### Chapter #8 Solutions - An Introduction to Thermal Physics - Daniel V. Schroeder - 1st Edition

1. For each of the diagrams shown in equation, write down the corresponding formula in terms of f-functions, and explain why the symmetry factor gives the correct overall coefficient.Equation:... Get solution

2. Draw all the diagrams, connected or disconnected, representing terms in the configuration integral with four factors of fij. You should find 11 diagrams in total, of which five are connected. Get solution

3. Keeping only the first two diagrams in equation , and approximating N ≈ N − l ≈ N − 2 ≈ ∙ ∙ ∙, expand the exponential in a power series through the third power. Multiply each term out, and show that all the numerical coefficients give precisely the correct symmetry factors for the disconnected diagrams.Equation:... Get solution

4. Draw all the connected diagrams containing four dots. There are six diagrams in total; be careful to avoid drawing two diagrams that look superficially different but are actually the same. Which of the diagrams would remain connected if any single dot were removed? Get solution

5. By changing variables as in the text, express the diagram in equation 8.18 in terms of the same integral as in equation 8.31. Do the same for the last two diagrams in the first line of equation. Which diagrams cannot be written in terms of this basic integral?Equation:... Get solution

6. You can estimate the size of any diagram by realizing that f(r) is of order 1 out to a distance of about the diameter of a molecule, and f ≈ 0 beyond that. Hence, a three-dimensional integral of a product of f’s will generally give a result that is of the order of the volume of a molecule. Estimate the sizes of all the diagrams shown explicitly in equation, and explain why it was necessary to rewrite the series in exponential form. Equation:... Get solution

7. Show that, if you don’t make too many approximations, the exponential series in equation includes the three-dot diagram in equation 8.18. There will be some leftover terms; show that these vanish in the thermodynamic limit.Equation:... Get solution

8. Show that the nth virial coefficient depends on the diagrams in equation that have n dots. Write the third virial coefficient, C(T), in terms of an integral of f-functions. Why it would be difficult to carry out this integral?Equation:... Get solution

9. Show that the Lennard-Jones potential reaches its minimum value at r = r0, and that its value at this minimum is –u0 At what value of r does the potential equal zero? Get solution

10. Use a computer to calculate and plot the second virial coefficient for a gas of molecules interacting via the Lennard-Jones potential, for values of κT/u0 ranging from 1 to 7. On the same graph, plot the data for nitrogen given in Problem, choosing the parameters r0 and u0 so as to obtain a good fit.Problem:Even at low density, real gases don’t quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial expansion,...where the functions B(T), C(T), and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations it’s sufficient to omit the third term and concentrate on the second, whose coefficient B(T) is called the second virial coefficient (the first coefficient being 1). Here are some measured values of the second virial coefficient for nitrogen (N2): T (K)B (cm3/mol)100−160200− 35300− 4.24009.050016.960021.3(a) For each temperature in the table, compute the second term in the virial equation, B(T)/(V/n), for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.(b) Think about the forces between molecules, and explain why we might expect B(T) to be negative at low temperatures but positive at high temperatures.(c) Any proposed relation between P, V, and T , like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,...where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients (B and C) for a gas obeying the van der Waals equation, in terms of a and b. (Hint: The binomial expansion says that ... provided that |px| ≪ 1. Apply this approximation to the quantity [1 − (nb/V)]−1.)(d) Plot a graph of the van der Waals prediction for B(T), choosing a and b so as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.) Get solution

11. Consider a gas of “hard spheres,” which do not interact at all unless their separation distance is less than r0, in which case then interaction energy is infinite. Sketch the Mayer f-function for this gas, and compute the second virial coefficient. Discuss the result briefly. Get solution

