1) For a hydrogen plasma find:

(P

_{e}N_{i + 1}/N_{i}) = P_{e}N_{HII}/N_{HI}
at a temperature
of 5040 K, given the hydrogen partition functions are:

Z

_{ i + 1}= Z_{2}= 1 and: Z_{i}= Z_{2}= 2 with c_{ i }= 13.6 eV*Solution*:

By the Saha
equation: log(P

_{e}N_{i + 1}/N_{i}) =
-0.48 + log (2 Z

_{ i + 1}/ Z_{i }) + 2.5 log T – 5040 c_{ i}/ T
= -0.48
+ log (1) + 2.5 log (5040) – 5040 (13.6)/ 5040

= -0.48 + 0
+ 9.25 – 13.6 = -4.83

Antilog (-4.83)
= 1.5 x 10

^{-5 }Ans.
2)
For the temperature and conditions of problem (1) of the
previous set find the ratio of the probability that the system will be found in
any of the eight degenerate states of energy level E2 to the probability the
system will be in any of the two degenerate states of energy level E1.

Solution:

We have:

P(E2)

**/ P(E1)**_{ }**= [g**_{ }**/ g**_{2}**]**_{1 }**exp (- E2 – E1) / kT**_{ }
With g

_{1 }= 2(1)^{2}= 2
And: g

_{2 }= 2(2)^{2}= 8
For that problem also: N

**= N**_{2 }**and:**_{1 }
E2
= - 13.6 eV and E1 = -3.4 eV, therefore:

[g

**/ g**_{2}**]**_{1 }**exp (- E2 – E1) / kT**_{ }
= 4 exp [- 13.6eV – (-3.4 eV) ]/ kT

N

**/ N**_{2 }**= 1 = 4 exp (-10.2 eV)/ kT**_{1 }
Taking
natural logs:

ln
(4) =
(10.2 eV)/ kT

Then:

P(E2)

**/ P(E1)**_{ }**= ln (4) = 1.38**_{ }
Thus, the probability that the system will be found in any of
the eight degenerate states of energy level E2 is 1.38 times that of the probability
the system will be in any of the two degenerate states of energy level E1.

Or: P(E2)

**= 1.38 P(E1)**_{ }_{ }
3)
An
H-alpha line undergoes triplet splitting in the vicinity of a sunspot. The
undisturbed line is measured at l

_{o }= 6.62 x 10^{-}^{5}cm. The line shift on either side is: 0.0025 A. Use this information to find the strength of the magnetic field in: a) gauss and b)Tesla
We
have: D l
= (l

_{o})^{2 }e H/ 4 π m_{e}c^{2 }
Solve
for H:

H
= 4 π
(D l) m

_{e}c^{2 }/ (l_{o})^{2 }e
Where: c = 3 x 10

^{10}cm/s
m

_{e}= 9.1 x 10^{-28}g
(D l) = 2 (0.0025 A) = 0.005 A

= 5.0 x 10

^{-}^{11}cm
e
= 4.8 x 10

^{-10}e.s.u.
H
=

4 π(5.0 x 10

^{-}^{11}cm) (9.1 x 10^{-28}g)( 3 x 10^{10}cm/s)^{ 2}/ (6.62 x 10^{-}^{5}cm)^{ 2}e
= 2.44 x
10

^{2}G
But
1 T = 10,000 G, so:

H
= 2.44 x 10

^{2}G/ (10^{4}G/ T) = 0.024 T
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