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***and thence find the***l**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:

*E**l**= 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 m_{e}kT/ h^{2}]^{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

**= 0.264 cm**_{l }^{2}/g.
Take
the density of Earth’s atmosphere as:

**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

**= 0.264 cm**_{l }^{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

**= 0.0001 cm**_{e }^{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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