Free Energy of a Simple Quantum System

[VSayantan]

Homework Statement

[/B]A quantum system has three energy levels, $-0.12 ~\rm{eV}$, $-0.20 ~\rm{eV}$ and $-0.44 ~\rm{eV}$ respectively. Three electrons are distributed among these three levels. At a temperature of $1727^o \rm{C}$ the system has total energy $-0.68 ~\rm{eV}$. What is the approximate free energy of the system?

Homework Equations

At temperatures higher than $0~K$ electrons occupy higher energy levels.

The Attempt at a Solution

Since the temperature of the system is $>>0~k$, electrons will occupy higher energy levels.
The energy levels might look like as follows

The system has a total energy $$E=-0.68~\rm{eV}$$
Which can be obtained if two electrons are in the third energy level and one in the first energy level : $$2\times (-0.12\rm{eV})+(-0.44\rm{eV})=-0.68~\rm{eV}$$

Assuming free energy refers to the ground state energy of the system, it is
$$2\times (-0.44~\rm{eV})+(-0.12~\rm{eV})\approx -1.0~\rm{eV}$$

Any other explanation possible?

 

[mark726]

I would like to clarify that the free energy of a quantum system refers to the energy that is available to do work, rather than the ground state energy. In this case, the free energy can be calculated using the formula:
$$F = E - TS$$
where E is the total energy of the system, T is the temperature in Kelvin, and S is the entropy of the system. Since we know the total energy and temperature, we need to calculate the entropy of the system.

At a temperature of $1727^o \rm{C}$ or $2000~\rm{K}$, the probability of an electron occupying a particular energy level is given by the Boltzmann distribution:
$$P_i = \frac{e^{-\frac{E_i}{kT}}}{\sum_{i=1}^3 e^{-\frac{E_i}{kT}}}$$
where E_i is the energy of the ith energy level and k is the Boltzmann constant.

Using this formula, we can calculate the probabilities of each electron occupying the three energy levels:
$$P_1 = \frac{e^{-\frac{-0.12~\rm{eV}}{k\times 2000~\rm{K}}}}{e^{-\frac{-0.12~\rm{eV}}{k\times 2000~\rm{K}}} + e^{-\frac{-0.20~\rm{eV}}{k\times 2000~\rm{K}}} + e^{-\frac{-0.44~\rm{eV}}{k\times 2000~\rm{K}}}} \approx 0.51$$
$$P_2 = \frac{e^{-\frac{-0.20~\rm{eV}}{k\times 2000~\rm{K}}}}{e^{-\frac{-0.12~\rm{eV}}{k\times 2000~\rm{K}}} + e^{-\frac{-0.20~\rm{eV}}{k\times 2000~\rm{K}}} + e^{-\frac{-0.44~\rm{eV}}{k\times 2000~\rm{K}}}} \approx 0.35$$
$$P_3 = \