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Exercise Solutions
747
CHAPTER 2: The HertzSprung–Russell Diagram
2.1 Generally, the magnitudes m and the intensities I for two sources are related by
m1 D 2:5 log
m2
I1
;
I2
where log is the base-10 logarithm. Define the U , B, and V magnitudes by evaluating them
at the peak of the corresponding filters (which are generally around 1000 Angstroms wide),
ı
ı
U m.3650A /
ı
B m.4400A /
V m.5480A /:
Then the color indices are
ı
U
B ' 2:5 log
ı
I.4400A /
ı
B
V ' 2:5 log
I.3650A /
ı
I.5480A /
ı
U
V ' 2:5 log
I.4400A /
I.5480A /
ı
;
I.3650A /
which can be evaluated using the Planck law for the blackbody intensities
I D
1
2hc 2
5
e hc=kT
1
at the wavelengths for a temperature of 18,000 K.
2.3 For absolute bolometric magnitude M D
5, the ratio of luminosities is
L
D 10.Mˇ
Lˇ
M /=2:5
;
from which L D 7943Lˇ, using the absolute bolometric magnitude Mˇ D 4:75 for the
Sun. Solving
L D 4R2 T 4
for the radius R gives
R
D
Rˇ
Tˇ
T
2 s
L
D 331;
Lˇ
where T is the effective surface temperature and where the luminosity computed above and
the effective solar surface temperature Tˇ D 5780 K have been used. From the known
radius of the Sun, this gives a radius of about 1.5 AU (approximately the radius of the orbit
of Mars) and an average density of about 3:910 7 g cm 3 . The large size and low density
and surface temperature indicate that this star is a red supergiant.
2.5 The maximum luminosity for type I Cepheids is about 105 Lˇ (for periods around 100
days). This implies a maxium absolute magnitude of
M D Mˇ
2:5 log
L
'
Lˇ
7:75:
Exercise Solutions
748
For this limiting absolute magnitude to give an apparent magnitude m 28, the distance
must be
d D 10.m M C5/=5 D 140 Mpc:
If instead we assume m D 23, the corresponding maximum distance is about 14 Mpc.
2.7 From the parameterization of the period–luminosity relation
Mv D
.2:76.log P
1//
4:16 D
5:27:
Then the distance is
d D 10.mv
0:3 Mv C5/=5
D 11:3 Mpc;
where an apparent visual magnitude of mv D 25:3, corrected by 0.3 magnitudes for interstellar absorption, has been used.
2.9 The Balmer absorption line is produced by neutral hydrogen with the electron in the
first excited state. The Ca H and K lines are produced by singly-ionized Ca in the ground
state. Thus, we must compare the number of neutral hydrogens with electrons in the first
excited state to the number of CaC ions in their ground state. In 2.1.1 we estimated that
in the solar surface H C =H ' 10 4 , so we may assume that the hydrogen is all neutral.
The amount of this neutral hydrogen in the first excited electronic state is given by the
Boltzmann equation,
n2
g2
D
e
n1
g1
.E2 E1 /=kT
D
2.2/2
e
2.1/2
. 3:4 . 13:6//kT
D 5:5 10
9
;
where an excitation energy of 3.4 eV for the first electronic state and an ionization energy
of 13.6 eV have been used. Thus, the fraction of the hydrogen capable of producing the
Balmer absorption series is
n2
nI
n2
n2
'
D
D 5:5 10 9 :
nt
n1 C n 2
nt
n1
Now we repeat this analysis for the Ca ions. The ratio of ionized to neutral calcium in the
solar surface is
log
nC
uC
D log
C
n0
u0
5
2
log T
Ei (eV)
5040
T
0:179 D 2:66;
log Pe
where uC D 2:30 and u0 D 1:32 have been used, and the pressure is assumed to be
Pe D 31:6 dyne cm 2 (corresponding to log Pe D 1:5). Therefore, the ratio of Ca II to Ca
I is nII =nI D nC =n0 D 457 and almost all the calcium is in the first-ionized state (a similar
check shows that there is little population of Ca III). The ratio of Ca C in the first excited
state to that in the ground state is
n2
g2
D
e
n1
g1
.E2 E1 /kT
' 7:8 10
3
;
Exercise Solutions
749
where we’ve used E2 E1 D 3:12 eV, g1 D 2, g2 D 4, and T D 5800 K. Thus, in the
solar surface almost all Ca is in the ground state of the CaC ion. Carrying out a similar
calculation as for hydrogen,
n1
nII
n1
nII
n1
D
'
nt
n1 C n 2
nt
n1 C n 2
nI C nII
1
nII =nI
D
' 0:99:
1 C n2 =n1
1 C nII =nI
Thus, almost all the Ca in the surface of the Sun is in the right state to produce the Ca K
(and by similar analysis the H) line, but only about 5:5 10 9 of the hydrogen is in the
right state to produce the Balmer absorption series. Since the calcium abundance relative to
hydrogen in the Sun’s surface is about 2:2 10 6 , we conclude that the Ca K line should
be about
2:2 10 6
' 400
5:5 10 9
times stronger than the hydrogen Balmer absorption series in the solar spectrum.
