Consider a free Fermi gas in two dimensions, confined to a square area A=L2

(a) Find the Fermi energy (in terms of Nand A), and show that the average energy of the particles is F2.

(b) Derive a formula for the density of states. You should find that it is a constant, independent of .

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when role="math" localid="1650186338941" kTFand when Tis much higher.

(d) Because gis a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for μas a function of N. Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, kTF, the chemical potential of this system is the same as that of an ordinary ideal gas.

Short Answer

Expert verified

(a) Fermi energy is, 12NFand the average energy relation is given below.

(b) Density of states is, g=NF

(c) As the density state is constant, so the behavior of chemical properties is the same as the Fermi-Dirac distribution i.e. it will decrease continuously.

(d) Value of μis, μ=kTlneFkT-1

(e) Since, μ=-kTlnAN×2lQ2. Therefore, we can say that the chemical potential of this system is the same as that of an ordinary ideal gas.

Step by step solution

01

Part(a) Step 1: Given information

We have been given that,

A=L2

02

Part (a) Step 2: Simplify

In two dimensional case,

N=20nmax0π/2ndndθ=π2nmax2

We know that,

F=h2nmax28mL2=h28mA×2NπF=h24πmAN

Now, we can find the total energy U,

U=20nmax0π2ndndθ=π2nmax2U=0Fπ8mAh28mAh212dU=0F4πmAh2dU=0FNFdU=12NF

So, we can easily get average energy of the particles is UN=F2

03

Part (b) Step 1: Given information

We need to find out the density of states.

04

Part (b) Step 2: Simplify

From part (a),

Density of states is,

g=NF

That is independent of

05

Part (c) Step 1: Given information

We need to find out behavior of chemical properties

06

Part (c) Step 2: Simplify

Since the density of state is constant, the behavior of the chemical potential is the same as that of the Fermi-Dirac distribution. As we know that Fermi-Dirac distribution has some symmetrical property but when the energy difference from μis higher than the chemical potential, it gets broken because the occupancy is zero when <0. So, the electrons with energy less than the Fermi energy are excited to a higher level than the Fermi energy with broken symmetry, which means that the number of electron loss with energy <Fis lower than the number of electron gain with energy >F.

To be sensible with the difference in no. of electrons, chemical potential should move i.e. Fermi-Dirac distribution has to move totally. As the number of electrons with energy >Fis larger than <F, so Fermi-Dirac distribution has to move on the left side with fixed F, which means that the chemical potential decreases. When the temperature slightly grows up, then the difference in the number of electrons also grows up. So, the chemical potential will decrease continuously and at some point, it will have a negative value.

07

Part (d) Step 1: Given information

We need to find out value ofμ

08

Part (d) Step 2: Simplify

Since g()is a constant, we can analytically integrate the equation,

N=0g1e-μkT+1d=NF0e--μkT1+e--μkTd=NF-kTln1+e--μkT=0==NFkTln1+eμkT

From this,

μ=kTlnFkT-1=FwherekTFkTlnFkTwherekTF

The second case is approximately and it will be negative value.

09

Part(e) Step 1: Given information

We need to find out that the chemical properties of this system is the same as that of an ordinary ideal gas

10

Part (e) Step 2: Simplify 

From the answer of part (d),

μ=kTlnFkT=kTlnAN×4πmkTh2=-kTlnAN×2lQ2

It seems to be the same form with the equation of chemical potential of an ordinary ideal gas.

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Most popular questions from this chapter

Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, a gas of approximately 2×1010atoms was trapped and cooled until its peak density was1.8×1014atoms/cm3. Calculate the condensation temperature for this system, and compare to the measured value of50μK.

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.)

The argument given above for why CvTdoes not depend on the details of the energy levels available to the fermions, so it should also apply to the model considered in Problem 7.16: a gas of fermions trapped in such a way that the energy levels are evenly spaced and non-degenerate.

(a) Show that, in this model, the number of possible system states for a given value of q is equal to the number of distinct ways of writing q as a sum of positive integers. (For example, there are three system states for q = 3, corresponding to the sums 3, 2 + 1, and 1 + 1 + 1. Note that 2 + 1 and 1 + 2 are not counted separately.) This combinatorial function is called the number of unrestricted partitions of q, denoted p(q). For example, p(3) = 3.

(b) By enumerating the partitions explicitly, compute p(7) and p(8).

(c) Make a table of p(q) for values of q up to 100, by either looking up the values in a mathematical reference book, or using a software package that can compute them, or writing your own program to compute them. From this table, compute the entropy, temperature, and heat capacity of this system, using the same methods as in Section 3.3. Plot the heat capacity as a function of temperature, and note that it is approximately linear.

(d) Ramanujan and Hardy (two famous mathematicians) have shown that when q is large, the number of unrestricted partitions of q is given approximately by

p(q)eπ2q343q

Check the accuracy of this formula for q = 10 and for q = 100. Working in this approximation, calculate the entropy, temperature, and heat capacity of this system. Express the heat. capacity as a series in decreasing powers of kT/η, assuming that this ratio is large and keeping the two largest terms. Compare to the numerical results you obtained in part (c). Why is the heat capacity of this system independent of N, unlike that of the three dimensional box of fermions discussed in the text?

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.

Suppose that the concentration of infrared-absorbing gases in earth's atmosphere were to double, effectively creating a second "blanket" to warm the surface. Estimate the equilibrium surface temperature of the earth that would result from this catastrophe. (Hint: First show that the lower atmospheric blanket is warmer than the upper one by a factor of 21/4. The surface is warmer than the lower blanket by a smaller factor.)

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