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Starting from equation 7.83, derive a formula for the density of states of a photon gas (or any other gas of ultra relativistic particles having two polarisation states). Sketch this function.

Short Answer

Expert verified

Hence, the formula for density of states of a photon gas isg(ϵ)=8πVϵ2(hc)3

Step by step solution

01

Given information

The equation 7.83 is

UV=8π(hc)30ϵ3eϵ/kT-1dϵ

02

Explanation

The equation 7.83 is:

UV=8π(hc)30ϵ3eϵ/kT-1dϵ

We can write the equation as:

localid="1647752992962">U=0ϵ8πVϵ2(hc)31eϵ/kT-1dϵ(1)

Distribution function for Planck's constant is given as:

n¯Pl=1eϵ/kT-1

Substituting this into (1)

U=0ϵ8πVϵ2(hc)3n¯Pldϵ

Hence the energy density for Planck's constant is

g(ϵ)=8πVϵ2(hc)3

Using Python to solve this function, the code is:

The graph is:

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Most popular questions from this chapter

Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation 7.121 numerically, without approximating it as an integral.*

(a) Write equation 7.121 for this system as a sum over energy levels, taking degeneracy into account. Replace Tandμwith the dimensionless variables t=kT/hfandc=μ/hf.

(b) Program a computer to calculate this sum for any given values of tandc. Show that, for N=2000, equation 7.121 is satisfied at t=15provided that c=-10.534. (Hint: You'll need to include approximately the first 200 energy levels in the sum.)

(c) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.

(d) Now reduce tto 14 , and adjust the value of cuntil the sum again equals 2000. Plot the number of particles as a function of energy.

(e) Repeat part (d) for t=13,12,11,and10. You should find that the required value of cincreases toward zero but never quite reaches it. Discuss the results in some detail.

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?

The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan's law.

(a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of2.4eV, while Sirius A is approximately 24times as luminous as the sun. How does the radius of Sirius A compare to the sun's radius?

(b) Sirius B, the companion of Sirius A (see Figure 7.12), is only role="math" localid="1647765883396" 3%as luminous as the sun. Its spectrum, plotted as a function of energy, peaks at about7eV. How does its radius compare to that of the sun?

(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of 0.8eV, while Betelgeuse is approximately10,000times as luminous as the sun. How does the radius of Betelgeuse compare to the sun's radius? Why is Betelgeuse called a "red supergiant"?

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic ϵmc2so that their energies are ϵ=pc(where p is the magnitude of the momentum vector).

(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by =

μ=hc(3N/8πV)1/3

(b) Find a formula for the total energy of this system in terms of N and μ.

Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is

N=kTZZμ

where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is

N2¯=(kT)2Z2Zμ2

Use these results to show that the standard deviation of Nis

σN=kTN/μ,

in analogy with Problem6.18Finally, apply this formula to an ideal gas, to obtain a simple expression forσNin terms ofN¯Discuss your result briefly.

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