SOLUTIONS to Midterm #2 for Astronomy 7A (Fall 2020)
This mid-term exam is a timed, “open-book” take-home exam. Open-book means that you
may refer to the textbooks, lecture slides, lecture recordings, your notes, your past problem
sets, discussion handouts, and course materials posted on this year’s Astro 7A bCourses
page, but that you may not consult anything or anyone else, including the Internet.
The exam is due by 11:59 pm on Monday, November 2, 2020. You should upload
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appropriate, and that you use the unit system specified in the question. Also be sure to
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1
Constants and Some Useful Formulae
c = 3.00 × 1010 cm/s = 3.00× 108 m/s
h = 6.626 × 10−27 erg s = 6.626× 10−34 J s = 4.14 ×10−15 eV s
σ = 5.67× 10−8 W m−2 K−4 = 5.67× 10−5 erg s−1 cm−2 K−4
kB = 1.38× 10−23 J/K = 1.38× 10−16 erg/K = 8.62× 10−5 eV/K
G = 6.67× 10−8 dyne cm2/g2 = 6.67× 10−11 m3 kg−1 s−2
mp = 1.673× 10−24 g = 1.673× 10−27 kg
mn = 1.675× 10−24 g = 1.675× 10−27 kg
me = 9.11× 10−28 g = 9.11× 10−31 kg
L = 3.90× 1026 W = 3.90× 1033 erg s−1
Solar mass: M = 2.0× 1033 g = 2.0× 1030 kg
R = 7.0× 1010 cm = 7.0× 108 m
T ≈ 5800 K (surface temperature)
Absolute Magnitude of Sun: M = +4.74
Apparent Magnitude of Sun: m = −26.83
M⊕ = 6.0× 1027 g = 6.0× 1024 kg
R⊕ = 6.4× 108 cm = 6.4× 106 m
MJupiter = 1.9× 1030 g = 1.9× 1027 kg
RJupiter = 7.1× 109 cm = 7.1× 107 m
1 AU = 1.5× 1013 cm = 1.5× 1011 m
1 pc = 3.09× 1018 cm = 3.09× 1016 m
1 eV = 1.602× 10−12 erg = 1.602× 10−19 J
1 radian = 206265 arcsec
1 Angstrom = 10−8 cm = 10−10 m
1 year = 12 months = 365 days = 3.15× 107 s
1
S
=
{
1/P − 1/P⊕, (inferior)
1/P⊕ − 1/P, (superior)
T = (JD − 2451545.0)/36525
b2 = a2(1− e2)
r =
a(1− e2)
1 + e cos θ
(for 0 ≤ e < 1)
Fnet = m
dv
dt
=
dp
dt
2
F = G
Mm
r2
U = −GMm
r
K =
1
2
mv2
vesc =
√
2GM/r
R =
m1r1 +m2r2
m1 +m2
µ ≡ m1m2
m1 +m2
r1 = − µ
m1
r
r2 =
µ
m2
r
L = r× p
r =
L2/µ2
GM(1 + e cos θ)
dA
dt
=
1
2
L
µ
P 2 =
4pi2
G(m1 +m2)
a3
〈U〉 = −2〈K〉
D (in pc) = 1/p
′′
F =
L
4pir2
m1 −m2 = −2.5 log10
(
F1
F2
)
3
m−M = 5log10(d)− 5 = 5log10
(
d
10pc
)
c = λν
λpeakT = 2.8977721× 10−3 m K
hνpeak ≈ 3kBT
Fblackbody = σT
4
L = AσT 4
Bλ(T ) =
2hc2
λ5
1
ehc/(λkBT ) − 1
Bν(T ) =
2hν3
c2
1
ehν/(kBT ) − 1
Bλ(T ) ≈ 2ckBT
λ4
BC = mbol − V = Mbol −MV
E = hν =
hc
λ
= pc
rn =
4pi0h¯
2
µe2
n2 (for hydrogen)
En = − µe
4
32pi220h¯
2
1
n2
= −13.606 eV 1
n2
(for hydrogen)
n1λ sin θ1 = n2λ sin θ2
1
fλ
= (nλ − 1)
(
1
R1
+
1
R2
)
dθ
dy
=
1
f
4
θdiff = 1.22
λ
D
m =
fobj
feye
a = a1 + a2
α = α1 + α2
α˜ = α cos i =
a
d
cos i
m1
m2
=
α2
α1
m1 +m2 =
4pi2
G
(αd)3
P 2
(λobserved − λrest)
λrest
= vr/c
m1
m2
=
v2
v1
=
v2r/ sin i
v2r/ sin i
=
v2r
v1r
m1 +m2 =
P
2piG
(v1r + v2r)
3
sin3 i
m32
(m1 +m2)
2 sin
3 i =
P
2piG
v31r
rs =
v
2
(tb − ta)
rl =
v
2
(tc − ta) = rs + v
2
(tc − tb)
B0 −Bp
B0 −Bs =
(
Ts
Tl
)4
δ ≈
(
rp
R?
