1)Consider the ionization
of helium, with three stages of ionization for which the partition
functions are: Z I
= Z 3 =
1, and Z 2
= 2
With ionization energies:
c 1 = 24.58 eV and c 2 = 13.6 eV
For a temperature of T =
5040 K write out two versions of the Saha equation for:
log( P e
N 2 /N 1)
and log( P e N 3
/N 2)
And hence, compute: (P e N 2 /N 1) and (P e
N 3 /N 2)
Solution:
Using the information given and the Saha eqn. we may write:
a) log( P e N 2 /N 1) = -0.48 + log (2 Z i + 1 / Z i ) + 2.5 log T – 5040 c i/
T
= -0.48 + log
(2 Z 2 / Z 1 ) + 2.5 log T – 5040
c 1 /
T
= -0.48 + log
(2 (2) / 1 )
+ 2.5 log (5040) – 5040 (24.85 eV ) / 5040
b) log( P e N 3 /N 2) = - 0.48 +
log (2 Z 3
/ Z 2 ) +
2.5 log T – 5040 c 2 /
T
= - 0.48 + log (2 (1) / 2 ) + 2.5 log (5040) – 5040 (13.6 ) /
5040
Now,
simplify each equation and solve for the respective quantities:
(P e N 2 /N 1) and (P e
N 3 /N 2)
a) log( P e N 2 /N 1) = -0.48 + log(2) + 9.25 - 24.85
= -0.48 + 0.30 + 9.25 – 24.85 = -15.78
(P e N 2 /N 1) = Antilog(-15.78) = 1.6 x 10 -16
b) log(
P e N 3 /N 2)
= - 0.48 + log (1) +
9.25 - 13.6
= -0.48 + 0
+ 9.25 - 13.6 = -
4.83
(P e
N 3 /N 2)
= Antilog(-4.83) = 1.5 x 10 -5
Ans.
2)The
diagram below was sketched by a first year astrophysics student.
It represents part of the emission spectrum of atomic hydrogen. It
contains a series of lines, and the wavelengths of some (in nm) are marked.
There are no lines in the series less than 91.2 nm
(a) In which region of the
E-M spectrum would these lines occur?
(b)
Obtain a relation between E and l and thence find the photon energies
equivalent for all the wavelengths marked.
(c)Use
this information to map a partial energy level diagram for hydrogen. Show and
label clearly the electronic transitions responsible for the emission lines
labeled.
Solutions:
a)the lines occur in the UV (ultraviolet)
part of the EM spectrum (Lyman series)
b) We
have: El =
1.99 x 10 -16 J nm
Then
the energy for the given wavelength can be obtained from:
E = 1.99 x 10 -16 J nm /
(l nm)
The energies for each wavelength, starting
from the shortest, are then:
E1 = 2.18 x
10 -18 J
E2 = 2.09 x
10 -18 J
E3 = 2.04 x 10 -18 J
E4 = 1.93 x 10 -18 J
E5 = 1.63 x
10 -18 J
For each of the above, an energy difference
(i.e. between energy levels) can be confirmed such that the wavelength of the photon emitted can be
found. Since E = hf = h (c/ l):
l =
hc/ (E2 – E1)
Here: E(n=2)
= - 13.6/4 (eV) =
-(3.4) x 1.6 x 10-19 J
=
-5.4 x 10-19 J
E(n=1) = -(13.6)
x 1.6 x 10-19 J =
-21.8 x 10-19 J
Then the energy
difference is:
E2 – E1 = [- 5.4 –
(-21.8)] x 10-19 J = 16.4 x 10 -19 J
l =
hc/ (E2 – E1)
l =
(6.626 x 10- 34 J-s)(3 x 10 8 m/s)/ 16.3 x 10-19
J
l =
1.21 x 10 -7 m
Which is the 121.7
nm value.
In similar fashion
all the other wavelengths can be computed using the energy difference, so that:
l4 = hc/ E4
= 102.7 nm
l3 =
hc/ E3 = 97.3 nm
l2 =
hc/ E2 = 95.0 nm
l1 =
hc/ E1 = 91.2
nm
The sketch of the line transitions is shown
below with the shortest wavelength (l1 ) corresponding to
the largest energy level difference, i.e. E1 =
(n5 – n1).
3)a) A
stellar atmosphere is composed of pure hydrogen, at a temperature of 9600 K and
with electron pressure, P e = 200 dyne cm -2 .
Use the Saha equation to compute the fraction of atoms that are ionized
in relation to the total, i.e. find: N II /( N I + N II )
The
partition functions are: Z II = 1 and
Z I = 2
Hint:
Check to see whether (E 2
- E 1 ) >> kT
with
k
= 8.6174 x 10 -5 eV/K
So the
Boltzman factor exp (- E2 – E1) /
kT<< 1
a)Solution:
Note
that (E2 – E1) = 10.2 eV and 10.2 eV >> kT
Where
kT= (8.62 x 10 -5 eV/K) (9600K) = 0.82 eV
Then the Boltzmann factor:
exp (- E2 – E1) / kT is much less than 1
Then:
(N i + 1 /N i) = (kT)/ P e [2 π me kT/ h2 ]3/2 ( e - c i/ kT)
= ( N II )/ N I
e - c i/ kT = exp (-13.6/ 0.82 eV)
= 6.2 x 10 -8
Then we find:
(kT)/ 200 dyne cm -2 )[ 2 π ((9.1 x 10
-28 g) (0.82)/ (6.62 x 10
-27 erg-s )2
]3/2 ( e - c i/ kT)
= 0.933
But: N II / N total =
N II /( N I + N II
) =
{ N II / N I /
(1 + N II / N I ) }
= [0.933 / 1 + 0.933]
= 0.933/ 1.933 = 0.482
Or nearly half of hydrogen is ionized at 9600K
For (b) 0ne will find by comparing the computed fractions for
ionization at the other temperatures, the one for 9600K is at a higher
proportion for ionized fraction, thus showing max. intensity is obtained near
that temp. (actually at 9520K which can be verified by computation)
(4)
Calculate how far you could see through the Earth’s atmosphere if it had the
same opacity as the solar atmosphere, where:
k l
= 0.264 cm 2 /g.
Take
the density of Earth’s atmosphere as:
r = 0.0012 g cm -3
r = 0.0012 g cm -3
The
density of the solar atmosphere is:
r s = 2.5 x 10
-7
g cm -3
And we
know the intensity declines over a characteristic distance (length scale):
ℓ = 1/ k l r
Solution:
We can
use the last relation: ℓ = 1/ k l r
But
substitute the solar opacity: k l = 0.264 cm 2 /g
So:
ℓ = 1/ (0.264 cm 2 /g) (0.0012 g cm -3 ) = 3.1 x 10
3
cm =
3.1 x
10 1 m = 31m
Earth’s
opacity is normally: k e = 0.0001 cm 2 /g
So
one would theoretically be able to see:
ℓ = 1/ k e r = 8.3
x 10 6 cm = 8.3 x 10
4
m or 83 km
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