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The Fermi energy in a quantum gas depends inversely on the volume. Basing your answer on simple Chapter S-type quantum mechanics (not such quaint notions as squeezing classical particles of finite volume into a container too small), explain why.

Short Answer

Expert verified
The Fermi energy depends inversely on volume due to the change in the number of available quantum states when volume changes. A larger volume leads to more available quantum states, thereby reducing the energy difference between these states and consequently the Fermi energy.

Step by step solution

01

Understanding Fermi Energy

The Fermi energy refers to the maximum energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature.
02

Understanding the effect of Volume

When the volume of the container changes, the number of quantum states that the particles can occupy also changes. A larger volume means there are more spatial quantum states available for the particles to occupy.
03

Drawing the connection between Fermi energy and Volume

The Fermionic particles occupy these quantum states in order of increasing energy. At absolute zero, the Fermi energy is the energy of the highest occupied quantum state. If there are more quantum states available (i.e., if the volume is larger), then the energy difference between quantum states is smaller. This means the Fermi energy—the energy of the highest occupied quantum state—is lower. In other words, with a larger volume, the Fermi energy decreases, and vice versa. Therefore, the Fermi energy depends inversely on the volume.

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Key Concepts

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

Quantum Mechanics and its Role in Understanding Fermi Energy
Quantum mechanics, the cornerstone of modern physics, provides a fundamental framework for describing the behaviour of particles at the microscopic scale. It departs from classical physics by introducing the concept of quantization, where energy and other properties of particles can only take on certain discrete values.

Fermi energy is an essential concept arising from this quantum terrain and pertains to the behavior of electrons, which are fermions — particles that follow Pauli's exclusion principle. This principle states that no two fermions can occupy the same quantum state simultaneously. In a solid material, electrons fill available energy states from the lowest up until all are occupied, leading to a 'Fermi sea' of electron states. The Fermi energy marks the top of this sea at absolute zero temperature.

The relationship between Fermi energy and volume in a quantum gas makes more sense when viewed through a quantum mechanics lens. As volume changes, so does the quantization of energy levels. More volume means more space for energy levels, leading to a decrease in energy differences between levels and thereby a lower Fermi energy.
Quantum Gas: A State of Matter Defined by Quantum Stats
A quantum gas is a collection of particles that abide by the rules of quantum statistics. Fermions and bosons represent the two categories of particles, which behave according to Fermi-Dirac and Bose-Einstein statistics, respectively.

In the context of a fermionic quantum gas, like an electron gas in a metal, the effects of quantum statistics are profound at very low temperatures. This encompasses scenarios close to absolute zero, where the Fermi-Dirac distribution comes into play. Under such conditions, the behavior of particles differs greatly from classical ideal gases. The quantization of energy levels means that the electrons will fill these levels in a way that is distinct from classical particles, which is why squeezing more 'quantum' electrons into the same volume increases the Fermi energy.

The quantum description is necessary to accurately explain the properties of these gases, as classical mechanics cannot account for the discrete occupation of energy levels by electrons, a characteristic trait that profoundly influences electric and thermal properties of materials.
Absolute Zero Temperature and Fermi Energy
Absolute zero, a theoretical state, is the lowest temperature possible, where a system reaches its minimum possible energy, known as the ground state in quantum mechanics. It is defined as 0 Kelvin (-273.15°C or -459.67°F).

At absolute zero, all classical motion of particles ceases, and particles settle into their lowest energy states due to quantum rules. For a fermionic system, this implies that the particles fill up the lowest available energy states up to the Fermi energy. Beyond this, there are no more filled states at absolute zero. While it's impossible to reach this temperature due to the Third Law of Thermodynamics, it provides a useful theoretical point to understand the distribution of particle energies in a quantum gas.

Therefore, when we consider the Fermi energy in relation to volume, the concept of absolute zero temperature helps us to understand how a quantum gas would behave under idealized conditions. With no thermal energy to excite particles to higher states, the Fermi energy becomes the ceiling for electron energies within a material at this lower bound of temperature.

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Most popular questions from this chapter

Example 9.2 obtains a ratio of the number of particles expected in the \(n=2\) state to that in the ground state. Rather than the \(n=2\) state, consider arbitrary \(n\). (a) Show that the ratio is \(\frac{\text { number of energy } E_{n}}{\text { number of energy } E_{1}}=n^{2} e^{-13.6 \mathrm{cV}\left(1-n^{-2}\right) / k_{\mathrm{B}} T}\) Note that hydrogen atom energies are \(E_{n}=\) \(-13.6 \mathrm{eV} / \mathrm{r}^{2}\) (b) What is the limit of this ratio as \(n\) becomes very large? Can it exceed \(1 ?\) If so, under what condition(s)? (c) In Example 9.2, we found that even at the temperature of the Sun's surface \((\sim 6000 \mathrm{~K})\), the ratio for \(n=2\) is only \(10^{-8}\). For what value of \(n\) would the ratio be \(0.01 ?\) (d) Is it realistic that the number of atoms with high \(n\) could be greater than the number with low \(n\) ?

We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of 11 oscillators sharing a total energy of just \(5 h \omega_{0}\). In the symbols of Section \(9.3, N=11\) and \(M=5\). (a) Using equation \((9-9),\) calculate the probabilities of \(n\) being \(0.1,2 .\) and \(3 .\) (b) How many particles. \(N_{n}\), would be expected in each level? Round each other nearest integer. (Happily. the number is still 11 . and the energy still \(\operatorname{she}_{0}\) ) What you have is a distribution of the energy that is as close to expectations is possible, given that numbers at each level in a real case are integers. (c) Entropy is related to the number of microscopic ways the macrostate can be obtained. and the number of ways of permuting particle labels with \(N_{0}\). \(N_{1}, N_{2},\) an \(d N_{1}\) fixed and totaling 11 is \(11 ! /\left(N_{0} ! N_{1} !\right.\) \(\left.N_{2} ! N_{3}^{\prime}\right) .\) (See Appendix J for the proof.) Calculate the number of ways for your distribution. (d) Calculate the number of ways if there were 6 particles in \(n=0.5\) in \(n=1\), and none higher. Note that this also has the same total energy. (e) Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways. (f) What do your findings suggest?

What information would you need in order to specify the macrostate of the air in a room? What information would you need to specify the microstate?

A block has a cavity inside, occupied by a photon gas. Briefly explain what the characteristics of this gas should have to do with the temperature of the block.

The entropy of an ideal monatomic gas is \((3 / 2) N k_{B} \ln E+\) \(N k_{\mathrm{B}} \ln V-N k_{\mathrm{B}} \ln N,\) to within an additive constant. Show that this implies the correct relationship between internal energy \(E\) and temperature.

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