Consider a solid surface to be a two-dimensional lattice with \(N_{\mathrm{s}}\) sites. Assume that \(N_{\mathrm{a}}\) atoms \(\left(N_{a} \ll N_{s}\right)\) are adsorbed on the surface, so that each lattice site has either zero or one adsorbed atom. An adsorbed atom has energy \(E=-\varepsilon\), where \(\varepsilon>0\). Assume the atoms on the surface do not interact with one another. If the surface is at temperature \(T\), compute the chemical potential of the adsorbed atoms as a function of \(T, \varepsilon\), and \(N_{\mathrm{a}} / N_{\mathrm{s}}\) (use the canonical ensemble).

Short Answer

Expert verified
The chemical potential \(\mu\) is \[ \mu = k_B T \ln \left( \frac{N_a/N_s}{1 - N_a/N_s} \right) - \varepsilon \]

Step by step solution

01

Define the system

The system consists of a two-dimensional lattice of sites with a total number of sites equal to \(N_{s}\) and the number of adsorbed atoms is \(N_{a}\). Each lattice site can be either unoccupied or occupied by a single adatom. The energy of an adsorbed atom is \(E=-\varepsilon\).
02

Express the partition function

The partition function for a single site can have two states: occupied with energy \(-\varepsilon\) or unoccupied with energy 0. The partition function for a single site is given by: \[ Z_1 = e^{\beta \varepsilon} + 1 \]where \(\beta = 1/(k_B T)\).
03

Total partition function

Given that the atoms do not interact, the total partition function is simply the product of the partition functions of all the sites. Thus, the total partition function becomes: \[ Z = (e^{\beta \varepsilon} + 1)^{N_s} \]
04

Determine the mean number of adsorbed atoms

The mean number of adsorbed atoms \(N_a\) is related to the probability \(p\) of a site being occupied by an atom: \[ p = \frac{e^{\beta \varepsilon}}{e^{\beta \varepsilon} + 1} \]The total mean number of adsorbed atoms is then: \[ N_a = N_s \cdot p = \frac{N_s e^{\beta \varepsilon}}{e^{\beta \varepsilon} + 1} \]
05

Extract the chemical potential

Using the relationship between chemical potential \(\mu\) and the configuration of the system, equate the number of adsorbed atoms to the expression depending on chemical potential. Given the number of adsorbed atoms and the partition function, find \(\mu\): \[ \frac{N_a}{N_s} = \frac{e^{\beta (\mu + \varepsilon)}}{1 + e^{\beta (\mu + \varepsilon)}} \]Solving for \(\mu\), we get: \[ \mu + \varepsilon = k_B T \ln \left( \frac{N_a/N_s}{1 - N_a/N_s} \right) \]Thus, the chemical potential is: \[ \mu = k_B T \ln \left( \frac{N_a/N_s}{1 - N_a/N_s} \right) - \varepsilon \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

canonical ensemble
In statistical thermodynamics, the canonical ensemble is a fundamental concept. It is a collection of systems in thermal equilibrium with a heat bath at a fixed temperature. Each system within this ensemble can exchange energy with the surroundings, but the total energy of the ensemble remains constant.

Key attributes of the canonical ensemble include:
  • A fixed number of particles.
  • A fixed volume and temperature.
The primary goal of using the canonical ensemble is to calculate the statistical properties of a system, like energy, pressure, or chemical potential, based on a probability distribution.

To work with the canonical ensemble, we often use functions like the partition function, which summarizes all possible states of the system and their probabilities.
partition function
The partition function is a cornerstone in statistical mechanics and thermodynamics. It is a mathematical function that encapsulates the statistical properties of a system in thermodynamic equilibrium. For a system with multiple possible energy states, the partition function, denoted as \(Z\), is given by:

Z = \sum_i e^{-\beta E_i}\, where:
  • \(E_i\) is the energy of the \(i\)th state.
  • \(\beta = 1/(k_B T)\), with \(k_B\) being the Boltzmann constant and \(T\) the temperature.

In the problem above, the partition function for a single site is:
\(Z_1 = e^{\beta \varepsilon} + 1 \)

For a system with non-interacting particles, the total partition function is the product of the individual partition functions of all sites. This multiplication simplifies the calculation of various thermodynamic quantities. The partition function is crucial for deriving properties such as energy, entropy, and chemical potential.

