Physics 220
Homework #1
Spring 2016
Due Wednesday 4/6/16
1. Using Excel or Mathematica or another plotting software, plot the thermal spectrum
(the Rayleigh-Jeans law with Planck’s correction) for a temperature of T = 5900K .
What feature(s) does your graph show? What happens to the maximum wavelength
of emission from the blackbody as the temperature of the object increases? Is this as
expected? What does the area under each curve represent? Now on the same plot,
add just the Rayleigh-Jeans law (without Planck’s correction) for a temperature of
T = 5900K . What do you notice? (For this question provide the fully commented
code from the software package that you used along with the graph that you produce.)
As the temperature increases the peak in the emission spectrum shifts towards lower
wavelengths as is shown by Wein’s law. The area under each curve represents the
total intensity, given by the Stefan-Boltzmann law.
2. We derived the Rayleigh-Jeans law with Planck’s correction in terms of the frequency
of light. What does the Rayleigh-Jeans law with Planck’s correction look like in
terms of the wavelength of light?
dS 2π hυ 3 ⎛
=
The Planck function is
⎜
df
c 2 ⎜⎝
⎞
⎟ and
⎟
− 1⎠
1
hυ
kT
e
c
c
c
c = f λ → υ = → dυ = − 2 d λ = 2 d λ . Thus the Planck function in terms of the
λ
λ
λ
wavelength is
⎞
3⎛
2π h ( λc )
dS
dS
1
⎜
⎟→
=
=
c
dυ λc2 d λ
c2
⎜ hkT( λ ) ⎟
⎝ e − 1⎠
dS 2π hc ⎛ 1 ⎞
∴
=
⎜
⎟
⎟
dλ
λ 5 ⎜⎝ λhckT
e − 1⎠
.
2
3. The rate at which energy from the Sun reaches a unit area on the Earth is called the
solar constant ( S ). The solar constant is measured to be S = 1350 mW2 on the earth at a
distance rE→S = 1.5 × 1011 m . What is the energy output of the Sun? If the Sun has an
average radius of RSun = 6.96 × 10 8 m , what is the approximate surface temperature of
the Sun?
To determine the energy output of the sun, we know the intensity of the sun at the
earth’s orbit. The total energy output of the sun is spread over a sphere centered on
the sun with radios of the distance between the earth and the sun. We have:
(
)
Esun = AS = ( 4π rES2 ) SE = 4π (1.5 × 1011 m ) × 1350 mW2 = 3.82 × 10 26 W . To determine
2
the surface temperature of the sun, we use the Stefan-Boltzmann law. Thus
S
E
3.82 × 10 26 W
T = 4 sun = 4 sun = 4
= 5800K .
2
8
−8 W
σ
Asunσ
4π 6.96 × 10 m × 6.57 × 10 2 4
(
(
))
m K
4. A light bulb emits 40W of energy every second at a wavelength of 6 × 10 −7 m . What
is the total number of photons emitted every second and what is the energy of one of
these photons?
The number of photons per second is determined from the power and the energy of
one photon. We have
× Eper photon
E N
P = = photons
→
t
t
N photons
P
P
40W × 6 × 10 −7 m
=
= λ=
= 1.2 × 10 20 photons
s
−34
8 m
s
Eper photon hc
6.6 × 10 Js × 3 × 10 s
5. Estimate the temperature of a “white” hot light bulb filament. If the light bulb has a
radiated power of 100W , what is the surface area of the filament? How many
photons are radiated per second?
“White” hot means that it is visible light. Your eye is most sensitive to green light, so
assuming a wavelength of say, λ = 500nm we have a filament temperature of
2.9 × 10 −3 mK
2.9 × 10 −3 mK
→T =
= 5800K . To determine the
approximately λmax =
T
500 × 10 −9 m
surface area, we use the Stefan-Boltzmann law. We have
P
P
100W
−6
2
S = = σT 4 → A =
=
4 = 1.6 × 10 m . The number of
4
−8 W
A
σT
5.67 × 10 m2 K 4 ( 5800K )
photons radiated per second is based off of the answer to question #6. So,
N photons
P
P
100W × 5 × 10 −7 m
.