12. Consider a gas of molecules whose interaction energy u(r) is infinite for r r0 and negative for r > r0, with a minimum value of –u0. Suppose further that κT ≫ u0, so you can approximate the Boltzmann factor for r > r0 using ex ≈ 1 + x. Show that under these conditions the second virial coefficient has the form B(T) = b − (a/κT), the same as what you found for a van der Waals gas in Problem. Write the van der Waals constants a and b in terms of r0 and u(r), and discuss the results briefly. Problem:Even at low density, real gases don’t quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial expansion,...where the functions B(T), C(T), and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations it’s sufficient to omit the third term and concentrate on the second, whose coefficient B(T) is called the second virial coefficient (the first coefficient being 1). Here are some measured values of the second virial coefficient for nitrogen (N2): T (K)B (cm3/mol)100−160200− 35300− 4.24009.050016.960021.3(a) For each temperature in the table, compute the second term in the virial equation, B(T)/(V/n), for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.(b) Think about the forces between molecules, and explain why we might expect B(T) to be negative at low temperatures but positive at high temperatures.(c) Any proposed relation between P, V, and T , like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,...where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients (B and C) for a gas obeying the van der Waals equation, in terms of a and b. (Hint: The binomial expansion says that ... provided that |px| ≪ 1. Apply this approximation to the quantity [1 − (nb/V)]−1.)(d) Plot a graph of the van der Waals prediction for B(T), choosing a and b so as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.) Get solution

13. Use the cluster expansion to write the total energy of a monatomic nonideal gas in terms of a sum of diagrams. Keeping only the first diagram, show that the energy is approximately...Use a computer to evaluate this integral numerically, as a function of T, for the Lennard-Jones potential. Plot the temperature-dependent part of the correction term, and explain the shape of the graph physically. Discuss the correction to the heat capacity at constant volume, and compute this correction numerically for argon at room temperature and atmospheric pressure. Get solution

14. In this section I’ve formulated the cluster expansion for a gas with a fixed number of particles, using the “canonical” formalism of Chapter 6. A somewhat cleaner approach, however, is to use the “grand canonical” formalism introduced in Section 7.1, in which we allow the system to exchange particles with a much larger reservoir.(a) Write down a formula for the grand partition function (Z) of a weakly interacting gas in thermal and diffusive equilibrium with a reservoir at fixed T and µ. Express Z as a sum over all possible particle numbers N, with each term involving the ordinary partition function Z(N).(b) Use equations to express Z(N) as a sum of diagrams, then carry out the sum over N, diagram by diagram. Express the result as a sum of similar diagrams, but with a new rule 1 that associates the expression ... with each dot, where λ = eβµ Now, with the awkward factors of N(N − 1) ∙ ∙ ∙ taken care of, you should find that the sum of all diagrams organizes itself into exponential form, resulting in the formula...Note that the exponent contains all connected diagrams, including those that can be disconnected by removal of a single line.(c) Using the properties of the grand partition function (see Problem), find diagrammatic expressions for the average number of particles and the pressure of this gas.(d) Keeping only the first diagram in each sum, express ... arid P(µ) in terms of an integral of the Mayer f-function. Eliminate µ to obtain the same result for the pressure (and the second virial coefficient) as derived in the text.(e) Repeat part (d) keeping the three-dot diagrams as well, to obtain an expression for the third virial coefficient in terms of an integral of f-functions. You should find that the Λ-shaped diagram cancels, leaving only the triangle diagram to contribute to C(T).Problem: In Section 6.5 I derived the useful relation F = − kT ln Z between the Helmholtz free energy and the ordinary partition function. Use an analogous argument to prove thatɸ = − κΤ ln Zwhere Z is the grand partition function and ɸ is the grand free energy introduced in Problem 1.Problem 1:By subtracting μN from U, H, F, or G, one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),...(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives of Φ with respect to T, V, and μ.(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φ tends to decrease.(c) Prove that Φ = – PV.(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in below Problem 2, the prediction for such a small system is only a probabilistic one.)Problem 2:The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if wc take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = κ ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.) Get solution

15. For a two-dimensional Ising model on a square lattice, each dipole (except on the edges) has four “neighbors” —above, below, left, and right. (Diagonal neighbors are normally not included.) What is the total energy (in terms of ϵ) for the particular state of the 4 × 4 square lattice shown in Figure?Figure:. One particular state of an Ising model on a 4 × 4 square lattice.... Get solution

16. Consider an Ising model of 100 elementary dipoles. Suppose you wish to calculate the partition function for this system, using a computer that can compute one billion terms of the partition function per second. How long must you wait for the answer? Get solution

