Consider a gas of neutral hydrogen (H). Using the Boltzmann equation and the information in the table above, 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
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