv
dB
m
er
dv
dB
m
erB
v
m
B
E
E
r
(c)
(
29
(
ϖ

∆
∆
=
=
=
×
×
=×
19
31
10
1 T
1.6 10
C
9.1 10
kg
8.8 10
rad sec
22
e
v
eB
rm
(
(
π
ϖ
λ

×
=
=
=
×
8
15
9
3.0 10
m s
2
2
2
3.8 10
rad sec
500 10
m
c
f
;
ϖ
ϖ

∆
5
2.3 10
(d)
For the
ϖ
0
line the electrons’ plane is parallel to
B
, therefore, the magnetic flux,
Φ
B
is always zero. This means that
F
and
E
are zero and as a consequence, there is no
force on the electrons and there will be no
∆
v
for the electrons. The
ϖ
ϖ
0
is the
case calculated in parts (a)–(c). The
ϖ
ϖ
∆
0
will have the same magnitude for
F
,
B
,
and
∆
v
as in (a)–(c) but the direction will be opposite.
e
B
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CHAPTER 3
THE QUANTUM THEORY OF LIGHT
32
Assume that your skin can be considered a blackbody. One can then use Wien’s displacement
law,
λ

=×⋅
2
max
0.289 8 10
m K
T
with
==
0
35 C
308 K
T
to find
λ


×⋅
=
=
×=
2
6
max
0.289 8 10
m K
9.41 10
m
9 410 nm
308 K
.
33
(a)
The total energy of a simple harmonic oscillator having an amplitude
A
is
2
2
kA
,
therefore,
(
29
(29
=
2
2
0.4 m
25 N m
2.0 J
22
kA
E
. The frequency of oscillation will be
( 29
ππ
===
12
1
1
25
0.56 Hz
2
k
f
m
.
(b)
If energy is quantized, it will be given by
=
n
E
nhf
and from the result of (a) there
follows
( 29
(
29
=
=
34
6.63 10
J s 0.56 Hz
2.0 J
n
E
nhf
n
. Upon solving for
n
one obtains
=×
33
5.4 10
n
.
(c)
The energy carried away by one quantum of charge in energy will be
( 29
(
29

×
34
34
6.63 10
J s 0.56 Hz
3.7 10
J
E
hf
.
34
(a)
From Stefan’s law, one has
σ
=
4
P
T
A
. Therefore,
(

82
4
62
4
5.7 10
W m
3000 K
4.62 10
W m
P
K
A
.
(b)
××
2
75 W
16.2 mm
4.62 10
W m
4.62 10
W m
P
A
.
35
(a)
Planck’s radiation energy density law as a function of wavelength and temperature is
given by
λ
π
λ
λ
=

5
8
,
1
B
h
cT
hc
uT
e
. Using
λ
∂
=
∂
0
u
and setting
λ
=
max
B
hc
x
kT
, yields an
extremum in
λ
,
with respect to
λ
. The result is
(
(
λλ
λ

=

+
max
max
1
max
0
51
BB
hc
k T
hc
k T
B
hc
ee
or
( 29

=
x
xe
.
(b)
Solving for
x
by successive approximations, gives
2245
4.965
x
or
λ

=
=
3
max
4.965
2.90 10
m K
B
hc
T
k
.
36
Planck length

=
35
4.05 10
m
hG
c
Planck time

=
45
5
1.35 10
s
hG
c
MODERN PHYSICS
31
Planck mass

=
=×
12
8
5.46 10
kg
hG
c
37
(a)
In general,
λ
=
2
n
L
where
=
K
1, 2, 3,
n
defines a mode or standing wave pattern
with a given wavelength. As we wish to find the number of possible values of
n
between 2.0 and 2.1 cm, we use
λ
=
2
L
n
(
29 (29
(
==
∆=
200
2.0 cm
2
200
2.0
200
2.1 cm
2
190
2.1
10
n
n
n
As
n
changes by one for each allowed standing wave, there are 10 standing waves of
different wavelength between 2.0 and 2.1 cm.
2 m
(b)
The number of modes per unit wavelength per unit length is
(29
λ

∆
∆
2
10
200
0.5 cm
0.1
n
L
.
(c)
For short wavelengths
n
is almost a continuous function of
λ
. Thus one may use
calculus to approximate
( 29
γλ
∆
=
∆
1
n
dn
L
Ld
. As
λ
=
2
L
n
,
λ
λ
=
2
2
d
nL
d
and
λ
λ
=
2
dn
.
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This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.
 Fall '09
 Staff

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