)2
5
T0 ≡ R?P
pia
= 13 hr
(
P
1 yr
)1/3(
ρ?
ρ
)−1/3
T ≈ T0
√
1− b2
τ ≈ T0√
1− b2
rp
R?
K1 =
28.4329 m s−1√
1− e2
m2 sin i
MJ
(
m1 +m2
M
)−2/3(
P
1 yr
)−1/3
nvdv = n
( m
2pikT
)3/2
e−mv
2/2kT4piv2dv
vmp =
√
2kT
m
vmp =
√
3kT
m
Nb
Na
=
gbe
−Eb/kT
gae−Ea/kT
=
gb
ga
e−(Eb−Ea)/kT
gn = 2n
2 (for neutral hydrogen)
Z =
∞∑
j=1
gje
−(Ej−E1)/kT
Ni+1
Ni
=
2Zi+1
neZi
(
2pimekT
h2
)3/2
e−χi/kT
Pe = nekT
Iλ ≡ ∂I
∂λ
≡ Eλ dλ
dλ dt dA cos θ dΩ
〈Iλ〉 ≡ 1
4pi
∫
IλdΩ =
1
4pi
∫ 2pi
φ=0
∫ pi
θ=0
Iλ sin θ dθ dφ
uλ dλ =
1
c
∫
Iλ dλ dΩ =
4pi
c
〈Iλ〉dλ
6
Fλdλ =
∫
Iλ dλ cos θ dΩ =
∫ 2pi
φ=0
∫ pi
θ=0
Iλ dλ cos θ sin θ dθ dφ
δτλ = −κλρds
τ =
∫ s
0
nσλds =
∫ s
0
κλρds
Iλ = Iλ,0 e
−τλ
dIλ = −κλρIλds+ jλρds
λmfp = 1/(nσ) = 1/(ρκ)
7
Questions (100 points total)
Please remember to write and sign the honor code pledge on the first page of your exam. See
the first page of this document for instructions.
1 Conceptual Understanding [20 pts]
a. [5 pts] You’ve just arranged flowers in a clear vase. After you filled the vase with water,
you noticed that the flower stems appear broken at the water line. Using physics, explain
why the flower stems no longer look like straight lines even though you carefully selected
the flowers so that all of them had perfectly straight stems. Is there a way to predict
how the flower stems appear in the glass of water? If so, write down any relevant
equations and state any information that you would need to know about the system in
order to predict the appearance of the flower stems. Be sure to label any variables or
equations that you use.
The stems appear bent because the water and the air have different indices of refraction
(nwater, nair). If we know the angle at which the flowers enter the water, we can use
Snell’s law (n1,λ sin θ1 = n2,λ sin θ2) to predict the degree of apparent bending.
b. [7 pts] In class, we learned a formula for the change in specific intensity accounting
for both absorption and emission. Write down that equation and explain each variable.