17. Consider an Ising model of just two elementary dipoles, whose mutual interaction energy is ±ε. Enumerate the states of this system and write down their Boltzmann factors. Calculate the partition function. Find the probabilities of finding the dipoles parallel and antiparallel, and plot these probabilities as a function of κT/ε. Also calculate and plot the average energy of the system. At what temperatures are you more likely to find both dipoles pointing up than to find one up and one down? Get solution

18. Starting from the partition function, calculate the average energy of the one-dimensional Ising model, to verify equation. Sketch the average energy as a function of temperature.... Get solution

19. The critical temperature of iron is 1043 K. Use this value to make a rough estimate of the dipole-dipole interaction energy ϵ, in electron-volts. Get solution

20. Use a computer to plot ... as a function of κT/ϵ, as predicted by mean field theory, for a two-dimensional Ising model (with a square lattice). Get solution

21. At T = 0, equation says that ... Work out the first temperature-dependent correction to this value, in the limit βεn ≫ 1. Compare to the low-temperature behavior of a real ferromagnet, treated in Problem.Equation:...Problem:A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At T = 0 the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are N atoms, the total magnetization is typically ~2µBN, where µB is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h/2π, and therefore reduces the magnetization by ~2µB. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin wave is proportional to the square of 1/λ (in the limit of long wavelengths). Therefore, since ϵ = hf and p = h/λ for any “particle,” the energy of a magnon is proportional to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write ϵ = p2/2m*, where m* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24 × 10−29 kg, about 14 times the mass of an electron. Another difference between magnons and phonons is that each magnon (or spin wave mode) has only one possible polarization.(a) Show that at low temperatures, the number of magnons per unit in a three-dimensional ferromagnet is given by...Evaluate the integral numerically.(b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(0) −M(T))/M(0). Write your answer in the form (T/T0)3/2, and estimate the constant T0 for iron.(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cy/Nk = (T/Ti)3/2, where TL differs from To only by a numerical constant. Estimate T1 for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470 K.)(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (three- dimensional) direction, so spin waves are still passible. Show that, the integral for the total number of magnons diverges in this case. (This re­sult is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section 8.2 we will consider a different two-dimensional model in which magnetization docs occur.)Figure: In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy, because the difference in direction between neighboring dipoles is very small.... Get solution

22. Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of −µBB if it points up and +µBB if it points down (where fm is the dipole’s magnetic moment). Analyze this system using the mean field approximation to find the analogue of equation. Study the solutions of the equation graphically, and discuss the magnetization of this system as a function of both the external field strength and the temperature. Sketch the region in the T-B plane for which the equation has three solutions. Equation:... Get solution

23. The Ising model can be used to simulate other systems besides ferromagnets; examples include antiferromagnets, binary alloys, and even fluids. The Ising model of a fluid is called a lattice gas. We imagine that space is divided into a lattice of sites, each of which can be either occupied by a gas molecule or unoccupied. The system has no kinetic energy, and the only potential energy comes from interactions of molecules on adjacent sites. Specifically, there is a contribution of −u0 to the energy for each pair of neighboring sites that arc both occupied. (a) Write down a formula for the grand partition function for this system, as a function of u 0, T, and µ. (b) Rearrange your formula to show that it is identical, up to a multiplicative factor that does not depend on the state of the system, to the ordinary partition function for an Ising ferromagnet in the presence of an external magnetic field B, provided that you make the replacements u 0 → 4ϵ and µ → 2µBB −8ϵ. (Note that µ is the chemical potential of the gas while µB is the magnetic moment of a dipole in the magnet.)(c) Discuss the implications. Which states of the magnet correspond to low-density states of the lattice gas? Which states of the magnet correspond to high-density states in which the gas has condensed into a liquid? What shape does this model predict for the liquid-gas phase boundary in the P-T plane? Get solution

