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Chapter 38
p. 1
CHAPTER 38 – Early Quantum Theory and Models of the Atom
Note:
At the atomic scale, it is most convenient to have energies in electronvolts and wavelengths in
nanometers.
A useful expression for the energy of a photon in terms of its wavelength is
E
=
hf
=
hc
/
λ
= (6.63
×
10
–34
J
·
s)(3.00
×
10
8
m/s)(10
–9
nm/m)/(1.60
×
10
–19
J/eV)
;
E
= (1.24
×
10
3
eV
·
nm)/
.
1.
We find the temperature for a peak wavelength of 440 nm:
T
= (2.90
×
10
–3
m
·
K)/
P
= (2.90
×
10
–3
m
·
K)/(440
×
10
–9
m) =
6.59
×
10
3
K
.
2. (
a
) The temperature for a peak wavelength of 25.0 nm is
T
= (2.90
×
10
–3
m
·
K)/
P
= (2.90
×
10
–3
m
·
K)/(25.0
×
10
–9
m) =
1.16
×
10
5
K
.
(
b
) We find the peak wavelength from
P
= (2.90
×
10
–3
m
·
K)/
T
= (2.90
×
10
–3
m
·
K)/(2800 K) = 1.04
×
10
–6
m =
1.04
µ
m
.
Note that this is not in the visible range.
3.
Because the energy is quantized,
E
=
nhf
, the difference in energy between adjacent levels is
?
E
=
hf
= (6.63
×
10
–34
J
·
s)(8.1
×
10
13
Hz) =
5.4
×
10
–20
J = 0.34 eV
.
4. (
a
) We find the peak wavelength from
P
= (2.90
×
10
–3
m
·
K)/
T
= (2.90
×
10
–3
m
·
K)/(273 K) = 1.06
×
10
–5
m =
10.6
m
.
This wavelength is in the
infrared
.
(
b
) We find the peak wavelength from
P
= (2.90
×
10
–3
m
·
K)/
T
= (2.90
×
10
–3
m
·
K)/(3300 K) = 8.79
×
10
–7
m =
879 nm
.
This wavelength is in the
near infrared
.
(
c
) We find the peak wavelength from
P
= (2.90
×
10
–3
m
·
K)/
T
= (2.90
×
10
–3
m
·
K)/(4 K) = 7.25
×
10
–4
m =
0.73 mm
.
This wavelength is in the
far infrared
.
5. (
a
) The potential energy on the first step is
U
1
=
mgh
= (58.0 kg)(9.80 m/s
2
)(0.200 m) =
114 J
.
(
b
) The potential energy on the second step is
U
2
=
mg
2
h
= 2
U
1
= 2(114 J) =
228 J
.
(
c
) The potential energy on the third step is
U
3
=
mg
3
h
= 3
U
1
= 3(114 J) =
342 J
.
(
d
) The potential energy on the
n
th step is
U
n
=
mgnh
=
nU
1
=
n
(114 J) =
114
n
J
.
(
e
) The change in energy is
?
E
=
U
2
–
U
6
= (2 – 6)(114 J) =
– 456 J
.
6.
We use a body temperature of 98°F = 37°C = 310 K.
We find the peak wavelength from
P
= (2.90
×
10
–3
m
·
K)/
T
= (2.90
×
10
–3
m
·
K)/(310 K) = 9.4
×
10
–6
m =
9.4
m
.
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View Full Document Chapter 38
p. 2
7. (
a
) To find the wavelength when
I
(
λ
,
T
) is maximal at constant temperature, we set d
I
/d
= 0:
d
I
d
=
d
d
2
π
hc
2
–5
e
hc
/
kT
–1
=2
π
hc
2
–6
e
hc
/
kT
–
–
hc
/
2
kT
e
hc
/
kT
e
hc
/
kT
2
=–
2
π
hc
2
6
5–(
hc
/
kt
)
e
hc
/
kT
e
hc
/
kT
– 1
2
= 0,which gives
hc
/
P
kT
)
e
hc
/
P
kT
–5=0, or 5–(
hc
/
P
kT
)=5
e
–
hc
/
P
kT
.
This equation will have a solution
P
T
= constant, which is the Wien displacement law.
(
b
) To find the value of the constant, we let
x
=
hc
/
P
kT
,
so the transcendental equation is
5
–
x
= 5
e
–
x
.
One way to solve this equation is to plot each side against
x
.
We see from the plot that the solution is very close to
x
= 5.
If we let
?
= 5 –
x
, we get
?
= 5
e
(
?
– 5)
˜ 5
e
– 5
= 0.034, so
x
= 4.966.
T
h
u
s
w
e
h
a
v
e
P
T
=
hc
/
xk
;
2.90
×
10
–3
m
·
K =
h
(3.00
×
10
8
m/s)/(4.966)(1.38
×
10
–23
J/K),
which gives
h
= 6.63
×
10
–34
J
·
s
.
(
c
) For the rate at which energy is radiated per unit surface area for all wavelengths we have
I
(
,
T
)d
0
∞
=
2
π
hc
2
e
hc
/
kT
d
0
∞
.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Energy, Photon

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