Label the term(s) corresponding to absorption and emission. Do the absorption and
emission components look symmetric? If not, explain why. As part of your answer,
define absorption and emission.
The formula is dIλ = −κλρIλds + jλρds. The first term corresponds to absorption,
which is the removal of light from the beam, and the second term corresponds to
emission, which is the addition of light to the beam. The components are:
• κλ: absorption coefficient
• ρ: gas density
• Iλ: intensity
• ds: path length
• jλ: emission coefficient
No, the components are not symmetric. The absorption term contains Iλ and the
emission term does not. The reason why Iλ appears in the absorption term is that you
8
can only absorb light that is present in the beam. In contrast, the level of emission
does not depend on the amount of light already in the beam.
c. [8 pts] You friends made slides showing spectra of different types of stars, but they
forgot to add labels. Looking at the spectra, you notice that the spectrum of Star 1
contains strong Balmer lines; the spectrum of Star 2 contains weaker Balmer lines and
strong Ca II H and K lines; the spectrum of Star 3 reveals molecular absorption by
TiO; and the spectrum of Star 4 contains the same lines as the spectrum of Star 2, but
the spectral lines are much narrower. Your friends say that they included spectra of an
A3 III star, a G2 V star, a G4 III star, and an M3 V star. Match the spectra to the
stellar classifications and explain your reasoning. As part of your answer, explain the
nomenclature used to classify each star (i.e., what is meant by “A3” and “III” in “A3
III”?) and rank the stars in order of decreasing temperature.
The correct answers are:
• Star 1: A3 III because Balmer lines peak near A0
• Star 2: G2 V because the Sun has strong Ca II H & K lines
• Star 3: M3 V because M dwarfs are cool enough that molecular features are
present
• Star 4: G4 III because giant stars have lower surface gravities than dwarf stars.
High surface gravities lead to broader spectral lines.
The first part of the nomenclature (e.g., “A3”) refers to the spectral type, which
indicates the temperature. The second part (e.g., “III”) refers to the luminosity class,
which corresponds to surface gravity.
In order of decreasing temperature, the correct order is Star 1, Star 2, Star 4, Star 3.
9
2 Telescopes [20 pts]
For this problem, you will consider a telescope that has a primary mirror with a diameter of
4 m and a focal ratio f/2.7.
a. [5 pts] What is chromatic aberration? Will you need to worry about it when acquiring
data with this telescope? Why or why not?
Chromatic aberration is an image distortion that occurs because lenses focus light
of different colors at different positions. This problem occurs because the index of
refraction depends on wavelength and therefore the focal length of a lens is different
for different colors of light. The telescope considered in Problem 2 has mirrors rather
than lenses, so you do not need to worry about chromatic aberration. Chromatic
aberration is an issue for refracting telescopes but not for reflecting telescopes.
b. [4 pts] What is the focal length of this telescope? State your answer in meters.
We know that the focal ratio F is defined as F = f/D. The focal length f is therefore:
f = DF
f = (4 m)(2.7) = 10.8 m
The focal length is 10.8 m .
c. [5 pts] Now consider a 6-m telescope with a focal length of 12-m. Determine the focal
ratio of the 6-m telescope and state which of the two telescopes has a faster optical
system. What does it mean for an optical system to be “fast”?
Begin by determining the focal ratio:
F =
f
D
F =
12 m
6 m
= 2
The focal ratio is f/2 . The telescope with the faster optical system is the 6-m telescope .
Faster optical systems require less time to collect the same amount of light.
10
d. [6 pts] Imagine that both of the 4-m telescope and the 6-m telescope are observing the
same stellar binary at the same time on the same clear night from the same location. If
the binary system is 200 pc from Earth and the two stars are separated by 300 AU, would
it be possible to resolve the system at a wavelength of 500 nm with either telescope?
To support your answer, compare the angular separation of the binary components to
the minimum distance that could be resolved with each telescope (i.e., compute the
ratios αbinary/α4m and αbinary/α6m). Assume that the stars are at maximum angular
separation, that the stars have circular orbits, and that the orbital plane of the binary
is perpendicular to our line of sight. The seeing is roughly 0.4” and neither telescope
is equipped with an adaptive optics system.
