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Chapter 2: Problem 6

Consider a simplified system having four energy levels of relative energy \(0,1,2\), and 3. The system contains eight particles and has a total relative energy of six. Determine the thermodynamic probability for every distribution consistent with the above constraints given the following system characteristics. a. The energy levels are nondegenerate and the particles obey Maxwell- Boltzmann statistics. b. Each energy level has a degeneracy of six and the particles obey either (i) Maxwell-Boltzmann or (ii) corrected Maxwell-Boltzmann statistics. c. Each energy level has a degeneracy of six and the particles obey either BoseEinstein or Fermi-Dirac statistics. d. Comment on your calculations for parts (b) and (c).

Short Answer

Expert verified
The thermodynamic probability for each distribution varies based on the statistics applied. Maxwell-Boltzmann (nondegenerate) has fewer microstates than degenerate versions. Bose-Einstein allows for the most microstates due to boson characteristics, while Fermi-Dirac has restrictions that can limit the number of microstates.

Step by step solution

01

Understanding the Basics

For any given energy distribution, the thermodynamic probability is determined by the number of ways the particles can be distributed among the energy levels while satisfying the given constraints (total number of particles and total energy). Maxwell-Boltzmann statistics assume distinguishable particles without quantum restrictions. Bose-Einstein statistics apply to indistinguishable bosons, allowing multiple occupation of the same state. Fermi-Dirac statistics apply to indistinguishable fermions, with at most one particle per state due to the Pauli exclusion principle.
02

Calculating for Maxwell-Boltzmann Statistics

For nondegenerate energy levels, we need to find the number of ways to distribute eight distinguishable particles among four energy levels to sum up to a total relative energy of six: (0,1,2,3).There are a limited number of ways to distribute the particles that add up to a total energy of six, given their individual energies can be 0, 1, 2, or 3 and the total number must be eight. Each distribution method represents a 'microstate'. We list out all possible microstates and calculate the thermodynamic probability, which is simply the count of microstates in this scenario.
03

Maxwell-Boltzmann Distributions with Degeneracy

When each energy level has a degeneracy of six, there are more possible microstates for each energy distribution because each energy level can now occupy up to six particles. The thermodynamic probability increases. For parts (i) Maxwell-Boltzmann and (ii) corrected Maxwell-Boltzmann, we must calculate the number of ways to distribute the particles across the degenerate energy levels. The corrected Maxwell-Boltzmann statistics account for the indistinguishability of particles in the same energy state.
04

Bose-Einstein and Fermi-Dirac Statistics with Degeneracy

For Bose-Einstein statistics, we consider the particles indistinguishable and allow multiple particles in the same energy state. For Fermi-Dirac statistics, particles are also indistinguishable, but each degenerate state can only be occupied by one particle. We calculate the distributions accordingly: for Bose-Einstein, we must enumerate the distributions that allow more than one particle per state up to the degeneracy limit; for Fermi-Dirac, we count distributions that have at most one particle per degenerate state, given that we have more degenerate states than particles.
05

Comments on Calculations for (b) and (c)

The calculations for part (b) highlight the increased thermodynamic probability due to degeneracy when compared to nondegenerate energy levels (part a). The numbers are larger due to more possible microstates. Part (c) calculations exhibit quantum statistics principles: Bose-Einstein increases the probability even more due to indistinguishability and no restriction on state occupation, while Fermi-Dirac statistics limit the distributions because of the Pauli exclusion principle, potentially reducing the thermodynamic probability in comparison to Bose-Einstein.

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Key Concepts

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

Maxwell-Boltzmann Statistics
Delving into the behavior of gases and particles often leads to understanding Maxwell-Boltzmann statistics—a fundamental concept in classical statistical mechanics. This statistical approach is used to describe the distribution of particles over various energy states in a system that complies with classical physics. Maxwell-Boltzmann statistics assume that particles are distinguishable from one another and that there are no quantum restrictions on the number of particles that can occupy a given energy level.

