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 =kln(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

Step by step solution

01

Given Information

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

02

Explanation

Helmholtz free energy of the first excited level is given by

$$\begin{gathered} F=U-TS \\ F=U-T\left[k\ln \right(4\left)\right] \end{gathered}$$

Substitute the given value and calculate

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

Rearrange

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

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

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

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