24. In this problem you will use the mean field approximation to analyze the behavior of the Ising model near the critical point.(a) Prove that, when x ≪ 1, ...(b) Use the result of part (a) to find an expression for the magnetization of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find M α (Tc − T)β, where β (not to be confused with 1/κT) is a critical exponent, analogous to the β defined for a fluid in Problem. Onsager’s exact solution shows that β = 1/8 in two dimensions, while experiments and more sophisticated approximations show that β ≈ 1/3 in three dimensions. The mean field approximation, however, predicts a larger value.(c) The magnetic susceptibility x is defined as x = (∂M/∂B)T. The behavior of this quantity near the critical point is conventionally written as χ α (T − Tc)-γ, where γ is another critical exponent. Find the value of γ in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Tc. (The exact value of γ in two dimensions turns out to be 7/4, while in three dimensions γ ≈ 1.24.)Probem:In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.(a) Expand the van der Waals equation in a Taylor series in (V − Vc) keeping terms through order (V − Vc)3. Argue that, for T sufficiently close to Tc, the term quadratic in (V − Vc) becomes negligible compared to the others and may be dropped.(b) The resulting expression for P(V) is antisymmetric about the point V = Vc Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary, dP/dT, at the critical point.(c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find (Vg −Vl) α (Tc −T)β, where β is known as a critical exponent. Experiments show that β has a universal value of about 1/3, but the van der Waals model predicts a larger value.(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function.(e) The shape of the T − Tc isotherm defines another critical exponent, called δ: (P −Pc) α (V − Vc)δ. Calculate δ in the van der Waals model. (Experimental values of δ are typically around 4 or 5.)(f) A third critical exponent describes the temperature dependence of the isothermal compressibility,...This quantity diverges at the critical point, in proportion to a power of (T−Tc) that in principle could differ depending on Whether one approaches the critical point from above or below. Therefore the clinical exponents γ and γ’ are defined by the relations....Calculate k on both sides of the critical point in the van der Waals model, and show that γ = γ’ in this model. Get solution

25. In Problem you manually computed the energy of a particular state of a 4 × 4 square lattice. Repeat that computation, but this time apply periodic boundary conditions.Problem: For a two-dimensional Ising model on a square lattice, each dipole (except on the edges) has four “neighbors” —above, below, left, and right. (Diagonal neighbors are normally not included.) What is the total energy (in terms of ϵ) for the particular state of the 4 × 4 square lattice shown in Figure?Figure:. One particular state of an Ising model on a 4 × 4 square lattice.... Get solution

26. Implement the ising program on your favorite computer, using your favorite programming language. Run it for various lattice sizes and temperatures and observe the results. In particular:(a) Run the program with a 20 × 20 lattice at T = 10, 5, 4, 3, and 2.5, for at least 100 iterations per dipole per run. At each temperature make a rough estimate of the size of the largest clusters.(b) Repeat part (a) for a 40 × 40 lattice. Are the cluster sizes any different? Explain.(c) Run the program with a 20 × 20 lattice at T = 2, 1.5, and 1. Estimate the average magnetization (as a percentage of total saturation) at each of these temperatures. Disregard runs in which the system gets stuck in a metastable state with two domains.(d) Run the program with a 10 × 10 lattice at T = 2.5. Watch it run for 100,000 iterations or so. Describe and explain the behavior.(e) Use successively larger lattices to estimate the typical cluster size at temperatures from 2.5 down to 2.27 (the critical temperature). The closer you arc to the critical temperature, the larger a lattice you’ll need and the longer the program will have to run. Quit when you realize that there are better ways to spend your time. Is it plausible that the cluster size goes to infinity as the temperature approaches the critical temperature? Get solution

27. Modify the ising program to compute the average energy of the system over all iterations. To do this, first add code to the initialize subroutine to compute the initial energy of the lattice; then, whenever a dipole is flipped, change the energy variable by the appropriate amount. When computing the average energy, be sure to average over all iterations, not just those iterations in which a dipole is actually flipped (why?). Run the program for a 5 × 5 lattice for T values from 4 down to 1 in reasonably small intervals, then plot the average energy as a function of T. Also plot the heat capacity. Use at least 1000 iterations per dipole for each run, preferably more. If your computer is fast enough, repeat for a 10 × 10 lattice and for a 20 × 20 lattice. Discuss the results. (Hint: Rather than starting over at each temperature with a random initial state, you can save time by starting with the final state generated at the previous, nearby temperature. For the larger lattices you may wish to save time by considering only a smaller temperature interval, perhaps from 3 down to 1.5.) Get solution

