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Astronomy 120: Galaxies and the Universe
Fall 2016
Homework # 2 Solutions
1) (15 points) Imagine a star the same size as the sun but with a spectrum that peaks at
0.100 microns.
a) What is the surface temperature of this star?
Km
Use Wien’s law for a blackbody: λmax = 0.0029
.
T
We are given that λmax = 0.100 microns.
Remember to convert microns to meters:
−6 10 m
0.100 micron
= 0.100 × 10−6 m
1 micron
Now plug into Wien’s law to find the surface temperature of the star given the peak wavelength
of its spectrum:
0.0029 Km
T =
λmax
0.0029 Km
T =
0.100 × 10−6 m
T = 29, 000 K
b) What color would the star appear?
The peak wavelength of the star’s spectrum, 0.100 microns, lies in the ultra-violet part of the
electromagnetic spectrum. Stars behave as blackbodies thus, despite peaking in the UV, it emits
light at all wavelengths. The amount of flux a star emits is directly proportional to T4 , thus for a
star with a high surface temperature will also emit more flux at all wavelengths than a similar star
with a lower surface temperature. Couple the high surface temperature with the peak emission
wavelength in the UV, this star will appear blue-ish as the flux in the blue part of the visible
spectrum will be much greater than the flux in the red part of the visible spectrum.
c) How much more or less energy is emitted each second from each square meter of the
surface of this star?
The trick here is to realize what quantity is being requested. Because the question asks about
energy per second per square meter, we must be talking about flux. Now, since we know the
temperature of the star from part (a), we can use the Stephan-Boltzmann Equation to find the
flux of the star and the sun.
F∗ =
σT∗4
=
F = σT4 =
5.67 × 10
−8
W
m2 K 4
5.67 × 10−8
W
m2 K 4
(29000 K)4 = 4.01 × 1010 W/m2
(5800 K)4 = 6.42 × 107 W/m2
Finally, we are asked to compare these numbers. If we take a ratio, we find
F∗
4.01 × 1010 W/m2
=
= 625
F
6.42 × 107 W/m2
while we can calculate the difference as
∆F = F∗ − F = 4.01 × 1010 W/m2 − 6.42 × 107 W/m2 = 4.01 × 1010 W/m2
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d) How many times more or less luminous than the sun would that star be?
Using the ratio method we can compare the luminosity of this star to the luminosity of the Sun.
The equation for luminosity is L = 4πR2 F , where F = σT 4 .
L∗
4πF∗ R∗ 2
=
L
4πF R 2
We are told that this star is the same size as the Sun so R∗ = R .
L∗
4πF∗ R 2
=
L
4πF R 2
L∗
F∗
=
L
F
L∗
σT∗ 4
=
L
σT 4
L∗
T∗ 4
=
L
T 4
The temperature of the Sun is 5800 K, and we know the temperature of the star from part (a).
L∗
29, 0004
=
L
58004
L∗
= 625.0
L
L∗ = 625.0L
2) a) The bright star Rigel (in the constellation of Orion) has a surface temperature of
12,000 K. What is the ratio of the energy flux of Rigel compared to the Sun?
Energy emitted each second from each square meter of a star’s surface is the star’s flux (usually
expressed in W/m2 ). Compare the flux from Rigel’s surface with the flux from the Sun’s surface
using ratios. There is no need to actually calculate the numerical value of the flux in W/m2 .
FD
σTD 4
=
F
σT 4
FD
TD 4
=
F
T 4
The temperature of the Sun is 5800 K, and the temperature of Rigel is given as 12, 000 K.
FD
120004
=
= (2.06897)4
F
58004
FD
= 18.32
F
Notice that Rigel, a star with a hotter surface temperature, emits more energy per square meter
than the Sun.
b) Rigel is a supergiant star with a luminosity 120,000 times greater than that of our
sun. What is the radius of Rigel, in solar units?
Again, there is no need to actually calculate the numerical value of the radius, as the question asks
for the radius in solar units.
L = 4πR2 F
2
Ratios come in handy again!
2
LD
4πRD
FD
=
2F
L
4πR
LD
RD 2 FD
=(
) ×
L
R
F
Solving for radius:
(
(
RD 2
LD
F
) =
×
R
L
FD
RD
LD
F 1/2
)=(
×
)
R
L
FD
We are told that Rigel has a luminosity 120,000 times that of the sun, so LD = 120, 000L . And
we know from part (a) that FD = 18.32F , thus
RD
120, 000L
F
=(
×
)1/2
R
L
18.32F
RD
120, 000 1/2
=(
)
R
18.32
RD
= (6550.22)1/2
R
RD = 80.9R
Note: “in solar units” for this problem means “How many solar radii are equivalent to the radius
of Rigel?” This is found by the ratio method, and then solve the ratio so that you get an answer
for RD in terms of R . You don’t need to solve for the radius of Rigel in meters or kilometers,
and be careful not to substitute 1 AU for the radius of the Sun. FYI: 1 AU is the distance from
the Earth to the Sun.