Begin by determining the angular separation α of the two stars:
α =
a
D
α =
(300 AU)(1.5× 1011 m/AU)
(200 pc)(3.09× 1016 m/pc) = 7.28 radians = 1.5”
Now determine the diffraction limits of each system:
θ = 1.22
λ
D
θ4−m = 1.22
500× 10−9 m
4 m
= 0.03”
θ6−m = 1.22
500× 10−9 m
6 m
= 0.02”
The diffraction limits are much smaller than the angular separation of the system, but
the telescopes will not achieve diffraction-limited images because of the presence of
the atmosphere. The question states that the seeing is 0.4”, which is roughly 3 times
smaller than the angular separation between the two stars. Accordingly, it should be
possible to resolve the system at 500 nm using either telescope.
11
3 Planetary System or Stellar Binary? [35 pts]
You’ve been conducting a photometric survey of the sky. When examining your most recent
data, you found that one of the stars in your sample (Star A) exhibits periodic dimming
events. In this problem, you’ll investigate whether the dimming events are more likely to be
caused by a transiting planet or an eclipsing binary star.
a. [10 pts] Star A has a bolometric apparent magnitude m = 9, a peak wavelength of
λmax = 600 nm, and a parallax of 23 mas. Determine the luminosity (in L), effective
temperature (in K), and radius (in R) of Star A.
We can find the luminosity from the absolute magnitude, which we can find by first
measuring the distance to the system:
d =
1
p
d =
1
23× 10−3 = 43.5 pc
We can now find the absolute magnitude from the distance modulus:
m−M = 5 log10(d)− 5
M = m− 5 log10(d) + 5
M = 9− 5 log10(43.5) + 5 = 5.8
Next, we can compute the luminosity by comparing this star to the Sun:
M1 −M2 = −2.5 log10
L1
L2
L1
L2
= 10−(M1−M2)/2.5
L = 10−(M−M)/2.5L
L = 10−(5.8−4.74)/2.5L
L = 0.373L
12
We can find the effective temperature from the peak wavelength:
λpeakT = 2.8977721× 10−3 m K
T =
2.8977721× 10−3 m K
λpeak
T =
2.8977721× 10−3 m K
600× 10−9 m
T = 4830 K
Finally, we can determine the radius from the Stefan-Boltzmann equation:
L = 4piR2σT 4eff
Comparing this star to the Sun, we find:
L
L
=
4piR2σT 4eff
4piR2σT 4eff,
L
L
=
R2T 4eff
R2T 4eff,
R
R
=
√
L
L
(
Teff,
Teff
)2
R
R
=
√
0.373
(
5800 K
4830 K
)2
R = 0.88R
The star has a luminosity L = 0.373L , an effective temperature Teff = 4830 K , and
a radius R = 0.88R .
b. [6 pts] If you were unable to solve Part (a), assume L = 0.22L, Teff = 4400 K, and
R = 0.8R for the rest of this problem. (Do not use those values when attempting
the previous question.) During the dimming events, Star A has a bolometric apparent
magnitude of m = 9.005. The initial decrease in brightness takes place over 10 minutes
and then the star remains at the minimum brightness for 3.0 hours before brightening
again over another 10 minutes. The total duration of the event (ingress, time at min-
imum brightness, and egress) is 3.3 hours. Assuming that the object responsible for
the dimming events does not contribute a significant amount of light to the system,
13
calculate the size of the object blocking the star during the dimming events. State your
answer in units of meters, solar radii (R), and Earth radii (R⊕).
We can measure the size of the object responsible for the dipping by converting the
magnitude change into a flux ratio and remembering that the decrease in flux is related
to the radius ratio.
min −mout = −2.5 log10
(
Fin
Fout
)
Fin
Fout
= 10−(min−mout)/2.5
Fin
Fout
= 10−(9.005−9)/2.5
Fin
Fout
= 0.995
The transit depth is therefore δ = 1 − 0.995 = 0.005. Solving for the radius of the
object responsible for the dimming, we find:
δ =
(
Rp
R?