28. Modify the ising program to compute the total magnetization (that is, the sum of all the s values) for each iteration, and to tally how often each possible magnetization value occurs during a run, plotting the results as a histogram. Run the program for a 5 × 5 lattice at a variety of temperatures, and discuss the results. Sketch a graph of the most likely magnetization value as a function of temperature. If your computer is fast enough, repeat for a 10 × 10 lattice. Get solution

29. To quantify the clustering of alignments within an Ising magnet, we define a quantity called the correlation function, c(r). Take any two dipoles i and j, separated by a distance r, and compute the product of their states: sisj. This product is 1 if the dipoles are parallel and −1 if the dipoles are antiparallel. Now average this quantity over all pairs that are separated by a fixed distance r, to obtain a measure of the tendency of dipoles to be “correlated” over this distance. Finally, to remove the effect of any over all magnetization of the system, subtract off the square of the average s. Written as an equation, then, the correlation function is ...where it is understood that the first term averages over all pairs at the fixed distance r. Technically, the averages should also be taken over all possible states of the system, but don’t do this yet.(a) Add a routine to the ising program to compute the correlation function for the current state of the lattice, averaging over all pairs separated either vertically or horizontally (but not diagonally) by r units of distance, where r varies from 1 to half the lattice size. Have the program execute this routine periodically and plot the results as a bar graph.(b) Run this program at a variety of temperatures, above, below, and near the critical point. Use a lattice size of at least 20, preferably larger (especially near the critical point). Describe the behavior of the correlation function at each temperature.(c) Now add code to compute the average correlation function over the duration of a run. (However, it’s best to let the system “equilibrate” to a typical state before you begin accumulating averages.) The correlation length is defined as the distance over which the correlation function decreases by a factor of e. Estimate the correlation length at each temperature, and plot a graph of the correlation length vs. T. Get solution

30. Modifiy the ising program to simulate a one-dimensional Ising model.(a) For a lattice size of 100, observe the sequence of states generated at various temperatures and discuss the results. According to the exact solution (for an infinite lattice), we expect this system to magnetize only as the temperature goes to zero; is the behavior of your program consistent with this prediction ? Row does the typical cluster size depend on temperature?(b) Modify your program to compute the average energy as in Problem 1. Plot the energy and heat capacity vs. temperature and compare to the exact result for an infinite lattice.(c) Modify your program to compute the magnetization as in Problem 2. Determine the most likely magnetization for various temperatures, and discuss your results.Problem 1:Modify the ising program to compute the average energy of the system over all iterations. To do this, first add code to the initialize subroutine to compute the initial energy of the lattice; then, whenever a dipole is flipped, change the energy variable by the appropriate amount. When computing the average energy, be sure to average over all iterations, not just those iterations in which a dipole is actually flipped (why?). Run the program for a 5 × 5 lattice for T values from 4 down to 1 in reasonably small intervals, then plot the average energy as a function of T. Also plot the heat capacity. Use at least 1000 iterations per dipole for each run, preferably more. If your computer is fast enough, repeat for a 10 × 10 lattice and for a 20 × 20 lattice. Discuss the results. (Hint: Rather than starting over at each temperature with a random initial state, you can save time by starting with the final state generated at the previous, nearby temperature. For the larger lattices you may wish to save time by considering only a smaller temperature interval, perhaps from 3 down to 1.5.)Problem 2:Modify the ising program to compute the total magnetization (that is, the sum of all the s values) for each iteration, and to tally how often each possible magnetization value occurs during a run, plotting the results as a histogram. Run the program for a 5 × 5 lattice at a variety of temperatures, and discuss the results. Sketch a graph of the most likely magnetization value as a function of temperature. If your computer is fast enough, repeat for a 10 × 10 lattice. Get solution

31. Modify the ising program to simulate a three-dimensional Ising model with a simple cubic lattice. In whatever way you can, try to show that this system has a critical point at around T = 4.5. Get solution

