The temperature of the surface of a certain star is 6000 K. Most hydrogen atoms

Question

The temperature of the surface of a certain star is 6000 K. Most hydrogen atoms at the surface of the star are in the electronic ground state. What is the approximate fraction of the hydrogen atoms that are in the first excited state (and therefore could emit a photon)? The energy of the first excited state above the ground state is (-13.6/22 eV) - (-13.6 eV) = 10.2 eV = 1.632e-18 J.
fraction=?

(In this estimate we are ignoring the fact that there may be several excited states with the same energies, for example the 2s and 2p states in hydrogen, because this makes only a small difference in the answer.)

Explanation / Answer

here,

From Boltzmann distribution :

Fraction = e^(- E/k*T)

Where,
E is the energy of the excited state , 1.632*10^-18 J.
k is the boltzman constant, which is 1.38*10^-23 J/K,
T is temperature, 6000K.

Fraction = e^(- (1.632*10^-18) / (1.38*10^-23*6000) )

Fraction = 2.75 * 10^-9

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