Team Quanta presents all possible short questions of Thermal & Statistical Physics’ Chapter#05: Stastical System.
Q.1: Show that the distinction between three types of statistics vanishes in limit of high temperature. Give a physical interpretation why this happens. Approximately how high should temperature be for distinction between three statistics to become unimportant?
Answer: Boltzmann distribution function is,
$$f\left(\epsilon\right)=g_ie^\frac{\mu-\epsilon}{k_BT}, g_i$$
is degeneracy of energy level.
Fermi-Dirac distribution function is,
$$f\left(\epsilon\right)=\frac{g_i}{e^\frac{\epsilon-\mu}{k_BT}+1}$$
Bose-Einstein distribution function is,
$$f\left(\epsilon\right)=\frac{g_i}{e^\frac{\epsilon-\mu}{k_BT}-1}$$
When $e^\frac{\mu}{k_BT}\gg1 \ \ \ or \ \ \ e^\frac{\mu}{k_BT}\ll1$, then
$$\frac{g_i}{e^\frac{\epsilon-\mu}{k_BT}\mp1} \approx gie^\frac{\mu-\epsilon}{k_BT}$$
In this case, the distinction among three types of statistics vanishes. From relation,
$$e^{\frac{\mu}{k_BT}=n\left(\frac{h^2}{2\pi mk_BT}\right)^\frac{3}{2}}$$
It is clear that above condition is satisfied when $T\gg\frac{n^{2/3}h^2}{2\pi mk_B}$. Therefore the distinction among three types of statistics becomes unimportant in limit of high temperatures.
When $e^{-\mu/k_BT}\gg 1$, then $\frac{f\left(\epsilon\right)}{g}\ll 1$. This shows that average number of particles in any quantum state is much less than 1. The reason is that the number of microstates available to particles is very large, much larger than total particle number. Hence probability for two particles to occupy same quantum state is very small and Pauli Exclusion Principle is satisfied naturally. As a result distinction between three types of statistics vanishes.
Q.2: Consider a system of two atoms, each having three energy states . The system is in contact with a heat reservoir at temperature . Calculate the partition function for system if particles obey Bose Einstein statistics.
Answer: When particles obey Bose-Einstein statistics, then partition function is,
$$Z=\left(1+e^{-\frac{\epsilon}{k_BT}}+e^{-\frac{2\epsilon}{k_BT}}\right)\left(1+e^{-\frac{2\epsilon}{k_BT}}\right)$$
Q.3: Consider a system of two atoms, each having three energy states . The system is in contact with a heat reservoir at temperature . Calculate the partition function for system if particles obey Fermi-Dirac statistics
Answer: When particles obey Fermi-Dirac statistics, then partition function is,
$$\mathbf{Z=\left(1+e^{-\frac{\epsilon}{k_BT}}+e^{-\frac{2\epsilon}{k_BT}}\right)e^{-\frac{\epsilon}{k_BT}}}$$
Q.4: Describe Boltzmann, Fermi and Bose statistics and distinguish between them especially from the point of view of Pauli Exclusion Principle and indistinguishability of identical particles.
Answer: In Boltzmann statistics, for a localized system, the particles are distinguishable and number of particles occupying a single quantum state is not limited. Boltzmann distribution function is,
$$\mathbf{f\left(\epsilon=g_ie^\frac{\mu-\epsilon}{k_BT}\right)}$$
In Fermi-Dirac statistics, for a system composed of fermions, the particles are indistinguishable and obey Pauli Exclusion Principle. Fermi-Dirac distribution is,
$$\mathbf{f\left(\epsilon\right)=\frac{g_i}{e^\frac{\epsilon-\mu}{k_BT}+1}}$$
In Bose-Einstein statistics, for a system composed of bosons, the particles are indistinguishable and number of particles occupying a single quantum state is not limited.
$$\mathbf{f(\left(\epsilon\right)=\frac{g_i}{e^\frac{\epsilon-\mu}{k_BT}-1}}$$
Q.5: Classify following between Bosons and Fermions and state the criterion which has been applied for classification: electron, proton, neutron, helium atom.
