Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 9

An ideal Bose-Einstein gas consists of noninteracting bosons of mass \(m\) which have an internal degree of freedom which can be described by assuming that the bosons are two-level atoms. Bosons in the ground state have energy \(E_{0}=p^{2} /(2 m)\), while bosons in the excited state have energy \(E_{1}\) \(=p^{2} /(2 m)+\Delta\), where \(p\) is the momentum and \(\Delta\) is the excitation energy. Assume that \(\Delta \gg k_{\mathrm{B}} T\). Compute the Bose-Einstein condensation temperature, \(T_{c}\), for this gas of two-level bosons. Does the existence of the internal degree of freedom raise or lower the condensation temperature?

Short Answer

Expert verified
The condensation temperature is lower by a factor of approximately 1.587 for two-level bosons.

Step by step solution

01

Define the Condensation Temperature for a Single State

The Bose-Einstein condensation temperature for non-interacting bosons of mass \(m\) without considering internal degrees of freedom is given by: \[ T_{c0} = \frac{2 \pi \hbar^2}{k_B m} \left( \frac{n}{g \zeta(3/2)} \right)^{2/3} \] where \(n\) is the number density, \(k_B\) is Boltzmann's constant, \(g\) is the degeneracy factor, and \(\zeta\) is the Riemann zeta function.
02

Apply the Two-Level Atom Condition

For two-level atoms, the energy of bosons in the excited state is \(E_1 = \frac{p^{2} }{2 m} + \Delta\). Given \( \Delta \gg k_{\text{B}} T \), the number of bosons in the excited state is negligible near the condensation temperature. Therefore, \( g = g_0 = 1 \) for the ground state.
03

Modify the Density of States

The total number density is evenly split between the ground state and the excited state because \( \Delta \gg k_{\text{B}} T \). Thus, the number density for the ground state bosons can be written as: \[\frac{n}{2} = \frac{1}{V} \sum_{\text{{ground state}}} n_{\text{{ground}}} \] where \(V \) is the volume and \(n_{\text{{ground}}} \) represents the number density in the ground state.
04

Find the New Condensation Temperature

Substitute this adjusted number density into the formula for condensation temperature and solve for \(T_c\): \[ T_{c} = \frac{2 \pi \hbar^2}{k_B m} \left( \frac{n/2}{ \zeta(3/2)} \right)^{2/3} \] This simplifies to: \[ T_{c} = \frac{T_{c0}}{2^{2/3}} \approx 0.63 T_{c0} \] Therefore, the condensation temperature \( T_c \) for two-level bosons is lower than the condensation temperature for single-state bosons by a factor of \[2^{2/3} \approx 1.587 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