32. Imagine taking a two-dimensional Ising lattice and dividing the sites into 3 × 3 “blocks”, as shown in Figure 8.11. In a block spin transformation, we replace the nine dipoles in each block with a single dipole, whose state is determined by “majority rule”: If more than half of the original dipoles point up, then the new dipole points up, while if more than half of the original dipoles point down, then the new dipole points down. By applying this transformation to the entire lattice, we reduce it to a new lattice whose width is 1/3 the original width. This transformation is one version of a renormalization group transformation, a powerful technique for studying the behavior of systems near their critical points. Get solution

### Chapter #7 Solutions - An Introduction to Thermal Physics - Daniel V. Schroeder - 1st Edition

1. Near the cells where oxygen is used, its chemical potential is significantly lower than near the lungs. Even though there is no gaseous oxygen near these cells, it is customary to express the abundance of oxygen in terms of the partial pressure of gaseous oxygen that would be in equilibrium with the blood. Using the independent-site model just presented, with only oxygen present, calculate and plot the fraction of occupied heme sites as a function of the partial pressure of oxygen. This curve is called the Langmuir adsorption isotherm (“isotherm” because its for a fixed temperature). Experiments show that adsorption by myoglobin follows the shape of this curve quite accurately. Get solution

2. In a real hemoglobin molecule, The tendency of oxygen to bind to a heme site increases as the other three heme sites become occupied. To model this effect in a simple way, imagine that a hemoglobin molecule has just two sites, either or both of which can be occupied. This system has four possible states (with only oxygen present). Take the energy of the unoccupied state to be zero, the energies of the two singly occupied states to be −0.55 eV, and the energy of the doubly occupied state to be −1.3 eV (so the change in energy upon binding the second oxygen is −0.75 eV). As in the previous problem, calculate and plot the fraction of occupied sites as a function of The effective partial pressure of oxygen. Compare to the graph from the previous problem (for independent sites) . Can you think of why this behavior is preferable for the function of hemoglobin? Get solution

3. Consider a system consisting of a single hydrogen atom/ ion, which has two possible states: unoccupied (i.e., no electron present) and occupied (i.e., one electron present, in the ground state). Calculate the ratio of the probabilities of these two states, to obtain the Saha equation , already derived in Section 5.6. Treat the electrons as a monatomic ideal gas, for the purpose of determining µ. Neglect the fact that an electron has two independent spin states. Get solution

4. Repeat the previous problem, taking into account the two independent spin states of the electron. Now the system has two “occupied” states, one with the electron in each spin configuration. However, the chemical potential of the electron gas is also slightly different. Show that the ratio of probabilities is the same as before: The spin degeneracy cancels out of the Saha equation. Get solution

5. 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 neighbouring 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 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, Nc/V.(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 Nc in terms of the number of donor atoms (Nd), eliminating μ. Solve for Nc using the quadratic formula. (Hint: It’s helpful to introduce some abbreviations for dimensionless quantities. Try x = Nc/Nd, t = kT/I, and so on.)(d) For phosphorus in silicon, the ionization energy is 0.044 eV. Suppose that there are 1017 P atoms per cubic centimeter. Using these numbers, calculate and plot the fraction of ionized donors as a function of temperature. Discuss the results. Get solution

6. Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is...where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is...Use these results to show that the standard deviation of N is...in analogy with Problem Finally, apply this formula to an ideal gas, to obtain a Simple expression for σN in terms of .... Discuss your result briefly.Problem:Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of E2 is...Then use this result and the results of the previous two problems to derive a formula for σE in terms of the heat capacity, ... You should find... Get solution

7. In Section 6.5, I derived the useful relation F = − kT ln Z between the Helmholtz free energy and the ordinary partition function. Use an analogous argument to prove thatФ = − kTlnZwhere z is the grand partition function and Ф is The grand free energy introduced in Problem 1.Problem 1:By subtracting μN from U, H, F, or G, one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),Φ ≡ U − TS − μN.(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives of Φ with respect to T, V, and μ.(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φ tends to decrease.(c) Prove that Φ = – PV.(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in below Problem 2, the prediction for such a small system is only a probabilistic one.)Problem 2:The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.) Get solution

