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An Introduction to Thermal Physics
Found in: Page 163
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

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 S=kln4, 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? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F=0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)

Helmholtz free energy is positive when temperature T < 8.5 x 104K and it is negative when temperature T < 8.5 x 104K .

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Step by Step Solution

Given Information

Entropy, S=kln(4)
Where value of k=8.62×10-5eVK-1Helmholtz free energy, F=0 (For ground state)First excitation of energy level, U=10.2eV

Explanation

Helmholtz free energy of the first excited level is given by

F=U-TSF=U-T[kln(4)]

Substitute the given value and calculate

F=U-T[kln(4)]

Rearrange

T=U-Fkln(4)

Since, for a ground state F=0,
After substituting the given values in the above equation we get

T=10.2eV-08.62×10-5eVK-1ln(4)=8.5×104 K

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