Answer: Electron is a fermion because it obeys Pauli Exclusion Principle and its wave function is skew-symmetric. Neutron also is a fermion because it obeys Pauli Exclusion Principle and its wave function is skew-symmetric. Maxwell-Boltzmann statistics is applicable to helium atom $\mathbf{{He}^4}$ because $\mathbf{n\lambda^3\ll1}$ .
Q.6: Why is the quantity $\frac{\mathbf{h}}{\sqrt{\mathbf{2\pi m}\mathbf{k}_\mathbf{B}\mathbf{T}}}$ identified with wavelength?
Answer: Here,
$$\mathbf{\frac{h}{\sqrt{2\pi mk_BT}}=\frac{Js}{\sqrt{kg \times \frac{J}{K}\times K}}=\frac{Js}{\sqrt{kg\times J}}=\frac{s\sqrt J}{\sqrt{kg}}=\frac{s\sqrt{kg\times m^2/s^2}}{\sqrt{kg}}=m}$$
Since given quantity has dimension of length, so it represents wavelength.
Q.7: Given that results $\mathbf{\mu=k_BTln(N\lambda^3/V) }$ for a perfect gas, what physical interpretation can be given to the limiting condition ?
Answer: From given,
$$\mathbf{\mu=k_BTln\left(\frac{N\lambda^3}{V}\right)\ \ \ \Rightarrow\frac{\mu}{k_BT}=\ln{\left(\frac{N\lambda^3}{V}\right)}\ \ \ \ \Rightarrow\frac{\mu}{e^{k_BT}}=\frac{N\lambda^3}{V}}$$
$$\mathbf{\Rightarrow\ e^\frac{\mu}{k_BT}=\frac{N}{V}\left(\frac{h^2}{2\pi mk_BT}\right)^\frac{3}{2}}$$
For $\mathbf{e^\frac{\mu}{k_BT}\ll1 \ \ \ \ or \ \ \ e^{-\frac{\mu}{k_BT}}\gg1,}$ , we see that average number of particles in any quantum state is much less than 1. The reason is that the number of microstates available to particles is very large, much larger than total particle number. Hence probability for two particles to occupy same quantum state is very small and Pauli Exclusion Principle is satisfied naturally. As a result, distinction between three types of statistics vanishes.
Q.8: How is question of symmetry of wave functions related to basic distinction between Fermi Dirac and Bose Einstein systems?
Answer: In Fermi-Dirac statistics, wave function is skew-symmetric where as in Bose-Einstein statistic, wave function is symmetric. Fermions obey Pauli Exclusion Principle where as bosons do not obey this principle.
Q.9: Given that the classical potential for a perfect gas is defined by $\mathbf{e^{-\mu/k_BT}=V/N\lambda^3}$ , explain the physical significance of λ. If the right side of equation is nearly 10-4 for electron gas in a metal at room temperature, what conclusion do you show about electron gas from this statistics?
Answer: In given relation λ is thermal de-Broglie wavelength given by relation,
$$\mathbf{\lambda=\frac{h}{\sqrt{2\pi mk_BT}}}$$
If right side of equation $\mathbf{e^{-\mu/k_BT}=V/N\lambda^3}$ is nearly equal to $\mathbf{{10}^{-4}}$, then classical approximation is entirely invalid. The distribution is said to be degenerate when classical distribution fails.
Q.10: State which of three statistics would apply to a sample of gas at room temperature and pressure, why? Use the following approximate values of data if required: $p={10}^6\ dynes\ cm^{-2},\ k_B=1.4\times{10}^{-16}\ \ ergK^{-1},\ h=6.6\times{10}^{-27}\ \ erg\ s \ Mass\ of\ proton=1.67\times{10}^{-24}\ g$
Answer: Here we calculate the product $\mathbf{n\lambda^3}$ for application of statistics.
$$\boldsymbol{n\lambda^3=\frac{p}{k_BT}\left(\frac{h^2}{2\pi mk_BT}\right)^\frac{3}{2}~\ 3\times{10}^{-6}\ll1}$$
This shows that Maxwell-Boltzmann statistics is appropriate.
