Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles.

(a) Describe the ground state of this system, for each of these three cases.

(b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case?

(c) Repeat part (b) for two units of energy and for three units of energy.

(d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.

Short Answer

Expert verified

(a) The ground state of this system is.

(b) The are five possible system states in each case.

(c) For two-unit of energy the graph is

and for three-unit of energy is

.

(d) The way that the behavior of the bosonic system differs from that of the system of distinguishable particles is discussed below.

Step by step solution

01

Part (a) Step 1: Given Information

We have to describe the ground state of the system, for each of these three cases.

02

Part (a) Step 2: Simplify

As bosons do not follow the Pauli exclusion principle, particles in the ground state on the same level are distinguishable, but if they are fermions, each one will occupy a level starting from the lowest level, resulting in something like this:

03

Part (b) Step 1: Given Information

We have to describe the allowed states of the system, for each of the three cases and find the possible system states in each case.

04

Part (b) Step 2: Simplify

Consider a particle that has been promoted to the second lowest level, one energy unit above the ground state. There is only one way to do this for identifiable particles or bosons, which is to promote one particle from the lowest level to the second lowest level. There are five ways to promote indistinguishable particles or fermions, and the particle is promoted from the fifth level to the sixth level, as represented graphically:

05

Part (c) Step 1: Given Information

We have to repeat part (b) for two units of energy and for three units of energy.

06

Part (c) Step 2: Simplify

Consider a particle promoted to the second lowest level with two energy units above the ground state. We can promote one particle up to two energy levels, leaving four particles in the first level, or we can promote two particles to the second lowest level. There is only one way to do this for either, which is to promote one particle from the lowest level to the second lowest level. There are ten ways to do the first arrangement (where the two particles are promoted to the second and third lowest levels above the last filled level) and five ways to do the second arrangement (where the last particle is promoted to the third level above the last filled level) for indistinguishable particles or fermions, as shown graphically:

Consider a particle promoted to the second lowest level with three energy units above the ground state. We can promote one particle up to three energy levels for distinguishable particles or bosons, leaving four particles in the first level, or promote one particle to the second lowest level and one to the third lowest level, or promote three particles to the second lowest level, but there is only one way to do this for each. There are ten ways to make the first arrangement, twenty ways to do the second arrangement, and five ways to do the third arrangement for indistinguishable particles or fermions, and these arrangements are depicted graphically as:

07

Part (d) Step 1: Given Information

We have to find the behavior of the bosonic system differ from that of the system of distinguishable particles.

08

Part (d) Step 2: Simplify

The Boltzmann factor is proportional to the likelihood that a system with a temperature of T, i.e.

Pe-E/kT

The energy of any system is the same, therefore this factor is the same, but when we factor in degeneracy (from the previous section, the degeneracy of the bosons is 3 and the degeneracy of the fermions is 35), we can see that at low temperatures, the fermions are more likely to be found than the bosons. The bosons are also likely to be discovered in the ground state.

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Most popular questions from this chapter

For a system of particles at room temperature, how large must ϵ-μbe before the Fermi-Dirac, Bose-Einstein, and Boltzmann distributions agree within 1%? Is this condition ever violated for the gases in our atmosphere? Explain.

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.

The tungsten filament of an incandescent light bulb has a temperature of approximately 3000K. The emissivity of tungsten is approximately 13, and you may assume that it is independent of wavelength.

(a) If the bulb gives off a total of 100watts, what is the surface area of its filament in square millimetres?

(b) At what value of the photon energy does the peak in the bulb's spectrum occur? What is the wavelength corresponding to this photon energy?

(c) Sketch (or use a computer to plot) the spectrum of light given off by the filament. Indicate the region on the graph that corresponds to visible wavelengths, between400and700nm.

(d) Calculate the fraction of the bulb's energy that comes out as visible light. (Do the integral numerically on a calculator or computer.) Check your result qualitatively from the graph of part (c).

( e) To increase the efficiency of an incandescent bulb, would you want to raise or lower the temperature? (Some incandescent bulbs do attain slightly higher efficiency by using a different temperature.)

(f) Estimate the maximum possible efficiency (i.e., fraction of energy in the visible spectrum) of an incandescent bulb, and the corresponding filament temperature. Neglect the fact that tungsten melts at 3695K.

In Problem 7.28you found the density of states and the chemical potential for a two-dimensional Fermi gas. Calculate the heat capacity of this gas in the limit role="math" localid="1650099524353" kTεF· Also show that the heat capacity has the expected behavior when kTεF. Sketch the heat capacity as a function of temperature.

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.

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