Brane Space: 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