## Wednesday, December 12, 2018

### Solutions To Stellar Emission & Absorption Revisited Problems (2)

1)     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)   =     [g2 / g1 ]   exp (- E2 – E1) / kT

With  g 1 = 2(1)2 = 2

And:   g 2 = 2(2)2 = 8

For that problem also:  N2 =  N1     and:

E2 =  - 13.6 eV and E1 = -3.4 eV, therefore:

[g2 / g1 ] exp (- E2 – E1) / kT

4 exp [- 13.6eV – (-3.4 eV) ]/ kT

N2 /  N1 = 1 = 4 exp (-10.2 eV)/ kT

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)

2)     An H-alpha line undergoes triplet splitting in the vicinity of a sunspot. The undisturbed line is measured at   lo  =   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  =     (lo)e H/ 4 π  me c2

Solve for H:

H =   4 π  (D  l)  me c2   /  (lo)e

Where: c =  3 x 10  10 cm/s

me  =    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