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Chapter 7: Q 7.40 (page 293)

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(系)=8蟺V系2(hc)3

Step by step solution

01

Given information

The equation 7.83 is

UV=8蟺(hc)3鈭�0鈭�系3e系/kT-1d系

02

Explanation

The equation 7.83 is:

UV=8蟺(hc)3鈭�0鈭�系3e系/kT-1d系

We can write the equation as:

localid="1647752992962">U=鈭�0鈭�系8蟺V系2(hc)31e系/kT-1d系(1)

Distribution function for Planck's constant is given as:

n炉Pl=1e系/kT-1

Substituting this into (1)

U=鈭�0鈭�系8蟺V系2(hc)3n炉Pld系

Hence the energy density for Planck's constant is

g(系)=8蟺V系2(hc)3

Using Python to solve this function, the code is:

The graph is:

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Most popular questions from this chapter

Use the results of this section to estimate the contribution of conduction electrons to the heat capacity of one mole of copper at room temperature. How does this contribution compare to that of lattice vibrations, assuming that these are not frozen out? (The electronic contribution has been measured at low temperatures, and turns out to be about40% more than predicted by the free electron model used here.)

Carry out the Sommerfeld expansion for the energy integral (7.54), to obtain equation 7.67. Then plug in the expansion for 渭to obtain the final answer, equation 7.68.

Consider a system consisting of a single impurity atom/ion in a semiconductor. Suppose that the impurity atom has one "extra" electron compared to the neighboring atoms, as would a phosphorus atom occupying a lattice site in a silicon crystal. The extra electron is then easily removed, leaving behind a positively charged ion. The ionized electron is called a conduction electron because it is free to move through the material; the impurity atom is called a donor, because it can "donate" a conduction electron. This system is analogous to the hydrogen atom considered in the previous two problems except that the ionization energy is much less, mainly due to the screening of the ionic charge by the dielectric behavior of the medium.

(a) Write down a formula for the probability of a single donor atom being ionized. Do not neglect the fact that the electron, if present, can have two independent spin states. Express your formula in terms of the temperature, the ionization energy I, and the chemical potential of the "gas" of ionized electrons.

(b) Assuming that the conduction electrons behave like an ordinary ideal gas (with two spin states per particle), write their chemical potential in terms of the number of conduction electrons per unit volume,NcV.

(c) Now assume that every conduction electron comes from an ionized donor atom. In this case the number of conduction electrons is equal to the number of donors that are ionized. Use this condition to derive a quadratic equation for Ncin terms of the number of donor atoms Nd, eliminating碌. Solve for Ncusing the quadratic formula. (Hint: It's helpful to introduce some abbreviations for dimensionless quantities. Tryx=NcNd,t=kTland so on.)

(d) For phosphorus in silicon, the ionization energy is localid="1650039340485" 0.044eV. Suppose that there are 1017patoms per cubic centimeter. Using these numbers, calculate and plot the fraction of ionized donors as a function of temperature. Discuss the results.

Number of photons in a photon gas.

(a) Show that the number of photons in equilibrium in a box of volume V at temperature T is

N=8蟺VkThc3鈭�0鈭�x2ex-1dx

The integral cannot be done analytically; either look it up in a table or evaluate it numerically.

(b) How does this result compare to the formula derived in the text for the entropy of a photon gas? (What is the entropy per photon, in terms of k?)

(c) Calculate the number of photons per cubic meter at the following temperatures: 300 K; 1500 K (a typical kiln); 2.73 K (the cosmic background radiation).

In Section 6.5 I derived the useful relation F=-kTln(Z)between the Helmholtz free energy and the ordinary partition function. Use analogous argument to prove that 蠒=-kT脳ln(Z^), where Z^ is the grand partition function and 蠒is the grand free energy introduced in Problem 5.23.
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