Q.11: A system has two non-degenerate energy levels. The band gap energy is 0.1eV. Calculate probability that the system is in higher energy level when it is in thermal equilibrium with a heat reservoir of absolute temperature 300 K.
Answer: The probability of system being in higher energy level is,
$$\mathbf{P=\frac{e^{-E_g/k_BT}}{1+e^{-E_g/k_BT}}=\frac{e^{-\frac{0.1\times1.6\times{10}^{-9}}{1.38\times{10}^{-23}\times300}}}{1+e^{-\frac{0.1\times1.6\times{10}^{-9}}{1.38\times{10}^{-23}\times300}}}=0.0205}$$
Q.12: Density of states in phase is given by, $\mathbf{\rho=\frac{4\pi gp^2}{h^3}\frac{E}{pc^2}=\frac{4\pi gp^2}{h^3\nu}}$ is number of internal states due to intrinsic angular momentum. Find density of states for non-relativistic electron gas.
Answer: Relation between energy and momentum is,
$$\mathbf{E=\frac{p^2}{2m}\ \ \ \ \Rightarrow p=\sqrt{2mE}}$$
Degeneracy is $\mathbf{g=2s+1=2\left(\frac{1}{2}\right)+1=2.}$ . So density is,
$$\mathbf{\rho=\frac{4\pi gp^2}{h^3\nu}=\frac{4\pi\times2\times2mE}{h^3\sqrt{2E/m}}=\frac{4\pi\left(2m\right)^\frac{3}{2}\sqrt E}{h^3}}$$
Q.13: If the universe is an impenetrable cavity of radius and inside temperature 3K, estimate the total number of photons in the universe. Note: $\mathbf{\int_{-\infty}^{\infty}{\frac{x^2}{e^x-1}dx}\ \cong2.4, \ \ \ k_B=1.38\times10^{-16}\frac{erg}{K},\ \ \ \ \ h=1.05\times10^{-27}\ erg\ s}$ ,
Answer: Total number of photons is,
$$\mathbf{N=\frac{2\times 1.2}{\pi^2}\times V\times\left(\frac{k_BT}{\hbar c}\right)^3=\frac{2.4}{\pi^2}\times\frac{4}{3}\pi\left({10}^{28}\right)^3\times\left(\frac{1.38\times{10}^{-16}\times3}{1.05\times{10}^{-27}\times3\times{10}^{10}}\right)^3}$$
$$\mathbf{\Rightarrow N=25.\times{10}^{87}}$$
Q.14: Why do we say that number of photons inside an enclosure may not remain constant?
Answer: This is because a photon may be completely absorbed on striking the walls of enclosure or the hot wall may emit a new photon or that a photon of energy equivalent to 2hv may be absorbed and two photons of energy hv may be emitted i. e. photons may be created or absorbed. However total energy of photons inside enclosure remains constant.
Q.15: What is wavelength at which human body radiates maximum energy?
Answer: Temperature of human body is $37^oC = 310K$ Using Wien displacement law,
$$\boldsymbol{\lambda_mT=2898\times{10}^{-6}m-K}$$
$$\mathbf{\Rightarrow\lambda_m=\frac{2898\times{10}^{-6}m-K}{310K}=9.35\times{10}^{-6}m}$$
Q.16: Does Fermi-energy depend upon size or volume of the conductor? Explain.
Answer: Fermi energy is,
$$\mathbf{\epsilon_F=\frac{h^2}{2m}\left(\frac{3N}{8\pi V}\right)^\frac{2}{3}}$$
From above expression, it is clear that Fermi energy is independent of size and volume of conductor as it only depends number of electrons per unit volume or electron concentration.