)2
Rp
R?
=
√
δ
Rp =
√
δR?
Rp =
√
0.005(0.88R)(7× 108 m/R) = 4.2× 107 m = 0.059R = 6.5R⊕
The radius of the object responsible for the dimming is 4.2× 107 m = 0.059R = 6.5R⊕ .
c. [3 pts] Given your answer to Part (b), do you think the dimming events are caused by
a transiting planet or an eclipsing binary star? Why?
The object responsible is a transiting planet because the radius is too small to be a
star. The object has an estimated radius larger than Neptune and smaller than Saturn.
d. [10 pts] You obtain radial velocity observations of Star A to learn more about the object
responsible for the dimming events. Inspecting your data, you find that the minimum
14
radial velocity is 112 m s−1, the maximum radial velocity is 142 m s−1, and the orbital
period is 8.23 days. Determine the mass (in M⊕) of the object responsible for the
dimming. Assume that the mass of the host star can be approximated using the mass-
luminosity relation L/L = (M/M)a where a = 3.5.
The radial velocity semiamplitude K is half of the difference between the maximum
radial velocity and the minimum radial velocity:
K =
RVmax −RVmin
2
K =
142 m s−1 − 112 m s−1
2
= 15 m s−1
Next, use the mass-luminosity relation to estimate the mass of the host star:
L/L =
(
M
M
)a
M/M =
(
L
L
)1/a
M =
(
L
L
)1/a
M
M = (0.373)1/3.5M = 0.755M
Finally, use the RV equation to determine the mass of the planet. We’ll make the
assumptions that e = 0, i = 90◦, and mP << M?.
K1 =
28.4329 m s−1√
1− e2
m2 sin i
MJ
(
m1 +m2
M
)−2/3(
P
1 yr
)−1/3
K ≈ 28.4329 m s−1 mp
MJ
(
M?
M
)−2/3(
P
1 yr
)−1/3
mp
MJ
≈ K
28.4329 m s−1
(
M?
M
)2/3(
P
1 yr
)1/3
mp
MJ
≈ 15 m s
−1]
28.4329 m s−1
(0.755M)
2/3
(
8.23 d
365.25 d
)1/3
mp ≈ 0.124MJ = 39M⊕
The object responsible for the dimming has a mass of approximately 39M⊕ .
15
e. [6 pts] As a final check, verify that the observed transit duration is consistent with the
orbital period revealed by the radial velocity data. Assume that the object has a circular
orbit (e = 0) and crosses directly across the center of the host star (b = 0).
The characteristic timescale of a planetary transit is given by
T0 ≡ R?P
pia
= 13 hr
(
P
1 yr
)1/3(
ρ?
ρ
)−1/3
The transit duration is T ≈ T0
√
1− b2, so the transit duration should be roughly T0
for a central transit (i.e., b = 0).
Writing the stellar density in terms of the stellar radius and mass, we find
ρ? =
M?
(4/3)piR3?
ρ?
ρ
=
M?
M
(4/3)piR3
(4/3)piR3?
ρ?
ρ
=
M?
M
(
R
R?
)3
The transit duration is therefore:
T0 = 13 hr
(
P
1 yr
)1/3(
ρ?
ρ
)−1/3
T0 = 13 hr
(
P
1 yr
)1/3(
M?
M
(
R
R?
)3)−1/3
T0 = 13 hr
(
P
1 yr
)1/3(
M?
M
)−1/3(
R
R?
)−1
T0 = 13 hr
(
P
1 yr
)1/3(
M
M?
)1/3(
R?