8. Suppose you have a “box” in which each particle may occupy any of 10 single-particle states. For simplicity, assume that each of these states has energy zero.(a) What is the partition function of this system if the box contains only one particle? (b) What is the partition function of this system if the box contains two distinguishable particles? (c) What is the partition function if the box contains two identical bosons? (d) What is the partition function if the box contains two identical fermions? (e) What would be the partition function of this system according to equation? (f) What is the probability of finding both particles in the same single-particle state, for the three cases of distinguishable particles, identical bosons, and identical fermions?Equation:... Get solution

9. Compute the quantum volume for an N2 molecule at room temperature, and argue that a gas of such molecules at atmospheric pressure can be treated using Boltzmann statistics. At about what temperature would quantum statistics become relevant for this system (keeping the density constant and pretending sthat the gas does not liquefy)? Get solution

10. 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. Get solution

11. For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is(a) 1 eV less than µ(b) 0.01 e V less than µ(c) equal to µ(d) 0.01 eV greater than µ(e) 1 eV greater than µ Get solution

12. Consider two single-particle states, A and B, in a system of fermions, where ϵA = µ − x and ϵB = µ + x that is, level A lies below B same amount that level B lies above µ. Prove that the probability of level B being occupied is the same as the probability of level A being unoccupied. In other words, the Fermi-Dirac distribution is “symmetrical” about the point where ϵ = µ. Get solution

13. For a system of bosons at room temperature, compute the average occupancy of a single-particle state and the probability of the state containing 0, 1, 2, or 3 bosons, if the energy of the state is(a) 0.001 eV greater than µ(b) 0.01 e V greater than µ(c) 0.1 eV greater than µ(d) 1 eV greater than µ Get solution

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

15. For a system obeying Boltzmann statistics, we know what µ is from Chapter 6. Suppose, though, that you knew the distribution function (equation) 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 = − kT ln( z1/N). (This is normally how µ is determined in quantum statistics, although the math is usually more difficult.)Equation:... Get solution

16. Consider an isolated system of N identical fermions, inside a container where the allowed energy levels are nondegcnerate 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 η be the spacing between energy levels, and let q be the number of energy units (each of size η) in excess of the ground-state energy. Assume that q 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, and q = 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 q is 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 µ and T that you would have to plug into the Fermi-Dirac distribution to best fit the graph you drew in part (b). (d) Calculate the entropy of this system for each value of q from 0 to 6, 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).Figure: A representation of the system states of a fermionic system with evenly spaeed, nondegen-erate energy levels. A filled dot represents an occupied single-particle state, while a hollow dot represents an unoccupied single-particle state.... Get solution

17. In analogy with the previous problem, consider a system of identical spin-0 bosons trapped in a region where the energy levels are evenly spaced. Assume that N is a large number, and again let q be the number of energy units.(a) Draw diagrams representing all allowed system states from q = 0 up to q = 6. Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level. (b) Compute the occupancy of each energy level, for q = 6. Draw a graph of the occupancy as a function of the energy of the level. (c) Estimate the values of µ and T that you would have to plug into the Bose-Einstein distribution to best fit the graph of part (b). (d) As in part (d) of the previous problem, draw a graph of entropy vs Energy and estimate the temperature at q = 6 from this graph. Get solution

18. Imagine that there exists a third type of particle which can share a single-particle state with one other particle of the same type but no more. Thus the number of these particles in any state can be 0, 1, or 2. Derive the distribution function for the average occupancy of a state by particles of this type, and plot the occupancy as a function of the state ’s energy, for several different temperatures. Get solution

19. Each atom in a chunk of copper contributes one conduction electron. Look up the density and atomic mass of copper, and calculate the Fermi energy, the Fermi temperature, the degeneracy pressure, and the contribution of the degeneracy pressure to the bulk modulus. Is room temperature sufficiently low to treat this system as a degenerate electron gas? Get solution

20. At the center of the sun, the temperature is approximately 107 K and the concentration of electrons is approximately 1032 per cubic meter. Would it be (approximately) valid to treat these electrons as a “classical” ideal gas (using Boltzmann statistics), or as a degenerate Fermi gas (with T ≈ 0), or neither? Get solution