Q.17: What is meant by thermal radiation spectrum? Prove that energy per volume is, $\mathbf{\frac{du}{d\epsilon}=\frac{8\pi}{h^3c^3}\left(\frac{\epsilon^3}{e^\frac{\epsilon}{k_BT}-1}\right)}$
Answer: Thermal radiation is an example of photon gas. Its spectrum is described by Bose-Einstein distribution along with density of states. The number of photons per volume per energy, the density of occupied states, is density of states times Bose-Einstein factor;
$$\mathbf{\frac{dn}{d\epsilon}=\rho\times f\left(\epsilon\right)=\frac{8\pi\epsilon^2}{h^3c^3}\times\frac{1}{e^\frac{\epsilon}{k_BT}-1}}$$
The energy per volume is,
$$\mathbf{\frac{du}{d\epsilon}=\epsilon\frac{dn}{d\epsilon}=\epsilon\times\frac{8\pi\epsilon^2}{h^3c^3}\times\frac{1}{e^{\epsilon/k_BT}-1}=\frac{8\pi}{h^3c^3}\left(\frac{\epsilon^3}{e^\frac{\epsilon}{k_BT}-1}\right)\ }$$
Note:- I will find density of states for photon gas in Q. 3.23.
Q.18: Do electrons have zero kinetic energy at 0K? If not, why?
Answer: The energy of system of N electrons at 0K is given by,
$$\mathbf{E=\frac{3}{5}N\epsilon_F}$$
The electrons at 0K do not have zero energy at because according to Pauli Exclusion Principle only one electron can occupy zero energy level. All other electrons will occupy higher energy levels and will not, therefore, have zero energy.
Q.19: Compare three types of statistics with reference to energy quantization.
Answer: Maxwell-Boltzmann statistics assumes a continuous distribution of energy where as Bose-Einstein and Fermi-Dirac statistics are based on assumption that energy is quantized.
Q.20: Consider a photon gas enclosed in a volume and in equilibrium at temperature . The photon is massless particle so that What is chemical potential of gas?
Answer: The chemical potential of photon gas is zero. Since number of photons is not conserved at a given temperature and volume, the average photon number is determined by relation,
$$\mathbf{\mu=\left(\frac{\partial F}{\partial N}\right)_{T,V}=0}$$
Q.21: Write down simple expression for internal part of partition function for a single isolated hydrogen atom in weak contact with a reservoir at temperature . Does your answer diverge for T=0 , for T≠0?
Answer: The internal energy levels of hydrogen atom are given by $(-E_\circ/n^2)$ with degeneracy n2 , where n=1,2,… . So
$$\mathbf{Z=\sum_{n=1}^{\infty}{2n^2e^\frac{E_\circ}{n^2k_BT}}}$$
WhenT=0 , it is meaningless and for T≠0 , it diverges.
Q.22: Given that mass of sun is 2 x 1033g , estimate number of electrons in sun. Assume sun is largely composed of atomic hydrogen.
Answer: The number of electrons is,
$$\mathbf{N=\frac{2\times{10}^{33}}{1.67\times{10}^{-24}}=1.2\times{10}^{57}}$$
Q.23: Density of states in phase is given by, $\mathbf{\rho=\frac{4\pi gp^2}{h^3}\frac{E}{pc^2}=\frac{4\pi gp^2}{h^3\nu}}$ is number of internal states due to intrinsic angular momentum. Find density of states for photon gas.
Answer: Relation between energy and momentum is,
E=pc
Degeneracy is g=2 since there are two polarization states of photon. So density is,
$$\mathbf{\rho=\frac{4\pi gp^2}{h^3\nu}=\frac{4\pi\times2\times\left(\frac{E}{c}\right)^2}{h^3c}=\frac{8\pi E^2}{h^3c^3}}$$
Q.24: Write an expression for Fermi energy of conduction electrons. Does Fermi energy depend upon size or volume of conductor?