When we apply Maxwell-Boltzmann statistics to thermodynamic problems like the provided exercise, we consider various 'microstates', or ways the particles can be distributed between energy levels, while respecting the total energy and particle constraints. For a nondegenerate system as in part (a), the idea is to count all possible microstates without considering identical arrangements, as each particle is unique and the order of particle arrangement makes a difference in each configuration. Furthermore, no energy state has multiple occupancy restrictions, rendering the solution simpler than those required for quantum statistics.
Bose-Einstein Statistics
Transitioning to the remarkable world of quantum mechanics, Bose-Einstein statistics come into play when dealing with bosons—particles that have integer spin. These particles are indistinguishable from one another and are not subject to the Pauli exclusion principle which governs fermions. Consequently, unlike Maxwell-Boltzmann statistics, Bose-Einstein statistics allow multiple bosons to occupy the same energy state, also known as energy level degeneracy.

The fascinating aspect of Bose-Einstein statistics is that they give rise to phenomena like Bose-Einstein condensates, where particles coalesce into the lowest energy state at very low temperatures. In our given exercise, this translates to a significantly higher number of potential microstates compared to classical statistics because multiple indistinguishable particles can share the same degenerate energy level, leading to a higher thermodynamic probability.
Fermi-Dirac Statistics
Exploring another facet of quantum mechanics brings us to Fermi-Dirac statistics, which govern the distribution of fermions—particles with half-integer spin, such as electrons, protons, and neutrons. These particles are also indistinguishable, but they adhere to the Pauli exclusion principle, meaning no two identical fermions can occupy the same quantum state simultaneously.

This exclusive behavior drastically influences the distribution of particles across energy levels in a system. For example, the exercise provided illustrates that for a degenerate system, each energy state could hold only one fermion. As a result, Fermi-Dirac statistics can limit the possible microstates since occupation of energy levels is capped by their degeneracy, directly impacting the thermodynamic probability. The 'exclusivity' of fermions shapes a wide array of physical phenomena, including the properties of conductors and semiconductors.
Energy Level Degeneracy
Energy level degeneracy is a concept that deeply influences the computation of thermodynamic probabilities. Degeneracy refers to the number of different quantum states of a system that have the same energy level. Simply put, when an energy level is degenerate, multiple particles can possess the same energy without violating any physical laws.

In scenarios where each energy level has a degeneracy, as in parts (b) and (c) of our exercise, the thermodynamic probability increases because there are more ways to distribute the particles while maintaining the same total energy. Under Maxwell-Boltzmann statistics, degeneracy increases the number of distinguishable configurations. On the flipside, quantum statistics consider the particles indistinguishable; Bose-Einstein statistics take advantage of degeneracy to permit multiple occupancies, while Fermi-Dirac statistics enforce a strict one-particle-per-state rule up to the degeneracy limit, providing a compelling contrast in the resultant thermodynamic probabilities.

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Most popular questions from this chapter

A thermodynamic system consists of \(N\) independent, distinguishable particles. Each particle has three energy levels at \(0, \varepsilon\), and \(2 s\), with degeneracies of 1,3 , and 5 , respectively. The system is in thermal equilibrium with a heat reservoir of absolute temperature \(T=\varepsilon / k\), where \(k\) is Boltzmann's constant. a. Calculate the partition function for this system. b. What fraction of the particles resides in each energy level? c. Determine the average particle energy and the associated mean particle entropy- d. At what temperature would the population of the energy level having \(2 \varepsilon\) be equal to that at \(\varepsilon\) ?

We have shown that the entropy for Bose-Einstein and Fermi-Dirac statistics is given by $$ S(E, V, N)=k(\beta E+\alpha N) \mp k \sum_{j} g_{j} \ln \left(1 \mp e^{-\alpha} e^{-\beta \varepsilon_{j}}\right), $$ where \(N=\sum_{j} N_{j}\) and \(E=\sum_{j} N_{j} \varepsilon_{j}\). Similarly, from classical thermodynamics, $$ d S(E, V, N)=\frac{1}{T} d E+\frac{P}{T} d V-\frac{\mu}{T} d N $$ for a single-component system. a. Prove that \(\beta=1 / k T\). b. Prove that \(\alpha=-\mu / k T\). c. Show that the pressure is given by $$ P=-\sum_{j} N_{j}\left(\partial \varepsilon_{j} / \partial V\right)_{E, N} . $$ d. We will soon demonstrate that the internal energy is independent of volume whereas the translational energy \(\varepsilon_{j, t r}=C_{j} V^{-2 / 3}\), where \(C_{j}\) is a constant for each translational energy level. Utilizing this information, show that $$ P V=\frac{2}{3} E_{t r}, $$ where \(E_{t r}\) is the translational portion of the total energy (external plus internal) for the particle assembly. e. Is this result limited to the dilute limit? Does it assume independent particles? Why should the pressure be related only to the translational energy? Explain.