It helps us understand how the probability of a system occupying a certain energy state changes with temperature and other factors.
chemical potential
Chemical potential is a vital concept in thermodynamics and statistical mechanics. It represents the change in the free energy of a system when the number of particles changes, at constant temperature and volume.

For the system in the problem, the chemical potential \(\mu\) of the adsorbed atoms can be obtained using the formula:
\(\frac{N_a}{N_s} = \frac{e^{\beta (\mu + \varepsilon)}}{1 + e^{\beta (\mu + \varepsilon)}} \). Solving for \(\mu\), we get:

\(\mu = k_B T \, \ln \, \left(\frac{N_a/N_s}{1 - N_a/N_s}\right) - \varepsilon\).

This equation demonstrates how the chemical potential depends on temperature \(T\), energy \(\varepsilon\), and the ratio of the number of adsorbed atoms \(N_a\) to the total lattice sites \(N_s\).

The chemical potential plays a key role in determining the distribution of particles in different energy states, and it is crucial for understanding many phenomena in both classical and quantum systems. It helps in predicting how a system will evolve when particles are added, removed, or when other external conditions change.

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Most popular questions from this chapter

Consider a two-dimensional lattice in the \(x-y\) plane with sides of length \(L_{x}\) and \(L_{y}\) which contains \(N\) atoms ( \(N\) very large) coupled by nearest-neighbor harmonic forces. (a) Compute the Debye frequency for this lattice. (b) In the limit \(T \rightarrow 0\), what is the heat capacity?

The vibrational frequency of the \(\mathrm{I}_{2}\) molecule is \(f=6.42 \times 10^{12} \mathrm{~s}^{-1}\). The vibrational temperature is \(\theta_{1_{2}}^{\text {vib }}=\frac{h f}{k_{B}}=308 \mathrm{~K}\). The rotational temperature is \(\theta_{1_{2}}^{\text {rot }}=0.0538 \mathrm{~K}\). Consider a gas of \(N\) \(\mathrm{I}_{2}\) molecules at temperature \(T=300 \mathrm{~K}\). (a) What fraction of the molecules is in the vibrational ground state and what fraction have one vibrational quantum of energy? (b) What percentage of the total internal energy of the gas is: (1) translational?; (2) vibrational?; (3) rotational?

The CO molecule has a rotational temperature \(\theta=\hbar^{2} /\left(2 I k_{\mathrm{B}}\right)=2.8 \mathrm{~K}\), where \(I\) is the moment of inertia of the CO molecule. The rotational partition function for one molecule is \(Z_{1}^{\mathrm{rot}}=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-l l(1) \theta / T}\) (a) If one mole of CO molecules could freely rotate at temperature \(T=\) \(3.2 \mathrm{~K}\), what is their total rotational entropy? (b) What is the rotational entropy of one mole of \(\mathrm{CO}\) molecules at temperature \(T=320 \mathrm{~K}\) ? (Hint: At high temperature, where many angular momenta contribute \(Z_{1}^{\text {rot }} \approx \int_{0}^{\infty} \mathrm{d} l(2 l+1) \mathrm{e}^{-\| l+1) \theta / T}\).) (c) What is the translational entropy of one mole of \(\mathrm{CO}\) molecules in a box of volume \(V=1.0 \mathrm{~m}^{3}\) at temperature \(T=320 \mathrm{~K}\) ?

Consider a magnetic system whose free energy, near the critical point, scales as \(\lambda^{5} g(\epsilon, B)=g\left(\lambda^{2} \epsilon, \lambda^{3} B\right)\). Compute (a) the degree of the coexistence curve, (b) the degree of the critical isotherm, (c) the critical exponent for the magnetic susceptibility, and (d) the critical exponent for the heat capacity. Do your results agree with values of the critical exponents found in experiments?

A system has three distinguishable molecules at rest, each with a quantized magnetic moment which can have \(z\)-components \(+1 / 2 \mu\) or \(-1 / 2 \mu\). Find an expression for the distribution function, \(f_{i}\) (i denotes the ith configuration), which maximizes entropy subject to the conditions \(\sum_{i} f_{i}=1\) and \(\sum_{i} M_{i, z} f_{i}=\gamma \mu\), where \(M_{i, z}\) is the magnetic moment of the system in the ith configuration. For the case \(y=1 / 2\), compute the entropy and compute \(f_{i}\).

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