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At the center of the sun, the temperature is approximately \(10^{7} \mathrm{K}\) and the concentration of electrons is approximately \(10^{32}\) per cubic meter. Would it be (approximately) valid to treat these electrons as a "classical" ideal gas (using Boltzmann statistics), or as a degenerate Fermi gas (with \(T \approx 0\) ), or neither?

Short Answer

Expert verified
Electrons should be treated as a degenerate Fermi gas.

Step by step solution

01

Understanding Classical Ideal Gas

To determine if the electrons can be treated as a classical ideal gas, we use Boltzmann statistics. We need to check if the thermal de Broglie wavelength, \(\lambda\), is significantly smaller than the mean separation between electrons. \(\lambda\) can be estimated as \( \lambda = \frac{h}{\sqrt{2 \pi m_e k T}} \) where \( h \) is Planck's constant, \( m_e \) is the electron mass, \( k \) is Boltzmann's constant, and \( T \) is the temperature.
02

Calculating Parameters for Classical Gas

Substitute the known values: \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m_e = 9.109 \times 10^{-31} \text{ kg} \), \( k = 1.381 \times 10^{-23} \text{ J K}^{-1} \), and \( T = 10^7 \text{ K} \). Calculate \( \lambda \) to find: \[\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \pi \times 9.109 \times 10^{-31} \times 1.381 \times 10^{-23} \times 10^7}} \approx 2.38 \times 10^{-12} \text{ meters}\] The mean separation \(a\) between electrons is about \(a = (n^{-1})^{1/3}\) where \(n\) is the electron concentration: \[ a \approx (10^{32})^{-1/3} \approx 10^{-11} \text{ meters}\]
03

Comparing De Broglie Wavelength and Mean Separation

We compare the thermal de Broglie wavelength \(\lambda\) and the mean separation \(a\). Since \(\lambda \approx 2.38 \times 10^{-12}\) m is much smaller than \(a \approx 10^{-11}\) m, the criteria for using Boltzmann statistics are satisfied.
04

Understanding Fermi-Dirac Degenerate Gas

For the electrons to behave as a degenerate Fermi gas, the condition \(T \ll T_F\) must hold, where \(T_F\) is the Fermi temperature. The Fermi temperature can be estimated with \(T_F = \frac{E_F}{k}\), where \(E_F = \frac{\hbar^2}{2m}(3\pi^2n)^{2/3}\). With the concentration \(n = 10^{32}\), calculate \(E_F\) and then \(T_F\).
05

Calculating Fermi Energy and Temperature

Using \( \hbar = \frac{h}{2\pi} \), calculate \(E_F\): \[E_F = \frac{(6.626\times10^{-34}/2\pi)^2}{2\times9.109\times10^{-31}}(3\pi^2\times10^{32})^{2/3} \approx 7.63 \times 10^{-13} \text{ J} \approx 4.76 \text{ MeV}\]Then find \(T_F = \frac{E_F}{k} \approx \frac{7.63 \times 10^{-13}}{1.381 \times 10^{-23}} \approx 5.52 \times 10^{9} \text{ K}. \) Compare \(T\) and \(T_F\).
06

Comparing Actual and Fermi Temperature

Since the actual temperature \(T = 10^7 \text{ K}\) is substantially lower than the Fermi temperature \(T_F = 5.52 \times 10^{9} \text{ K}\), we should treat the electrons as a degenerate Fermi gas at approximately \(T \approx 0\). However, this significantly exceeds the classical ideal gas condition, making Fermi statistics appropriate for this high concentration.

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Key Concepts

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

Classical Ideal Gas
In the realm of statistical mechanics, a classical ideal gas is modeled by a perfectly random collection of non-interacting particles. These particles obey the well-known Boltzmann statistics. The classical ideal gas model is applicable when thermal de Broglie wavelengths are much smaller compared to the average distance between particles. This ensures that quantum effects are negligible and each particle behaves independently.
This model successfully describes real gases at high temperatures or low concentrations, where quantum effects are less significant. However, in systems like metal electrons or astrophysical bodies, where quantum mechanics often play a crucial role due to high densities or low temperatures, the classical ideal gas model might fall short.
Boltzmann Statistics
Boltzmann statistics form the foundation of calculating probabilities in classical statistical mechanics. Named after the physicist Ludwig Boltzmann, this statistical approach helps determine the likelihood of a specific energy level being occupied by a particle within a system.
Boltzmann statistics assume that particles are distinguishable and do not obey the Pauli exclusion principle, making them suitable for modeling classical ideal gases. This assumption breaks down for systems where quantum effects are dominant, leading us to explore other statistical models like Bose-Einstein or Fermi-Dirac statistics.
Fermi-Dirac Statistics
Fermi-Dirac statistics are essential for understanding systems of particles known as fermions, which include electrons, protons, and neutrons. Unlike in classical systems, Fermi-Dirac statistics account for the indistinguishable and anti-symmetrical nature of fermions.
Fermions obey the Pauli exclusion principle, prohibiting two identical fermions from occupying the same quantum state. This significantly affects the behavior of a system with high densities, such as electrons in metals or stars. In such scenarios, Fermi-Dirac statistics dominate and provide a more accurate description of the system than classical statistics.
The concept of electron degeneracy, arising from these statistics, is crucial for understanding dense astrophysical objects and electronic structure in solids.
Electron Degeneracy
Electron degeneracy refers to a quantum mechanical state in which particles are delocalized across many identical states due to high densities. This occurs naturally in systems described by Fermi-Dirac statistics, such as electron behavior in white dwarf stars.
At very high densities, the electrons are packed closely enough that degeneracy pressure — a quantum mechanical force — becomes significant. This pressure is independent of temperature and can prevent further collapse of a stellar object, influencing its evolution as a white dwarf.
This quantum pressure is a direct result of the Pauli exclusion principle, demonstrating how degeneracy plays a pivotal role in cosmic structures.
Thermal de Broglie Wavelength
The thermal de Broglie wavelength is a fundamental concept in quantum mechanics, providing insight into the wave-particle duality of matter. It represents the wavelength of a particle at a specific temperature and becomes significant when comparing particle separations to quantum wavelengths.
For classical ideal gases, the thermal de Broglie wavelength must be much smaller than the mean separation between particles, ensuring classical behavior and serving as a criterion for the use of Boltzmann statistics.
This wavelength is inversely related to the square root of the temperature, indicating that as temperature increases, the wavelength becomes smaller, pushing the system towards classical behavior rather than quantum.