R
)
T0 = 13 hr
(
8.23 d
365.25 d
)1/3(
1M
0.755M
)1/3(
0.88R
1R
)
= 3.5 hr
The transit duration should be approximately 3.5 hr , which is fairly similar to the
value given in part b. The observed total transit duration is 3.3 hr. The fact that the
observed duration is slightly shorter suggests that the planet might not be transiting
across the exact center of the star (i.e., our assumption that b = 0 was invalid). It’s
also possible that the orbit is less circular, but eccentricity needs to be quite large to
cause an appreciable difference in transit duration.
16
4 Stellar Atmospheres [25 pts]
a. [10 pts] Imagine that a star contains a fictional element known as Ay. The ground
state energy of Ay is E1 = −7 eV, the energy of the first excited state (n = 2) has an
energy of E2 = −4 eV, and the second excited state has an energy of E3 = −2 eV. You
may ignore the higher order bound states. The degeneracy of each energy level is given
by the formula gn = 3n
2 − n. Assume ZI = 1 for Ay and consider only singly ionized
Ay (Ay II) and neutral Ay (Ay I). At a temperature of 7000 K and an electron density
Pe = 25N m
−2, what fraction of the Ay atoms should be ionized?
Note: There was a typo in this question. The question should have instructed you
to assume ZII = 1 not ZI = 1. We will therefore accept answers that made either
assumption.
The Saha equation tells us the fraction of atoms in a particular ionization state:
Ni+1
Ni
=
2Zi+1
neZi
(
2pimekT
h2
)3/2
e−χi/kT
Before using the Saha equation, we need to determine the partition functions. The
question tells us that ZII = 1 for ionized Ay and provides the degeneracy formula
needed to calculate ZI for neutral Ay. Considering the ground state (n = 1), first
excited state (n = 2), and second excited state (n = 3), we have:
gn = 3n
2 − n
g1 = 3(1
2)− 1 = 2
g2 = 3(2
2)− 2 = 10
g3 = 3(3
2)− 3 = 24
Considering the first three states, the partition function is therefore
ZI =
∞∑
j=1
gje
−(Ej−E1)/kT
ZI ≈ g1e−(E1−E1)/kT + g2e−(E2−E1)/kT + g3e−(E3−E1)/kT
ZI ≈ g1 + g2e−(E2−E1)/kT + g3e−(E3−E1)/kT
ZI ≈ 2 + 10e−(−4 eV+7 eV)/kT + 24e−(−2 eV+7 eV)/kT
where kT = (8.62× 10−5 eV K−1)(7000 K) = 0.6033 eV = 9.66× 10−20 J
17
The partition function for neutral Ay is therefore:
ZI ≈ 2 + 10e−(3 eV)/(0.6033 eV) + 24e−(5 eV)/(0.6033 eV)
ZI ≈ 2.075
Recall that
The problem should have stated that ZII = 1. Rewriting ne in terms of the electron
pressure using Pe = nekT and substituting ZI = 2.075 and = ZII = 1, we can use the
Saha equation to find the ionization fraction:
NII
NI
=
2ZII
neZI
(
2pimekT
h2
)3/2
e−χ1/kT
NII
NI
=
2ZIIkT
PeZI
(
2pimekT
h2
)3/2
e−χ1/kT
NII
NI
=
(2)(1)(9.66× 10−20 J)
(25N m−2)(2.075)
(
2pi(9.11× 10−31 kg)(9.66× 10−20 J)
(6.626× 10−34 J s−1)2
)3/2
e−(7 eV)/(0.6033 eV)
NII
NI
= 48.3
The ionization fraction is
NII
NI +NII
=
1
NI/NII + 1
NII
NI +NII
=
1
1/48.3 + 1
= 0.98
Approximately 98% of the Ay atoms should be ionized.
Given the typo in the question, students may have also assumed that ZII = 2.075 and
ZI = 1, which would lead to a ratio NII/NI = 208 and an ionization fraction of 99.5%.
If students assumed that ZII = 1 and ZI = 1, then they would have found a ratio
NII/NI = 100 and an ionization fraction of 99%.