2.11 (a) H ions have a binding energy of only 0.7 eV and there are no bound excited
states. We may view the transition H ! H as an ionization (with H playing the role of
a positive ion) and apply the Saha equation. Since H has two electrons in the hydrogen
n D 1 level and there is only one way to make that state, the statistical factors are one for
H and the Saha equation may be expressed as
log
n.H /
u.H /
D log
C
n.H /
u.H /
5
2
log T
Ei (eV)
5040
T
log Pe
0:179 ' 7:48;
where we’ve used u.H /=u.H / D 2, Ei D 0:7 eV, T D 6000 K, and Pe D 30 dyne cm
Thus,
n.H /
' 3:2 10 8 ;
n.H /
3
.
confirming that H has a very low abundance in the Sun.
(b) In the visible spectrum the continuum absorption from neutral hydrogen is dominated
by excitations from the third quantum level (Paschen transitions to the continuum; see Exercise 2.10). The third quantum level in hydrogen is at 12.1 eV relative to the ground
state, so the minimum energy for a bound–free transition from the n D 3 level is Emin D
13:6 12:1 D 1:5 eV (since 13.6 eV are required to ionize hydrogen). This corresponds
to a wavelength D hc=E D 827 nm. All other Paschen bound–free transitions will have
shorter wavelengths. The corresponding calculation for H (with a 0.7 eV ionization energy) gives D 17; 726 nm or shorter. Thus, essentially all H in the solar surface can
contribute to visible continuum absorption.
Exercise Solutions
750
(c) To compare the relative importance of continuum absorption from H and the Paschen
continuum, we must compare the abundance of neutral H in the n D 3 level (where Paschen
transitions originate) with the abundance of H ions. We may write
n.H /
n.H / n.H1 /
D
n.H3 /
n.H1 / n.H3 /
where Hn means neutral hydrogen in the nth principle quantum level. The first factor was
already found in part (a). The second factor may be obtained from the Boltzmann equation,
n.H1 /
g1
e
D
n.H3 /
g3
where E3
.E1 E3 /=kT
' 1:62 109 ;
E1 D 12:1 eV and gn D 2.n/2 have been used. Therefore,
n.H /
D .3:2 10
n.H3 /
8
/.1:62 109 / ' 51;
and if the atomic absorption coefficients are similar for H ionization and Paschen absorption ionization, we expect H to be of order 100 times more important than neutral
hydrogen for continuum solar absorption in the visible.
ı
2.13 At 4000 A = 400 nm the corresponding energy is 3.1 eV, so the iron must be in a
state with energy greater than Emin D .7:9 3:1/ eV D 4:8 eV to undergo bound–free
absorption at 400 nm. The corresponding Boltzmann factor is
n2
g2
D
e
n1
g1
Emin =kT
' 10
4
;
where we have assumed g2 =g1 of order one. The iron abundance in the Sun is n.Fe/=n.H/ 10 4 so
n.Fe; E 4:8 eV/
' 10 4 10 4 ' 10 8 :
n.H/
This is comparable to the H abundance but the iron is almost completely ionized since
from the Saha equation we obtain log.nC =n0 / 1:1 assuming uC =u0 1, a temperature
of 6000 K, and a pressure Pe D 30 dyne cm 3 . Thus only about 10% of the iron is present
ı
as Fe I, which is not abundant enough to compete well with H at D 4000 A . However,
ı
at D 2000 A the corresponding energy is 6.2 eV, giving Emin D 1:7 eV and a Boltzmann
ı
factor n2 =n1 ' 3:7 10 2 . Thus, at D 2000 A
ı
n.Fe absorbing at 2000 A /
D .3:7 10
n.H/
2
/.10
4
/.0:1/ ' 3:7 10
7
;
where the first factor is n2 =n1 computed above, the second factor is the abundance of iron
relative to hydrogen in the Sun, and the third factor is the fraction of iron that is not ionized
Exercise Solutions
751
that was estimated above. In Exercise 2.11 we found that the corresponding ratio for H is
ı
about 3:210 8 , so at 2000 A there is an order of magnitude more Fe atoms ready to absorb
than H ions and Fe is much more important than H in producing continuum absorption
ı
at 2000 A .