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Most popular questions from this chapter

Consider a free Fermi gas in two dimensions, confined to a square \(\operatorname{area} A=L^{2}\) (a) Find the Fermi energy (in terms of \(N\) and \(A\) ), and show that the average energy of the particles is \(\epsilon_{\mathrm{F}} / 2\) (b) Derive a formula for the density of states. You should find that it is a constant, independent of \(\epsilon\) (c) Explain how the chemical potential of this system should behave as a function of temperature, both when \(k T \ll \epsilon_{\mathrm{F}}\) and when \(T\) is much higher. (d) Because \(g(\epsilon)\) is a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for \(\mu\) as a function of \(N .\) Show that the resulting formula has the expected qualitative behavior. (e) Show that in the high-temperature limit, \(k T \gg \epsilon_{\mathrm{F}},\) the chemical potential of this system is the same as that of an ordinary ideal gas.

The planet Venus is different from the earth in several respects. First, it is only \(70 \%\) as far from the sun. Second, its thick clouds reflect \(77 \%\) of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light. (a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight. (b) Estimate the surface temperature again, taking the reflectivity of the clouds into account. (c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly 70 times that of earth's atmosphere. You can therefore model the atmosphere of Venus as 70 successive "blankets" of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of \(2^{1 / 4}\). The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At \(T=0\) the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are \(N\) atoms, the total magnetization is typically \(\sim 2 \mu_{\mathrm{B}} N,\) where \(\mu_{\mathrm{B}}\) is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure \(7.30 .\) Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of \(h / 2 \pi,\) and therefore reduces the magnetization by \(\sim 2 \mu_{\mathrm{B}} .\) However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin wave is proportional to the square of \(1 / \lambda\) (in the limit of long wavelengths). Therefore, since \(\epsilon=h f\) and \(p=h / \lambda\) for any "particle," the energy of a magnon is proportional to the square of its momentum. In analogy with the energy- momentum relation for an ordinary nonrelativistic particle, we can write \(\epsilon=p^{2} / 2 m^{*},\) where \(m^{*}\) is a constant related to the spin- spin interaction energy and the atomic spacing. For iron, \(m^{*}\) turns out to equal \(1.24 \times 10^{29} \mathrm{kg}\), about 14 times the mass of an electron. Another difference between magnons and phonons is that each magnon (or spin wave mode) has only one possible polarization. (a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by $$\frac{N_{m}}{V}=2 \pi\left(\frac{2 m^{*} k T}{h^{2}}\right)^{3 / 2} \int_{0}^{\infty} \frac{\sqrt{x}}{e^{x}-1} d x$$ Evaluate the integral numerically. (b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, \((M(0)-M(T)) / M(0) .\) Write your answer in the form \(\left(T / T_{0}\right)^{3 / 2},\) and estimate the constant \(T_{0}\) for iron. (c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find \(C_{V} / N k=\left(T / T_{1}\right)^{3 / 2},\) where \(T_{1}\) differs from \(T_{0}\) only by a numerical constant. Estimate \(T_{1}\) for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470 K.) (d) Consider a \(t w o\) -dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverges in this case. (This result is an indication that there can be no spontaneous magnetization in such a two- dimensional system. However, in Section 8.2 we will consider a different two- dimensional model in which magnetization does occur.)

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is (a) \(1 \mathrm{eV}\) less than \(\mu\) (b) \(0.01 \mathrm{eV}\) less than \(\mu\) (c) equal to \(\mu\) (d) \(0.01 \mathrm{eV}\) greater than \(\mu\) (e) \(1 \mathrm{eV}\) greater than \(\mu\)

For a system of bosons at room temperature, compute the average occupancy of a single-particle state and the probability of the state containing \(0,1,2,\) or 3 bosons, if the energy of the state is (a) \(0.001 \mathrm{eV}\) greater than \(\mu\) (b) \(0.01 \mathrm{eV}\) greater than \(\mu\) (c) \(0.1 \mathrm{eV}\) greater than \(\mu\) (d) \(1 \mathrm{eV}\) greater than \(\mu\)

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