ideal Bose-Einstein gas
An ideal Bose-Einstein gas consists of bosons that do not interact with each other. This simplification helps in understanding the fundamental properties of Bose-Einstein condensates. These bosons can occupy the same quantum state, and they follow Bose-Einstein statistics instead of classical Maxwell-Boltzmann statistics. The defining characteristic of an ideal Bose-Einstein gas is its ability to form a Bose-Einstein condensate at low temperatures, where a large fraction of bosons occupy the lowest quantum state.
noninteracting bosons
In the context of Bose-Einstein condensation, noninteracting bosons are particles that do not exert forces on each other. This assumption simplifies the mathematical treatment of the gas and allows us to derive properties such as the Bose-Einstein condensation temperature. Because the bosons do not interact, their behavior is governed by quantum statistics rather than any potential energy arising from interactions. This leads to interesting quantum effects like the formation of a condensate.
internal degrees of freedom
Internal degrees of freedom refer to the additional quantum states that bosons can occupy due to their internal properties, like spins or other intrinsic qualities. These degrees of freedom add complexity to the system because they provide more possible states for the bosons, affecting factors such as energy distribution and condensation temperature. For two-level atoms, the internal degree of freedom is crucial in determining the Bose-Einstein condensation temperature.
two-level atoms
Two-level atoms are a type of boson with two distinct energy states: a ground state and an excited state. The ground state has the energy given by \(E_{0} = p^2 / (2m)\), while the excited state has an additional energy \(\Delta\), i.e., \(E_{1} = p^2 / (2m) + \Delta\). Because \(\Delta \gg k_{\text{B}}T\), where \(k_{\text{B}}\) is Boltzmann's constant and \(T\) is the temperature, most bosons remain in the ground state near the condensation temperature. The existence of this excited state affects the overall boson distribution.
condensation temperature
The Bose-Einstein condensation temperature \(T_c\) is the temperature below which a significant fraction of bosons occupy the lowest quantum state. For a gas of noninteracting bosons, the condensation temperature depends on factors like mass \(m\), number density \(n\), and the degeneracy factor \(g\). When considering two-level atoms, this formula needs to be adjusted. Given that half of the bosons populate the ground state due to \(\Delta \gg k_{\text{B}}T\), the new condensation temperature is lower than for single-state bosons: \(T_c = T_{c0} / 2^{2/3}\), approximately 63% of the original temperature without internal states.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A two-dimensional electron gas can be formed at the interface of GaAs/AlGaAs semiconductors. The effective mass of the electrons is \(m=0.067 m_{e}\), where \(m_{\mathrm{e}}\) is the mass of the, electron in free space. Treat the electrons like an ideal Fermi gas of particles with spin-1/2 and mass \(m\) in a two-dimensional box with area \(A=L^{2}\). (a) What is the density of states of the electron gas? (b) If the electron density is \(\mathrm{N} / \mathrm{A}=2.0 \times 10^{13} / \mathrm{cm}^{2}\), what is the Fermi energy of the two- dimensional electron gas?

A monatomic dilute gas, in a box of volume \(V\), obeys the Dieterici equation of state \(P=\) \(n R T /(V-n b) \exp (-n a /(V R T)\) ), where \(n\) is mole number, \(a\) and \(b\) are constants determined by the type of molecules in the gas. (a) Find the second virial coefficient for this gas and express it in terms of \(a\). and \(b\). (b) Compute the constants \(a\) and \(b\) for a gas of particles that interacts via a two-body potential \(V\), (r) that has an infinite hard core and an attractive square-well potential region such that \(V(r)=\infty\) for \(r\) \(<\sigma, V(r)=-\epsilon\) for \(\sigma \leq r \leq \lambda \sigma\), and \(V(r)=0\) for \(r>\lambda \sigma\), where \(r\) is the relative displacement of a pair of particles. (c) The constants \(a\) and \(b\) for a gas of \(\mathrm{CO}_{2}\) molecules are \(a=0.3658 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\) and \(b=\) \(0.00004286 \mathrm{~m}^{3} / \mathrm{mol}\) (see Table 3.6. Section 3.8). Assuming that \(\lambda=2\) obtain estimates for the radius \(\sigma\) and binding energy \(\epsilon\) (in eVs) for \(\mathrm{CO}_{2}\).

One mole of a dilute gas of \(\mathrm{He}^{3}\) atoms (which are spin- \(1 / 2\) fermions) at temperature \(T\) \(=140 \mathrm{~K}\) is contained in a box of volume \(\mathrm{V}=1.0 \mathrm{~cm}^{3}\). (a) Compute the lowest order (in density) correction to the classical ideal gas pressure. (b) What fraction of the total pressure is due to the Fermi statistics of the atoms?

Electrons in a piece of copper metal can be assumed to behave like an ideal Fermi gas. Copper metal in the solid state has a mass density of \(9 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that each copper atom donates one electron to the Fermi gas. Assume the system is at \(T=0 \mathrm{~K}\). (a) Compute the Fermi energy, \(\varepsilon_{\mathrm{F}}\), of the electron gas. (b) Compute the Fermi "temperature", \(T_{\mathrm{F}}=\varepsilon_{\mathrm{P}} / \mathrm{k}_{\mathrm{B}}\).

Compute the magnetization of an ideal gas of spin-1/2 fermions in the presence of a magnetic field. Assume that the fermions each have magnetic moment \(\mu_{\mathrm{e}}\). Find an expression for the magnetization in the limit of weak magnetic field and \(T \rightarrow 0 \mathrm{~K}\).
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Nuclear Physics

Read Explanation

Fields in Physics

Read Explanation

Electricity

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free