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5.20

Expert-verifiedFound in: Page 163

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 10^{4}K and it is negative when temperature T < 8.5 x 10^{4}K .

Entropy, $S=k\mathrm{ln}\left(4\right)$

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

Helmholtz free energy of the first excited level is given by

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

Substitute the given value and calculate

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

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

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

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