Grand Canonical Partition Function and Adsorption Statistics

[Arjani]

Homework Statement

Consider a two dimensional surface on a three dimensional crystal. This surface has M positions that can adsorb particles, each of which can bind one particle only and an adsorption does not affect the adsorption on nearby sites. An adsorbed particle has energy ε and an empty site has energy 0.

(Question A and B come here, but I could answer those.)

The surface is now considered to be in diffusive and thermal equilibrium with a gas of temperature T and chemical potential μ, so the energy E and the number of adsorbed particles can now vary.

c) Calculate the grand canonical partitition function $\mathcal{Z_1} (T, \mu)$ of one adsorption position and then the grand canonical partition function for the entire surface $\mathcal{Z_M} (T, \mu)$.

d) Calculate the chance $P(T,\mu)$ that one adsorption position is taken.

Homework Equations

$\mathcal{Z} = \sum e^{\beta(\mu N_i - E_i)}$

The Attempt at a Solution

So for $\mathcal{Z_1} (T, \mu)$, $N = 1$ and $E_i = 0$ or $E_i = \epsilon$ and so

$$\mathcal{Z_1} = \sum e^{\beta(\mu - E_i)} = e^{\beta(\mu - \epsilon)} + e^{\beta(\mu - 0)} = e^{\beta \mu}(e^{-\beta \epsilon} + 1)$$

This is my solution. However, I read that you have to take $N = 0$ for $E_i = 0$ and $N = 1$ for $E_i = \epsilon$, resulting in $\mathcal{Z_1} = e^0 + e^{\beta(\mu - \epsilon)}$, so I'm confused. What is correct here?

As for $\mathcal{Z_M}$, I'm not sure how to go about that. Can you do something like this?

$\mathcal{Z_M} = \sum_{i=0}^{M} e^{\beta N_i} \sum_{i=0}^{\epsilon} e^{- \beta E_i}$

d) This is simply $P = \frac{e^{\beta(\mu - \epsilon)}}{\mathcal{Z_1}}$?

 

[TSny]

Homework Equations

$\mathcal{Z} = \sum e^{\beta(\mu N_i - E_i)}$

Think about what the summation index is here.

The Attempt at a Solution

So for $\mathcal{Z_1} (T, \mu)$, $N = 1$ and $E_i = 0$ or $E_i = \epsilon$ and so

$$\mathcal{Z_1} = \sum e^{\beta(\mu - E_i)} = e^{\beta(\mu - \epsilon)} + e^{\beta(\mu - 0)} = e^{\beta \mu}(e^{-\beta \epsilon} + 1)$$

This is my solution. However, I read that you have to take $N = 0$ for $E_i = 0$ and $N = 1$ for $E_i = \epsilon$, resulting in $\mathcal{Z_1} = e^0 + e^{\beta(\mu - \epsilon)}$, so I'm confused. What is correct here?

The summation is over all allowed values of $N$, namely $N = 0$ for no particle absorbed at the site and $N = 1$ for one particle absorbed at the site. You didn't quite handle the $N_i=0$ case correctly in the expression $\mathcal{Z} = \sum e^{\beta(\mu N_i - E_i)}$. You should get the result that you stated from your reading.