Physics 220 Homework #1 Spring 2015 Due Wednesday 4/8/15 1

Physics 220
Homework #1
Spring 2015
Due Wednesday 4/8/15
1. Using Excel or Mathematica, plot the thermal spectrum (the Rayleigh-Jeans law with
Planck’s correction) for temperatures of 300K , 1500K , 3000K , and 5900K .
Makes all of the curves on the same graph. What features do your graphs 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 a separate plot, plot the Rayleigh-Jeans law with
Planck’s correction and just the Rayleigh-Jeans law for a temperature of T = 5900K .
What do you notice?
Intensity, Wêm^ 3
5 ¥101 2
4 ¥101 2
3 ¥101 2
3000. K
2 ¥101 2
1 ¥101 2
0
1500. K
300.,Km
Wavelength
0
2.¥10- 6
4. ¥10- 6
6.¥10- 6
8.¥10- 6
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
wavelength of light. What does the Rayleigh-Jeans law with Planck’s correction look
like in terms of the frequency of light?
The Planck function is
dS 2π hc 2 ⎛
=
⎜
dλ
λ 5 ⎜⎝
1
hc
⎞
⎟ and
⎟
− 1⎠
e λ kT
c
c
c
c = f λ → λ = → d λ = − 2 df = 2 df . Thus the Planck function in terms of the
f
f
f
frequency is
dS 2π hc 2 ⎛ 1 ⎞
2π hc 2 f 5 ⎛ 1 ⎞
2 dS
=
→
f
=
⎜
⎟
⎜ hfc
⎟
⎜⎝ ckT
⎟
⎟⎠
dλ
λ 5 ⎜⎝ λhckT
cdf
c5
e − 1⎠
e −1
.
3⎛
⎞
dS 2π hf
1
∴ =
⎜ hf
⎟
2
⎟
df
c ⎜⎝ kT
e − 1⎠
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
T=
4
Ssun
=
σ
4
Esun
=
Asunσ
4
(
3.82 × 10 26 W
)
4π ( 6.96 × 10 8 m ) × 6.57 × 10 −8
2
= 5800K .
W
m2K 4
4. Suppose that the Sun had no internal source of energy and was formed as a hot ball of
hydrogen at some very hot initial temperature ( Ti ). How long would it take to cool to
a final temperature T f = 6000K ? Use the Stefan-Boltzmann law and assume that the
total kinetic energy of the particles in the Sun is proportional to the temperature and is
3
given by E = NkT , where N is the number of particles in the Sun. Assume that the
2
entire mass of the sun ( M Sun = 2 × 10 30 kg ) is composed of protons. (Of course the
sun does have an internal source of energy. Proton fusion heats the sun and this
converts matter into kinetic energy so the lifetime of the Sun is clearly much larger
than the value you obtained.)
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 sun is losing energy
Asun dt
3
3
as time increases. The energy is given by E = NkT , so dE = NkdT . Substituting
2
2
what we have, we get
3
NkdT
2
TF dT
dE
8π rsun
σ t
S=−
= σT 4 → − 2 2
= σT 4 → ∫
=
−
dt
4
∞ T
Asun dt
4π rsun dt
3Nk ∫0
∫
TF
∞
∴
T
dT
1 −3 f
1
=
−
T
=− 3
4
T
3
3T f
∞
2
1
8π rsun
σ
=
t
3
3T f
3Nk
t=
⎛ 2 × 10 30 kg ⎞
−23 J
⎜⎝ 1.67 × 10 −27 kg ⎟⎠ × 1.38 × 10 K
Nk
=
2
8π rsun
σ T f3 8π ( 6.96 × 10 8 m )2 × 5.67 × 10 −8
W
m 2T 4
× ( 6000K )
3
= 1.11× 1011 s = 3540yrs
5. Assume that you are a blackbody radiator. How much energy do you radiate in one
hour if your temperature is T = 310K ? At what maximum wavelength do you
radiate?
The energy you radiate in one hour is given from the Stefan-Boltzmann law. We
have
E
S=
= σT 4
At
.
E = Atσ T 4 = (1.02m 2 ) × 3600s × 5.67 × 10 −8
W
m2K 4
× ( 310K ) = 1.9 × 10 6 J
4
The surface are of a “body” was calculated by assuming a person of height 5' 7" and
about 1' wide, multiplying by two and assuming that the body is reasonably thin.
The maximum wavelength is obtained from Wein’s law.
2.9 × 10 −3 mK 2.9 × 10 −3 mK
λmax =
=
= 9.7 × 10 −6 m = 9.7 µ m which is in the infrared.
T
310K
6. 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
7. 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
8. 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
e
hc
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
⎜ ⎛⎜
⎟
⎟
−7
−23
⎜⎝ ⎝ 4.83×10 m×1.38×10 KJ ×6000 K ⎠
⎟⎠
e
−1
W
µ m⋅m 2
9. For thermal radiation, show that the relationship between the energy per volume per
du
dn
unit wavelength (
) and the number of photons per volume per wavelength (
)
dλ
dλ
du ⎛ hc ⎞ dn
=⎜ ⎟
is
.
dλ ⎝ λ ⎠ dλ
Let n be the number of oscillators per unit volume. The energy density is therefore
the energy per oscillator ( E ) times the number per unit volume, n . We have
therefore, u = E n , so
du
dn ⎛ hc ⎞ dn
= E
=⎜ ⎟
.
dλ
dλ ⎝ λ ⎠ dλ