Imagine that there exists a third type of particle, which can share a single-particle state with one other particle of the same type but no more. Thus the number of these particles in any state can be 0,1 or 2 . Derive the distribution function for the average occupancy of a state by particles of this type, and plot the occupancy as a function of the state's energy, for several different temperatures.

Short Answer

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The distribution function for the average occupancy of a state by given types of particles is derived.

The graph is as follows,

Step by step solution

01

Step 1. Given Information 

We are given a particle that can share a single-particle state with one other particle of the same type but no more and the number of these particles in any state can be 0,1or 2.

02

Step 2. Grand partition function

The grand partition function or Gibbs sum is,

Z=ne-n(ϵ-μ)kT

According to the given problem, the number of particles of third type in any state can be 0,1or 2. That is n can be 0,1or2.

Z=n=02exp-n(ε-μ)kTZ=exp(0)+exp-(ε-μ)kT+exp-2(ε-μ)kTZ=1+exp-(ε-μ)kT+exp-2(ε-μ)kT

03

Step 3. Probability of n- particals

Let x=exp-(ε-μ)kT,

Z=1+x+x2

The probability of state being occupied by n-particles is,

P(n)=1Zexp-n(ε-μ)kT=1ZxnP(n)=xn1+x+x2

04

Step 4. Average occupancy of state by particles 

The average occupancy of state by particles of this type is,

n¯=n=02nP(n)=0·P(0)+1·P(1)+2·P(2)=1·x11+x+x2+2·x21+x+x2n¯=x+2x21+x+x2

So, the distribution function for the number of these particles in any state can be 0,1or 2is,

n=x+2x21+x+x2x=exp-(ε-μ)kT

05

Step 5. Graphing the occupancy as a function of the state's energy 

The graph is as follows,

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Most popular questions from this chapter

When the attractive forces of the ions in a crystal are taken into account, the allowed electron energies are no longer given by the simple formula 7.36; instead, the allowed energies are grouped into bands, separated by gaps where there are no allowed energies. In a conductor the Fermi energy lies within one of the bands; in this section we have treated the electrons in this band as "free" particles confined to a fixed volume. In an insulator, on the other hand, the Fermi energy lies within a gap, so that at T = 0 the band below the gap is completely occupied while the band above the gap is unoccupied. Because there are no empty states close in energy to those that are occupied, the electrons are "stuck in place" and the material does not conduct electricity. A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature. Figure 7 .17 shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor, and defines some terminology and notation to be used in this problem.

(a) As a first approximation, let us model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas, with an appropriate zero-point: g(ϵ)=g0ϵ-ϵc, where go is the same constant as in equation 7.51. Let us also model the density of states near the top

Figure 7.17. The periodic potential of a crystal lattice results in a densityof-states function consisting of "bands" (with many states) and "gaps" (with no states). For an insulator or a semiconductor, the Fermi energy lies in the middle of a gap so that at T = 0, the "valence band" is completely full while the-"conduction band" is completely empty. of the valence band as a mirror image of this function. Explain why, in this approximation, the chemical potential must always lie precisely in the middle of the gap, regardless of temperature.

(b) Normally the width of the gap is much greater than kT. Working in this limit, derive an expression for the number of conduction electrons per unit volume, in terms of the temperature and the width of the gap.

(c) For silicon near room temperature, the gap between the valence and conduction bands is approximately 1.11 eV. Roughly how many conduction electrons are there in a cubic centimeter of silicon at room temperature? How does this compare to the number of conduction electrons in a similar amount of copper?

( d) Explain why a semiconductor conducts electricity much better at higher temperatures. Back up your explanation with some numbers. (Ordinary conductors like copper, on the other hand, conduct better at low temperatures.) (e) Very roughly, how wide would the gap between the valence and conduction bands have to be in order to consider a material an insulator rather than a semiconductor?

It's not obvious from Figure 7.19 how the Planck spectrum changes as a function of temperature. To examine the temperature dependence, make a quantitative plot of the functionu(ϵ) for T = 3000 K and T = 6000 K (both on the same graph). Label the horizontal axis in electron-volts.

Suppose you have a "box" in which each particle may occupy any of 10single-particle states. For simplicity, assume that each of these states has energy zero.

(a) What is the partition function of this system if the box contains only one particle?

(b) What is the partition function of this system if the box contains two distinguishable particles?

(c) What is the partition function if the box contains two identical bosons?

(d) What is the partition function if the box contains two identical fermions?

(e) What would be the partition function of this system according to equation 7.16?

(f) What is the probability of finding both particles in the same single particle state, for the three cases of distinguishable particles, identical bosom, and identical fermions?

Calculate the condensate temperature for liquid helium-4, pretending that liquid is a gas of noninteracting atoms. Compare to the observed temperature of the superfluid transition, 2.17K. ( the density of liquid helium-4 is 0.145g/cm3)

Consider any two internal states, s1 and s2, of an atom. Let s2 be the higher-energy state, so that Es2-Es1=ϵ for some positive constant. If the atom is currently in state s2, then there is a certain probability per unit time for it to spontaneously decay down to state s1, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

A = probability of spontaneous decay per unit time.

On the other hand, if the atom is currently in state s1 and we shine light on it with frequency f=ϵ/h, then there is a chance that it will absorb photon, jumping into state s2. The probability for this to occur is proportional not only to the amount of time elapsed but also to the intensity of the light, or more precisely, the energy density of the light per unit frequency, u(f). (This is the function which, when integrated over any frequency interval, gives the energy per unit volume within that frequency interval. For our atomic transition, all that matters is the value of u(f)atf=ϵ/h) The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coefficient:

B=probability of absorption per unit timeu(f)

Finally, it is also possible for the atom to make a stimulated transition from s2down to s1, again with a probability that is proportional to the intensity of light at frequency f. (Stimulated emission is the fundamental mechanism of the laser: Light Amplification by Stimulated Emission of Radiation.) Thus we define a third coefficient, B, that is analogous to B:

B'=probability of stimulated emission per unit timeu(f)

As Einstein showed in 1917, knowing any one of these three coefficients is as good as knowing them all.

(a) Imagine a collection of many of these atoms, such that N1 of them are in state s1 and N2 are in state s2. Write down a formula for dN1/dt in terms of A, B, B', N1, N2, and u(f).

(b) Einstein's trick is to imagine that these atoms are bathed in thermal radiation, so that u(f) is the Planck spectral function. At equilibrium, N1and N2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related by

B'=BandAB=8πhf3c3

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