18
b. [6 pts] At the same conditions as in part (a), what fraction of the neutral Ay atoms
would have electrons in the first excited state?
We should use the Boltzmann equation:
Nb
Na
=
gbe
−Eb/kT
gae−Ea/kT
=
gb
ga
e−(Eb−Ea)/kT
We’re instructed to consider only the ground state (n = 1), first excited state (n = 2),
and second excited state (n = 3), so we have
N2
N1
=
g2
g1
e−(E2−E1)/kT
N2
N1
=
10
2
e−(−4 eV+7 eV)/(0.6033 eV) =
10
2
e(−3 eV)/(0.6033 eV) = 0.03465
N3
N2
=
g3
g2
e−(E2−E1)/kT
N3
N2
=
24
10
e−(−2 eV+4 eV)/(0.6033 eV) =
24
10
e(−2 eV)/(0.6033 eV) = 0.08724
N3
N1
=
g3
g1
e−(E2−E1)/kT
N3
N1
=
24
2
e−(−2 eV+7 eV)/(0.6033 eV) =
24
2
e(−5 eV)/(0.6033 eV) = 0.00302
The fraction of neutral Ay atoms that have electrons in the first excited state is there-
fore:
N2
N1 +N2 +N3
=
1
N1/N2 + 1 +N3/N2
N2
N1 +N2 +N3
=
1
1/0.03465 + 1 + 0.08724
N2
N1 +N2 +N3
= 0.033
Roughly 3% of the neutral Ay atoms should have electrons in the first excited state.
19
c. [6 pts] If the temperature of the star were to increase or decrease, how would the fraction
of Ay atoms with electrons in the first excited state change? Make a rough sketch of
showing the fraction of Ay atoms with electrons in the first excited state on the y-axis
versus temperature on the x-axis. Be sure to label your axes and indicate which way
the temperature increases. You do not need to do any math for this problem or put
specific numbers on the axes, but write 2-3 sentences explaining the behavior shown on
your graph.
If the temperature were to increase, then there would be even fewer neutral atoms
so the fraction of Ay atoms with electrons in the first excited state would decrease .
Similarly, if the temperature were to decrease significantly, the neutral fraction would
increase but more of the electrons would be in the ground state so the fraction of Ay
atoms with electrons in the first excited state would also decrease . A slight decrease
in temperature to temperatures near 5300 K would lead to a higher fraction of atoms
with electrons in the first excited state because there would be more neutral atoms and
the temperature would still be high enough for the electrons to be excited into the first
excited state. Students can receive full credit for slightly different answers as long as
they mention the dueling effects of ionization and excitation and correctly identifying
the general trends with temperature. Most students will probably not have identified the
exact peak near 5300 K. This peak occurs roughly where half of the Ay is ionized.
A graph showing the combined effects of ionization and excitation is shown in Figure 1.
Student charts should show the general shape of a peak at intermediate temperatures
and low values at significantly higher and significantly lower temperatures, but the
exact shape and position of the peak does not need to match. The axes do not need to
have numbers but the axes should have titles and the direction of temperature increase
should be indicated.
d. [3 pts] In words, explain why some spectral features appear as jumps (e.g., the Balmer
jump) while others appear as absorption lines at particular wavelengths. What physical
processes are responsible for these features?
Spectral features that appear as jumps are caused by bound-free absorption while fea-
tures that appear as absorption lines are caused by bound-bound absorption . Bound-
bound transitions (i.e., electronic excitation) require exact amounts of energy, so the
feature occurs at a discrete range of wavelengths. In contrast, bound-free phenomena
like ionization require only a threshold energy level so they occur over a broader range
of wavelengths as long as the wavelength is a least as blue as the threshold level.
20
Figure 1: Fraction of ionized Ay (top left), fraction of neutral Ay with electrons in the first
excited state (top right), and fraction of Ay atoms that are both neutral and have electrons
in the first excited state (bottom) as a function of temperature. The dashed vertical line
indicates the temperature considered in the problem (T = 7000 K).
21