3) (21 points) One of the nearest stars is Sirius B star, a white dwarf star which orbits
Sirius A, the brightest star int he sky. Sirius B star has a radius of 0.0084R and a
luminosity of 0.026L .
a) What is its surface temperature?
Since the values given in the problem are relative to the solar values, it is straightforward to
compare the ratio of Sirius B’s surface temperature to that of the Sun:
LV M
L
=
4πσRV2 M TV4 M
2 T4
4πσR
0.026
=
(0.0084)2
368.48
=
TV M
=
TV4 M
T4
TV4 M
T4
4.38T
Assume
T = 5800K
TV M = 4.38 × 5800K = 25, 404K
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b) At what wavelength would that star emit most of its radiation?
Using Wien’s Law:
λmax
=
λmax
=
λmax
=
0.0029
T
0.0029
25, 404 K
1.14 × 10−7 m = 0.114 µm
Note: We use the temperature of Sirius B star found in part a. Even though we round to two
significant figures in part a, remember to use a more precise temperature in determining the answer
to part b. You don’t want to cause incorrect rounding later on. Also, remember to show as much
work as possible, so that if you used an incorrect answer to part a in order to find the answer to
part b, you might at least earn some partial credit for showing correct use of Wien’s Law.
c) What is the radius of the star compared to that of the earth? (express your answer
as a ratio)
Since the R = 6.955 × 105 km and R⊕ = 6.378 × 103 km, we can express the radius of the
earth with respect to the sun by taking the ratio. We then get that R⊕ = 0.00917R . Now, by
substituting this ratio into what we already know,
RSB
RSB
=
0.0084R
=
0.0084R
=
0.916R⊕
R⊕
0.00917R
or R⊕ = 1.09RSB
4) (13 points) Black holes are objects whose gravity is so strong that not even an object
moving at the speed of light can escape from them. Hence black holes themselves do
not emit light. But it is possible to detect radiation from material falling toward a black
hole. Some of this material is compressed and heated to temperatures around 106 K.
a) Calculate the wavelength of maximum emission for this temperature.
This requires a straightforward application of Wien’s Law:
λmax
=
λmax
=
λmax
=
0.0029 mK
T
0.0029 mK
106 K
2.9 × 10−9 m = 2.9 nm
b) In what part of the electromagnetic spectrum does this wavelength lie?
Comparing this wavelength to a chart of the electromagnetic spectrum, this light falls in what
is known as the “soft” X-ray band. (“Hard” X-rays just have shorter wavelengths, and therefore
more energy, and typically penetrate intervening material more easily.) Ultraviolet radiation was
also accepted as an answer.
5) (40 points) When an electron in a hydrogen atom drops from the 5th orbital level to the
2nd orbital level, it emits a photon with a frequency of 6.91 × 1014 Hz. When an electron
drops from the 4th orbital level to the 2nd orbital level, it emits a photon with a frequency
of 6.16 × 1014 Hz.
a) Calculate the energies of these 2 photons (in eV).
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For both transitions, a photon will be emitted whose energy is equal to the difference in energy
between the two orbital levels:
= E initial − E final
∆E initial−final
and
∆E
= hν,
where h is Planck’s constant, which is 4.1357 × 10−15 eV · s.
So, inserting the numbers,
∆E52
= hν
=
=
(4.1357 × 10−15 eV · s) × (6.91 × 1014 Hz)
2.86 eV
and
∆E42
= hν
=
(4.1357 × 10−15 eV · s) × (6.16 × 1014 Hz)
=
2.55 eV
b) Draw an energy level diagram (NOT simply a picture of the Bohr model of the atom)
showing these electronic transitions.
Your diagram should look something like the following:
!"#"$"
!"#"%"
!"#"&"
)*+,+!"
)*+,+!"
!"#"'"
!"#"("
c) Using the energies of the photons (NOT the Bohr formula!) determine the WAVELENGTH of the photon emitted when the electron drops from the 5th orbital level
to the 4th orbital level.
In order to find the wavelength λ of the emitted photon, we first need to calculate its energy. The
electron falls from orbital level 5 to level 4, so the energy of the photon will be ∆E54 . Fortunately
we calculated everything that we need in part (a): ∆E54 is just the amount of additional energy
released when the photon falls from level 5 rather than 4 as long as they’re both falling to the
5
same level (2 in this case).
So,
∆E54
=
∆E52 − ∆E42
=
2.86 eV − 2.55 eV
=
0.31 eV
Now we can get the photon’s wavelength by using the formula
∆E
λ54
hc
λ
hc
=
∆E54
(4.1357 × 10−15 eV · s)(3.00 × 108 m/s)
=
0.31 eV
= 4.00 × 10−6 m
=
=
4.00 microns
d) In what part of the electromagnetic spectrum is each of these 3 photons? (e.g., x-ray,
ultraviolet, visible, infrared, radio, etc.)
Note that the 4 microns photon has wavelength longer than 0.7micron = 700nm but shorter than
1000microns = 1mm. The electromagnetic spectrum in the Textbook indicates that photons in this
wavelength range are called infrared photons. The 2.86eV and 2.55eV photons have wavelengths
434nm and 487nm, which are both in the visible (optical) part of the electromagnetic spectrum.
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