Solutions To Stellar Line Formation and Transition Problems (2)

1) Find the frequency of the spectral line for which the hydrogen electron transits from the n=3 to the n= 2 energy level.

The energy at the n= 2 level is:

E(n=2) =  - 13.6/ n²   = - 13.6/ (2)²      =  - 13.6/4  (eV)

Now, 1 eV =  1.6 x 10^-19 J  so:

E(n=2) =  - 13.6/4  (eV) =  -(3.4) x 1.6 x 10^-19 J  =

 -5.4 x  10^-19 J 

 The n= 3 level has energy:

E(n=3) =  - 13.6/ n²   = - 13.6/ (3)²      =  - 1.51  (eV)

=  -(1.51)  x 1.6 x 10^-19 J  = -2.4 x 10^-19 J 

Then the energy difference is:

E3 – E2 = [- 2.4 – (-5.4)]  x 10^-19 J  = 3.0 x 10 ^-19 J 

From this, the wavelength of the photon emitted can be found. 

Since E = hf = h (c/ l):   l =   hc/ (E2 – E1) 

l =    (6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ 3.89 x 10^-19 J 

l =    1.21 x 10 ^-7 m 

The frequency can then be found from:

f = (c/ l) = (3 x 10 ⁸ m/s) / 1.21 x 10 ^-7 m  = 2.47 x 10 ¹⁵ Hz

 

 (1) 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   for T = 10 ⁴ K

(

       REM:    At the n=1 level the statistical weight is:  g ₁ = 2(1)² = 2

                   At the n=2 level the statistical weight is: g ₁ = 2(2)² = 8

b)  Obtain the frequency of the spectral line for this transition

Solutions:

(a) 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) E3 – E2 = 1.88 eV

But 1 eV =  1.6 x 10 ^-19 J 

Then:  E3 – E2 = 1.88 ( 1.6 x 10^-19 J )  = 3.0 x 10 ^-19 J 

From this, the wavelength of the photon emitted can be found. 

Since E = hf = h (c/ l):   l =   hc/ (E2 – E1) 

l =    (6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ 3.89 x 10^-19 J 

l =    1.21 x 10 ^-7 m 

The frequency can then be found from:

f = (c/ l) = (3 x 10 ⁸ m/s) / 1.21 x 10 ^-7 m  = 2.47 x 10 ¹⁵ Hz