The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-st

Bernard Mora

Bernard Mora

Answered question

2022-05-19

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of 8 = kIn 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative?

Answer & Explanation

Percyaehyq

Percyaehyq

Beginner2022-05-20Added 18 answers

Given data:
The energy of the first excited energy level of a hydrogen atom is U=10.2 eV
The entropy is S=kln4.
The expression for the Helmholtz free energy of the first excited level is given by:
F = U T S F = ( 10.2 e V ) T ( k ln 4 )
At low temperature this is positive, so the atom would rather be in the ground state which has F = 0; at high temperature, however, the value of the F for the excited level becomes negative, so this level becomes preferred over the ground state, the transition temperature is where F = 0.
Thus, the value of the temperature can be calculated as:
F = 0 ( 10.2 e V ) T ( k ln 4 ) = 0 T = ( 10.2 e V ) k ln 4 T = ( 10.2 e V ) ( 8.617 × 10 5 e V / K ) × ( 1.3863 ) T = 8.539 × 10 4
Thus, the value of the temperature is T = 8.539 × 10 4

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