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Found in: Page 163

### An Introduction to Thermal Physics

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

# 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 $S=k\mathrm{ln}4$, 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? (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 .

See the step by step solution

## Given Information

Entropy, $S=k\mathrm{ln}\left(4\right)$
Where value of $k=8.62×{10}^{-5}{\mathrm{eVK}}^{-1}$Helmholtz free energy, F=0 (For ground state)First excitation of energy level, $\mathrm{U}=10.2\mathrm{eV}$

## Explanation

Helmholtz free energy of the first excited level is given by

$F=U-TS\phantom{\rule{0ex}{0ex}}F=U-T\left[k\mathrm{ln}\left(4\right)\right]$

Substitute the given value and calculate

$F=U-T\left[k\mathrm{ln}\left(4\right)\right]$

Rearrange

$T=\frac{U-F}{k\mathrm{ln}\left(4\right)}$

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

$T=\frac{10.2\mathrm{eV}-0}{\left(8.62×{10}^{-5}{\mathrm{eVK}}^{-1}\right)\mathrm{ln}\left(4\right)}\phantom{\rule{0ex}{0ex}}=8.5×{10}^{4}\mathrm{K}$