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Chapter 7: Q. 7.69 (page 323)

If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate μfor T>Tc.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms dimensionless variables. So define t=T/Tc,c=μ/kTc,andx=ϵ/kTc. Express the integral that defines μ, equation N=∫0∞g(ϵ)1e(ϵ-μ)/kT-1dϵ, in terms of these variables, you should obtain the equation

2.315=∫0∞xdxe(x-c)/t-1

(b) According to given figure , the correct value of cwhen T=2Tc, is approximately -0.8. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary μ, holding Tfixed, to find the precise value of μfor T=2Tc. Repeat for values ofT/Tcranging from 1.2up to 3.0, in increments of 0.2. Plot a graph of μas a function of temperature.

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Most popular questions from this chapter

Consider an isolated system of Nidentical fermions, inside a container where the allowed energy levels are nondegenerate and evenly spaced.* For instance, the fermions could be trapped in a one-dimensional harmonic oscillator potential. For simplicity, neglect the fact that fermions can have multiple spin orientations (or assume that they are all forced to have the same spin orientation). Then each energy level is either occupied or unoccupied, and any allowed system state can be represented by a column of dots, with a filled dot representing an occupied level and a hollow dot representing an unoccupied level. The lowest-energy system state has all levels below a certain point occupied, and all levels above that point unoccupied. Let ηbe the spacing between energy levels, and let be the number of energy units (each of size 11) in excess of the ground-state energy. Assume thatq<N. Figure 7 .8 shows all system states up to q=3.

(a) Draw dot diagrams, as in the figure, for all allowed system states with q=4,q=5,andq=6. (b) According to the fundamental assumption, all allowed system states with a given value of q are equally probable. Compute the probability of each energy level being occupied, for q=6. Draw a graph of this probability as a function of the energy of the level. ( c) In the thermodynamic limit where qis large, the probability of a level being occupied should be given by the Fermi-Dirac distribution. Even though 6 is not a large number, estimate the values of μand T that you would have to plug into the Fermi-Dirac distribution to best fit the graph you drew in part (b).

A representation of the system states of a fermionic sytern with evenly spaced, nondegen erate energy levels. A filled dot rep- resents an occupied single-particle state, while a hollow dot represents an unoccupied single-particle state . {d) Calculate the entropy of this system for each value of q from 0to6, and draw a graph of entropy vs. energy. Make a rough estimate of the slope of this graph near q=6, to obtain another estimate of the temperature of this system at that point. Check that it is in rough agreement with your answer to part ( c).

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.

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

The heat capacity of liquid H4ebelow 0.6Kis proportional to T3, with the measured valueCV/Nk=(T/4.67K)3. This behavior suggests that the dominant excitations at low temperature are long-wavelength photons. The only important difference between photons in a liquid and photons in a solid is that a liquid cannot transmit transversely polarized waves-sound waves must be longitudinal. The speed of sound in liquid He4is 238m/s, and the density is 0.145g/cm3. From these numbers, calculate the photon contribution to the heat capacity ofHe4in the low-temperature limit, and compare to the measured value.

Consider a gas of nidentical spin-0 bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are ε=nhf, where nis any nonnegative integer and fis the classical oscillation frequency. The degeneracy of level nis(n+1)(n+2)/2.

(a) Find a formula for the density of states, g(ε), for an atom confined by this potential. (You may assume n>>1.)

(b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency f.

(c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order kT) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and πand so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.
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