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).

Short Answer

Expert verified

(a). The dot diagrams for all allowed systems are: q=4,q=5,andq=6are

For q=4

For q=5


For q=6

(b). The graph of probability as a function of energy level is


(c). The probability of a level being occupied is

n¯FD=1e(ϵ-9.5)/2.1+1

(d). The entropy for the system is

Sk=ln(Ω)kT=2.35η

Step by step solution

01

Part(a) Step 1: Given information

We have been given that q=4,5,6

02

Part(a) Step 2:Simplify

Using dot diagram

For q=4


For q=5

For localid="1651066047791" q=6

03

Part(b) Step 1: Given information

We have been given that n¯=0/11

04

Part(b) Step 2: Simplify

Probabilities are equal to the number of black dots in each row over the total states, therefore:10111011911811711611511411311211111111

Therefore, the energy of the level equals the spacing between levelsη.

05

part(c) Step 1: Given information

we have been given thatn¯FD=1e(ϵ-μ)/kT+1

06

part(c) Step 2: Simplify

Plotting the function and the points on the graph will be:

07

part(d) Step 1: Given information

we have been given thatSk=ln(Ω)

08

part(d) Step 2: Simplify

A plot between qandSk

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Most popular questions from this chapter

For a system obeying Boltzmann statistics, we know what μis from Chapter 6. Suppose, though, that you knew the distribution function (equation 7.31) but didn't know μ. You could still determine μ by requiring that the total number of particles, summed over all single-particle states, equal N. Carry out this calculation, to rederive the formula μ=-kTlnZ1/N. (This is normally how μ is determined in quantum statistics, although the math is usually more difficult.)

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define t=T/Tcc=μ/kTc,andx=ϵ/kTc. Express the integral that defines μ, equation 7.122, in terms of these variables. You should obtain the equation

(b) According to Figure 7.33, the correct value of cwhen T=2Tcis 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 . Repeat for values of T/Tcranging from 1.2up to 3.0, in increments of 0.2. Plot a graph of μas a function of temperature.

Change variables in equation 7.83 to λ=hc/ϵ and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and find a numerical formula for the wavelength where the spectrum peaks, in terms of hc/kT. Explain why the peak does not occur at hc/(2.82kT).

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)

The planet Venus is different from the earth in several respects. First, it is only 70% as far from the sun. Second, its thick clouds reflect 77%of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light.

(a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.

(b) Estimate the surface temperature again, taking the reflectivity of the clouds into account.

(c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly 70times that of earth's atmosphere. You can therefore model the atmosphere of Venus as 70successive "blankets" of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of 21/4. The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)

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