Key for AST3018 HMW 3
25 Points
Leah Simon
8 Oct 2009
3.1 (3 pts)
Here we simply need to use the Boltzmann Distribution:
N
2
N
1
=
g
1
g
2
e
−
E
2
−
E
1
/
kT
(1)
where g
n
is the degeneracy in the nth state. For Hydrogen this is g
n
= 2n
2
, and E
n
= 13.6 eV * 1/n
2
a.) n
2
/n
1
= 1, g
2
/g
1
= 4, therefore, using k = 8.617 × 10
−5
eV K
−1
yields T = 8.54 × 10
4
K.
b.) n
3
/n
1
= 1, g
3
/g
1
= 9, therefore T = 6.38 × 10
4
K.
Q 3.2 (1 pts)
We get this from the original classi±cation method where the stars were given a letter
designation based on their Hα strength. As such strongest to weakest would be: A, B, F, G, K, M, O.
Q 3.9 (2 pts)
When the temperature increases, first the average time between collisions is
decreased, increasing the chance for a collisional excitation, that is, the collisional cross section is
increased. Next, because the thermal energy of the gas goes as ≈ kT , there is more energy available per
collision for excitation.
Q 3.12 (2 pts)
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 Fall '08
 Lada
 pts, absolute magnitude, Leah Simon

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