## Friday, December 7, 2018

### SolutionsTo Stellar Emission - Absorption (Revisited) Problems

1) Consider a gas of neutral hydrogen. Using the Boltzmann equation and the information in the table (see April 2 blog post), compute the temperature at which one will expect equal numbers of atoms in the ground state and the first excited state.

Solution:

The Boltzmann equation is:

N2 / N1  =     [g2 / g1 ]   exp (- E2 – E1) / kT

And from the table, g2 = 8 and  g1 = 2

We require the condition that:  N2 =  N1     so:

1 =     [8/ 2 ] exp (- E2 – E1) / kT

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

1 = 4 exp [- 13.6eV – (-3.4 eV) ]/ kT

1 =  4 exp (-10.2 eV)/ kT

Taking natural logs:

ln (4)  =    (10.2 eV)/ kT

where: k =   8.6174 x 10 -5 eV/K

Solving for T:

T =    10.2 eV/  (ln 4) (8.6174 x 10 -5 eV/K)

T =  10.2 eV/ (1.3862) (8.6174 x 10 -5 eV/K)

T= 85 388 K or T = 8.54 x 10 4 K

2 ) For the Balmer a line (called H- alpha), we know:

E3 – E2 =  - 13.6 eV ( 1/ 3 2   -  1/ 2 2 )   = 1.88 eV

a)     From this information calculate the ratio N2 / N1

b)     Obtain the specific intensity from:

I u  =  2h u 3 / c 2  [1/ exp (hc/lkT]

Solution:

N2 / N1   =     [g2 / g1 ]   exp (- E2 – E1) / kT

From the table in the Dec. 5th  post;  g2 = 8 and  g1 = 2

k =   8.6174 x 10 -5 eV/K

N2 / N1    =     [8/ 2 ]   exp (- E2 – E1) / kT

N2 / N1      =  4 exp (-1.88 eV)/ (8.61 x 10 -5 eV/K) (10 4 K)

N2 / N1      =  4(0.113) =  0.452

b)I u  =  2h u 3 / c 2  [1/ exp (hc/lkT]

We need to use consistent cgs units.  Planck constant h = 6.62 x 10  -27 erg-s

c=  3 x 10  10 cm/s

l  =  hc/ E =  (6.62 x 10  -27 erg-s) (3 x 10  10 cm/s)/ 3.0 x 10  -12 erg

l  =  6.62 x 10  -5 cm

k=1.38 x 10  -16 erg/K

[1/ exp (hc/lkT]  =

1/ [exp (6.62 x 10  -27 erg-s) (3 x 10  10 cm/s)/ (6.62 x 10  -5 cm)  (1.38 x 10  -16 erg/K)( 10 4 K)

= 0.113

I u  =   2h u 3 / c 2  [0.113] erg cm -2/s

But  u =   E/h   =

3.0 x 10  -12 erg/ 6.62 x 10  -27 erg-s  =  4.53 x 10  14 /s

So:
I u  =    0.226(6.62 x 10  -27 erg-s) (4.53 x 10  14 /s) 3 / (3 x 10  10 cm/s) 2

I u  =     1.51  x 10  -4 erg cm -2/s

3)Calculate  the transition probability you get using the Einstein equation:
A 21=   6.67 x 10 16 [g f/g2  l2   Å]
What possible errors might cause the values to diverge? (Take g = f » 1)
l =  6.62 x 10  -7 m  =  6.62 x 10  -5 cm =  6620 Å

With g2 = 8  and g = f = 1

A 21=   1.93 x 10 8 [Å-2]

Compare to standard form (see e.g. Wikipedia, “Einstein coefficients”)  given in multiple physics papers as:

A 21=  [f g1/g2]{ 2 π  u 3  e 2 }/ e o  me c3

A 21 =    1.87 x 10 7    or:    0.187  (in defined units of    10 8  s)

Error sources: Imprecise oscillator frequency f

Error in one of the statistical weights.

Uncertainty in line measurements for absorption and extrapolating this for deduction of  A 21.