To illustrate the order of magnitude of the fluctuations in a macroscopic system, consider \(N\) distinguishable particles, each of which can be with equal probability in either of two available states; e.g., an "up" and a "down" state. a. Determine the total number of microstates and the entropy of this \(N\)-particle system. b. What is the number of microstates for which \(M\) particles are in the "up" state? Hint: Recall the binomial distribution. c. The fluctuation in the number of particles \(M\) is given by \(\sigma / \bar{M}\), where \(\sigma\) is the standard deviation and \(\bar{M}\) is the mean number of particles in the "up" state. Develop an expression for \(\sigma / \bar{M}\). d. Consider a macroscopic system for which \(N=6.4 \times 10^{23}\) spatially separated particles. Calculate the fluctuation for this system. What are the implications of your result?

Paramagnetism can occur when some atoms in a crystalline solid possess a magnetic dipole moment owing to an unpaired electron with its associated orbital angular momentum. For simplicity, assume that (1) each paramagnetic atom has a magnetic dipole moment \(\mu\) and (2) magnetic interactions between unpaired electrons can be neglected. When a magnetic field is applied, the magnetic dipoles will align themselves either parallel or antiparallel to the direction of the magnetic field. If the magnetic moment is parallel to a magnetic field of induction \(\vec{B}\), the potential energy is \(-\mu B\), when the magnetic moment is antiparallel to \(\vec{B}\), the potential energy is \(+\mu B\). a. Prove that the probability for an atomic magnetic dipole moment to point parallel to the magnetic field is given at temperature \(T\) by $$ P_{\mathrm{a}}=\left(1+e^{-2 t}\right)^{-1} $$ where \(x=\mu B / k T\). Give a physical explanation for the value of \(P\) as \(T \rightarrow 0\) and as \(T \rightarrow \infty\) Hint: Determine the partition function for the system. b. Show that, for \(N\) independent magnetic dipoles, the mean effective magnetic moment parallel to the magnetic field is $$ m=N \mu \tanh (x) . $$ c. Demonstrate that the mean magnetic moment at high temperatures and/or weak magnetic fields ( \(x \& 1\) ) is proportional to \(1 / T\). This is Curie's law. d. Show that the contribution from paramagnetism to the internal energy of a crystalline solid is \(U=-m B\). Determine this paramagnetic contribution at \(T=\) \(\infty\). Why should this result have been expected? e. Develop an expression for the entropy of this paramagnetic system.

Classical Maxwell-Boltzmann statistics considers particles to be distinguishable with no limit on the number of particles in each energy state. A physical example is a solid composed of localized atoms at distinguishable lattice sites. The thermodynamic probability in this case is given by $$ W_{M B}=N ! \prod_{j} \frac{g_{j}^{N_{i}}}{N_{j} !} $$ where \(N_{j}\) is the number of particles and \(g_{j}\) is the degeneracy of the \(j\) th energy level. a. Using the methods of statistical thermodynamics, show that the equilibrium particle distribution is $$ N_{j}=g_{j} e^{-\alpha} e^{-\beta \varepsilon_{j}} \text {. } $$ b. Defining the molecular partition function \(Z=\sum_{j} g_{j} e^{-\beta \varepsilon_{j}}\), show that $$ S=k(\beta E+N \ln Z) . $$ c. Using classical thermodynamics, verify that \(\beta=1 / k T\). Hence, the entropy for classical Maxwell-Boltzmann statistics becomes $$ S=\frac{U}{T}+k N \ln Z \text {. } $$ Compare this expression with that for corrected Maxwell-Boltzmann statistics. Explain the difference. d. Show that the probability of a particle being in the \(i\) th energy state is given by $$ P_{i}=\frac{N_{i}}{N}=\frac{e^{-\varepsilon_{i} / k T}}{Z}, $$ where the partition function \(Z=\sum_{i} e^{-\varepsilon_{i} / k T}\). e. Demonstrate that the entropy can be directly related to the probabilities \(P_{i}\) of the various energy states accessible to the system, i.e., $$ S=-k N \sum_{i} P_{i} \ln P_{i} . $$ Discuss the significance of this result.
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