Adsorbtion: Exercises

Introduction

The problems below require you to use what you have learnt about the canonical and grand canonical paritition functions to derive a simple model for adsorbtion of atoms/molecules on a surface.

Example problems

Click on the problems to reveal the solution

Problem 1

A surface has $M$ sites that can be either occupied by an atom or not. How many different ways are there of arranging $N \lt M$ atoms on this surface

Notice that we have two types of objects - empty sites and filled sites. Furthermore, there are $N$ filled sites and $M-N$ empty sites. The number of different arrangments of these two types of object can be calculated using the binomial coefficient: $$ \binom{M}{N} = \frac{M!}{N!(M-N)!} $$

A surface has $M$ sites where molecules can be adsorbed in two states: the ground state, which has energy $E = 0$ and an excited state with energy $E = U$. Calculate the canonical partition function for a single molecule adsorbed on a particular adsorption site, $Z(1)$. Explain why when $N$ molecules adsorb on the surface the canonical partition function for the whole system is given by: $$ Z_c^{(N)} = \frac{ M!}{N!(M-N)!} \left[ Z_c^{(1)} \right]^N $$ When performing the derivation for this second part you should assume that the adsorbed molecules do not interact with each other.

The canonical partition function is given by: \[ Z_c = \sum_i e^{-\frac{E_i}{k_B T}} \] where the sum runs over all possible microstates and $E_i$ is the energy of microstate $i$. For one molecule adsorbed on one site on the surface we thus have: \[ Z_c^{(1)} = e^{0} + e^{-\frac{U}{k_B T}} = 1 + e^{-\frac{U}{k_B T}} \] As it's energy can be either equal to 0 or $U$. For $N$ particles we thus have: \[ \begin{aligned} Z_c^{(N)} & = \frac{M!}{M!(M-N)!} \sum_{i_1} \sum_{i_2} \sum_{i_1} \dots \sum_{i_N} e^{-\frac{E_{i_1} + E_{i_2} + E_{i_3} + \dots + E_{i_N}}{k_B T}} \\ & = \frac{M!}{M!(M-N)!} \sum_{i_1} \sum_{i_2} \sum_{i_1} \dots \sum_{i_N} \prod_{j=1}^N e^{-\frac{E_{i_j}}{k_B T}} \\ & = \frac{M!}{M!(M-N)!} \sum_{i_1} e^{-\frac{E_{i_1}}{k_B T}} \sum_{i_2} e^{-\frac{E_{i_2}}{k_B T}} \sum_{i_3} e^{-\frac{E_{i_3}}{k_B T}} \dots \sum_{i_N} e^{-\frac{E_{i_N}}{k_B T}} \\ & = \frac{M!}{M!(M-N)!} \left(Z_c^{(1)}\right)^N = \frac{M!}{M!(M-N)!} \left(1 + e^{-\frac{U}{k_B T}}\right)^N \end{aligned} \] The factor of $\frac{M!}{M!(M-N)!}$ comes about because when $N$ particles adsorb there will be $N$ filled sites and $M$ empty sites. Hence, because we have to sum over all possible microstates when calculating the partition function, we have to consider all possible arrangements of the $N$ atoms on the $M$ absorbption sites. As discussed in the answer to the first part of this question we can calculate the number of orderings of $N$ filled sites and $M-N$ empty sites using the binomial coefficients.

Explain why the grand canonical partition function for this system can be calculated from the canonical partition function using: $$ Z_{gc} = \sum_{n=0}^M Z_c^{(n)} e^{\beta \mu n} $$ Hence, show that the grand canonical partition function for the system that was introduced in the previous part is given by: $$ Z_{gc} = ( 1 + e^{\beta \mu}\left[ 1 + e^{-\frac{U}{k_B T}} \right] )^M $$ where $\mu$ is the chemical potential. Then, by taking an appropriate derivative of this partition function, show that on average {\bf a fraction}: $$ \langle K \rangle =\frac{ e^{\beta \mu} [ 1 + e^{-\frac{U}{k_B T}} ] }{ 1 + e^{\beta \mu}\left[ 1 + e^{-\frac{U}{k_B T}} \right] } $$ of the available lattice sites are occupied.

The grand canonical partition is related to the canonical partition function by: \[ \begin{aligned} Z_{gc} & = \sum_{N=0}^M Z_c(N,V) e^{\beta \mu N} \\ & = \sum_{N=0}^M e^{\beta \mu N} \binom{M}{N} \left(1 + e^{-\frac{U}{k_B T}}\right)^N \\ & = \sum_{N=0}^M \binom{M}{N} \left[ e^{\beta \mu} \left(1 + e^{-\frac{U}{k_B T}}\right) \right]^N \\ & = ( 1 + e^{\beta \mu}[1 + e^{-\frac{U}{k_B T}}])^M \end{aligned} \] In deriving this formula we have used the binomial theorem $\sum_{i=1}^N a^i b^{N-i} = (a+b)^N$ with $a=e^{\beta \mu}[1 + e^{-\frac{U}{k_B T}}]$ and $b=1$.

To get the average fraction of occupied lattices sites, $\langle K \rangle$ we have to take the following derivative: \[ \langle K \rangle = \frac{1}{M} \frac{\partial \ln Z_{gc} }{\partial (\beta \mu) } \] when we take this derivative we find that: \[ \begin{aligned} \langle K \rangle & = \frac{M}{M} \frac{\partial }{\partial(\beta \mu)} \ln\left[ 1+e^{\beta \mu}[1 + e^{-\frac{U}{k_B T}}] \right] \\ & = \frac{ e^{\beta \mu}[1 + e^{-\frac{U}{k_B T}}] }{1+e^{\beta \mu}[1 + e^{-\frac{U}{k_B T}}]} \end{aligned} \]

Contact Details

School of Mathematics and Physics,
Queen's University Belfast,
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Email: g.tribello@qub.ac.uk
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