Solutions to Astrophysics 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, 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  m_e 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