Answer: Fermi energy is given by relation,
$$\mathbf{\epsilon_F=\frac{h^2}{2m}\left(\frac{3N}{8\pi V}\right)^\frac{2}{3}}$$
It is clear from the above expression that Fermi energy is independent of size and volume of conductor as it depends only on (N/V) , number of electrons per unit volume i.e. electrons concentration.
Q.25: Determine the fraction of total number of particles in ground state of degenerate Bose gas at 0.9To , where To is degeneracy temperature. What does degeneracy temperature signify?
Answer: For Bose-Einstein condensation,
$$N_\circ=N\left{1-\left(\frac{T}{T_\circ}\right)^\frac{3}{2}\right}$$
If T=0.9To , then
$$N_\circ=N\left{1-\left(\frac{0.9T_\circ}{T_\circ}\right)^\frac{3}{2}\right}=0.15\ N\ \cong\frac{1}{6}N$$
That is if we increase small amount of temperature in comparison with degeneracy temperature, a large number of particles (about $\mathbf{\frac{1}{6}th}$ ) will go up to ground state.
If we increase the temperature then particles which are in ground state will be shifted to excited state and ground state will become empty at a certain temperature called degeneracy temperature To . At temperature To , all particles are in excited state. Degeneracy temperature is defined by relation,
$$\mathbf{N=2.612\frac{V}{\lambda_\circ^3}=2.612V\left(\frac{2\pi mk_BT\ }{h^2}\right)^\frac{3}{2}}$$
Q.26: Is there a lowest temperature below which black body radiation is no longer given off by an object?
Answer: At a finite temperature, all bodies emit black body radiations. Only an object at absolute zero emits no black body radiation which is physically impossible by third law of thermodynamics which says that absolute zero temperature is not attainable.
Q.27: The black body spectrum from a black body P peaks at lower wavelength than that of black body Q. is temperature of both black bodies same?
Answer: From Wien displacement law,
$$\mathbf{\lambda_mT=2898\mu m\ K}$$
The body having longer wavelength is at lower temperature, so black body P is at lower temperature than Q.
Q.28: In a white dwarf star of one solar mass the atoms are all ionized and contained in a sphere of radius 2 x 109cm . Find Fermi energy of electrons in .
Answer: Use,
$$\mathbf{\epsilon_F=\frac{h^2}{2m}\left(\frac{3}{8\pi}\frac{N}{V}\right)^\frac{2}{3}=\frac{h^2}{2m}\left(\frac{3}{8\pi}\frac{N}{\frac{4}{3}\pi r^3} \right)^\frac{2}{3}=4 \times {10}^4eV}$$
Q.29: You are given a system of two identical particles which may occupy any of three energy levels;
∊n=n∊, n=0,1,2,3…..
The lowest energy state is doubly degenerate. The system is in thermal equilibrium at temperature . Determine partition function and energy if particles obey Bose-Einstein statistics.
Answer: Partition function is,
$$\mathbf{Z=3+2e^{-\beta\epsilon}+2e^{-2\beta\epsilon}+e^{-3\beta\epsilon}+e^{-4\beta\epsilon}}$$
Energy is,
$$\mathbf{E=-\frac{1}{Z}\frac{\partial Z}{\partial \beta}=\frac{\epsilon}{Z}e^{-\beta \epsilon}\left(2+6e^{-\beta \epsilon}+3e^{-2\beta \epsilon}+4e^{-3\beta \epsilon}\right)}$$
Q.30: What do you mean by electron gas? What kind of distribution is shown by conduction electrons at 0K?
Answer: Metals are very good conductors and have free valance electrons. The electrons in a metal have their energies quantized and being Fermi particles obey Pauli Exclusion Principle. The free electrons inside a metal move about freely and continuously collide with fixed atoms and thus behave like an electron gas. At T=o , there is complete cut off beyond $\mathbf{\epsilon=\epsilon_F=\mu_\circ}$
vibration, so total degrees of freedom are five.
