Solving Schottky's Defect Homework with 3 States

[kudoushinichi88]

Homework Statement

A ‘lattice gas’ consists of a lattice of N sites. Each of these sites can be empty, in which
case its energy is zero, or occupied by one particle, in which case its energy is e. Each particle
has a magnetic moment of magnitude μ which in the presence of an applied magnetic field
B, can adopt two orientations (parallel or anti-parallel to the field). Evaluate the mean energy
and mean magnetic moment of the system assuming that the particles are not interacting with
each other.

Homework Equations

$$Z=\sum_r e^{-\beta E_r}$$
$$p_r=\frac{1}{Z}e^{-\beta E_r}$$

For a system of n defects in a system of N sites,
$$\frac{n}{N}=\frac{1}{e^{\beta \epsilon}+1}$$
where ε is the energy associated with the defect

Mean energy,
$$\bar{E}=\frac{\partial \ln{Z}}{\partial\beta}$$

The Attempt at a Solution

My problem is that I'm not sure whether there are three separate states for each site, or are there only 2 energy states and the schottky's defect must be considered separately.

If I consider that there are 3 possible states, then the possible energies are
$$\epsilon+\mu\beta,0\ \textrm{and}\ \epsilon-\mu\beta$$

But if this is not right, then I'm not sure how to go about on this question.

 

[mark726]

Thank you for your post. It is a great question and I am happy to help you find a solution.

To start, let's clarify the number of energy states for each site in the lattice gas. In this system, each site can be either empty or occupied by a particle. Therefore, there are only two possible energy states for each site: zero energy for an empty site, and e for an occupied site. The magnetic moment of each particle can be either parallel or anti-parallel to the applied magnetic field, but this does not affect the energy levels of the system. Therefore, we can focus on the two energy states mentioned above.

Now, let's consider the total number of possible configurations for the system. Since there are N sites and each site can have two energy states, there are a total of 2^N possible configurations. However, we need to take into account the fact that the particles are not interacting with each other, so we can assume that each site is independent. This means that the probability of a site being occupied is the same for all sites, and similarly for the empty sites. Therefore, the total number of configurations can be simplified to N!/n!(N-n)!, where n is the number of occupied sites.

Using this information, we can now calculate the partition function Z:

Z = e^{-\beta n\epsilon}(1+e^{-\beta e})^{N-n}

Next, we can calculate the probability of a site being occupied, p_n, by dividing the number of configurations with n occupied sites by the total number of configurations:

p_n = \frac{N!/n!(N-n)!}{2^N}

This can be further simplified to:

p_n = \frac{1}{e^{\beta e}+1}

Finally, we can use the equations for mean energy and mean magnetic moment to calculate the desired values:

\bar{E} = -\frac{\partial \ln{Z}}{\partial \beta}

\bar{m} = \frac{\partial \ln{Z}}{\partial B}

By substituting the expression for Z and simplifying, we get:

\bar{E} = \frac{e^{-\beta e}\epsilon}{1+e^{-\beta e}} = \frac{\epsilon}{e^{\beta e}+1}

\bar{m} = \frac{n\mu}{N} = \frac{\