Grand Canonical Partition Function and Adsorption Statistics

AI Thread Summary
The discussion revolves around calculating the grand canonical partition function for a two-dimensional surface with M adsorption sites in thermal equilibrium with a gas. The correct expression for the partition function of one adsorption position, \mathcal{Z_1}(T, μ), is debated, with one participant suggesting it should include terms for both occupied and unoccupied states. The confusion arises from whether to include the case of no particles adsorbed correctly. The correct approach involves summing over the allowed values of N, leading to \mathcal{Z_1} = e^0 + e^{\beta(\mu - \epsilon)}. The overall partition function for the entire surface, \mathcal{Z_M}(T, μ), requires further clarification on its formulation.
Arjani
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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}}?
 
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Arjani said:

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.
 
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