The temperature of the surface of a certain star is 6500 K. Most hydrogen atoms

Question

The temperature of the surface of a certain star is 6500 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 second excited state (and therefore could emit a photon)? The energy of the second excited state above the ground state is (-13.6/3^2 eV) - (-13.6) eV = 12.1 eV = 1936e - 18). (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

Fraction of atoms in the excited state = e(-E/kT)

where,

E is the energy of the excited state in which the atom is trying to reach, in this case, the first excited state, which will be the given value of 1.936*10-18J.

k is the boltzman constant, which is 1.3807*10-23 J/K, and

T is the temperature, which for this problem is 6500 K.

fraction = e(-E/kT)

fraction = 4.2791*10-10

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