=
= λ=
= 2.5 × 10 20 photons
s
−34
8 m
s
Eper photon hc
6.6 × 10 Js × 3 × 10 s
6. A thermal light source has a temperature of T = 6000K and a total radiated power of
dS
100W . What is the power per unit area per µ m (
) at the peak of the thermal
dλ
spectrum?
At the given temperature, the maximum wavelength of emission is
2.9 × 10 −3 mK 2.9 × 10 −3 mK
λmax =
=
= 4.83 × 10 −7 m = 483nm . We use this value to
T
6000K
calculate the intensity per unit wavelength. Thus,
dS 2π hc 2 ⎛
=
⎜
dλ
λ 5 ⎜⎝
1
hc
e λ kT
⎞
⎟
⎟
− 1⎠
−34
8
dS 2π × 6.6 × 10 Js × ( 3 × 10
=
5
dλ
( 4.83 × 10−7 m )
m
s
)
2
⎛
⎞
1
⎜
⎟ = 9.94 × 1013 W = 9.94 × 10 7
m3
⎞
6.6×10 −34 Js×3×10 8 ms
⎜ ⎛⎜
⎟
⎜⎝ ⎝ 4.83×10−7 m×1.38×10−23 KJ ×6000 K ⎟⎠
⎟⎠
e
−1
W
µ m⋅m 2
7. Suppose that the Earth was formed as molten rock at some very high temperature.
a. About how much time would it take the Earth to cool to say 300K ? The radius
of the Earth is 6.4 × 10 6 m and the average kinetic energy of a particle in the Earth
3
is given by E = k BT .
2
The total intensity is given by the Stefan-Boltzmann law, and have
dE
S=−
= σ T 4 where the negative sign is due to the fact the Earth is losing
AEarth dt
3
3
energy as time increases. The energy is given by E = NkT , so dE = NkdT .
2
2
Substituting what we have, we get
3
NkdT
2
TF dT
dE
8π rEArth
σ
4
4
2
S=−
= σT → −
= σT → ∫
=−
2
4
∞
AEarth dt
4π rEarth dt
T
3Nk
∫
TF
∞
t
∫ dt
0
T
dT
1 −3 f
1
=
−
T
=− 3
4
T
3
3T f
∞
2
1
8π rEarth
σ
∴ 3=
t
3T f
3Nk
t=
⎛ 6 × 10 24 kg ⎞
−23 J
⎜⎝ 1.67 × 10 −27 kg ⎟⎠ × 1.38 × 10 K
Nk
=
8π r σ T f3 8π ( 6.4 × 10 6 m )2 ( 300K )3 × 5.67 × 10 −8
2
Earth
W
m2K 4
= 3.1× 1013 s = 9.9 × 10 5 yrs
b. Is this time consistent with what we know for the age of the earth ( ~ 4.5 × 10 9 yr
)? If this time is not consistent with the age of the earth, what could explain the
difference?
No, this is not consistent with the accepted age of the earth. This is due to
radioactive elements continually heating the earth as it cools, so the time is much
longer.
c. Speaking of the Earth, there are some crazy things that go on. Suppose that, for
some strange reason, a naked person at the South Pole gets out of a sauna (hot
tub) with a body temperature of 311K . The person encounters the outside air,
which is at a temperature very much below the freezing point of water. How
much energy per second does the person lose by radiation cooling if their surface
area is 1m 2 ?
From the Stefan-Boltzmann law we have
E
E
S=
= σ T 4 → = Aσ T 4 = 1m 2 × 5.67 × 10 −8
At
t
W
m 2T 4
× ( 311K ) = 530W
4
d. By approximately what factor would this person’s loss of energy to radiation
cooling differ from the previous case if the person now were to step out of the
sauna into a cool room with a temperature of 290K ?
From the Stefan-Boltzmann law we have
E
4
4
⎤⎦
S=
= σ ⎡⎣Tyou
− Troom
At
.
E
4
4
4
4
2
−8 W
⎡
⎤
= Aσ ⎡⎣Tyou − Troom ⎤⎦ = 1m × 5.67 × 10 m2T 4 × ⎣( 311K ) − ( 290K ) ⎦ = 130W
t
Therefore the factor is about four times less.