21. An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of 0.18 fm−3 (where 1 fm = 10−15 m). Because nucleons come in two different types (protons and neutrons), each with spin 1/2, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result. Get solution

22. Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic (ϵ ≫ mc2), so that their energies are ϵ = pc (where p is the magnitude of the momentum vector). (a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by µ = hc(3N/8πV)1/3. (b) Find a formula for the total energy of this system in terms of N and µ. Get solution

23. A white dwarf star (see Figure) is essentially a degenerate electron gas, with a bunch of nuclei mixed in to balance the charge and to provide the gravitational attraction that holds the star together. In this problem you will derive a relation between the mass and the radius of a white dwarf star, modeling the star as a uniform-density sphere. White dwarf stars tend to be extremely hot by our standards; nevertheless, it is an excellent approximation in this problem to set T = 0.(a) Use dimensional analysis to argue that the gravitational potential energy of a uniform-density sphere (mass M, radius R) must equal...where (constant) is some numerical constant. Be sure to explain tire minus sign. The constant turns out to equal 3/5; you can derive it by calculating the (negative) work needed to assemble the sphere, shell by shell, from the inside out. (b) Assuming that the star contains one proton and one neutron for each electron, and that the electrons are nonrelativistic, show that the total (kinetic) energy of the degenerate electrons equals The numerical factor can be expressed exactly in terms of π and cube roots and such, but it’s not worth it....(c) The equilibrium radius of the white dwarf is that which minimizes the total energy Ugrav + Ukinetic Sketch the total energy as a function of R, and find a formula for the equilibrium radius in terms of the mass. As the mass increases, does the radius increase or decrease? Does this make sense? (d) Evaluate the equilibrium radius for M = 2 × 1030 kg, the mass of the sun. Also evaluate the density. How does the density compare to that of water? (e) Calculate the Fermi energy and the Fermi temperature, for the case con­sidered in part (d). Discuss whether the approximation T = 0 is valid. (f) Suppose instead that the electrons in the white dwarf star are highly relativistic. Using the result of the previous problem, show that the total kinetic energy of the electrons is now proportional to 1/R instead of 1/R2. Argue that there is no stable equilibrium radius for such a star. (g) The transition from the nonrelativistic regime to the ultra relativistic regime occurs approximately where the average kinetic energy of an electron is equal to its rest energy, mc2. Is the nonrelativistic approximation valid for a one-solar-mass white dwarf? Above what mass would you expect a white dwarf to become relativistic and hence unstable?Figure: The double star system Sirius A and B. Sirius A (greatly overexposed in the photo) is the brightest star in our night sky. Its companion, Sirius B, is hotter but very faint, indicating that it must be extremely small—a white dwarf. From the orbital motion of the pair we know that Sirius B has about the same mass as our sun. (UCO/Lick Observatory photo.)... Get solution

24. A star that is too heavy to stabilize as a white dwarf can collapse further to form a neutron star: a star made entirely of neutrons, supported against gravitational collapse by degenerate neutron pressure. Repeat the steps of the previous problem for a neutron star, to determine the following: the mass-radius relation; the radius, density, Fermi energy, and Fermi temperature of a one-solar-mass neutron star and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse. Get solution

25. 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 about 40% more than predicted by the free electron model used here.) Get solution

26. In this problem you will model helium-3 as a noninteracting Fermi gas. Although 3He liquefies at low temperatures, the liquid has an unusually low density and behaves in many ways like a gas because the forces between the atoms are so weak. Helium-3 atoms are spin-1/2 fermions, because of the unpaired neutron in the nucleus.(a) Pretending that liquid 3He is a noninteracting Fermi gas, calculate the Fermi energy and the Fermi temperature. The molar volume (at low pressures) is 37 cm3 Get solution

27. The argument given above for why ...does not depend on i.e 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 nondegenerate. (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 ... 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? Get solution

28. Consider a free Fermi gas in two dimensions, confined to a square area A = L2.(a) Find the Fermi energy (in terms of N and A), and show that the average energy of the particles is ϵF/2(b) Derive a formula for the density of states. You should find that it is a constant, independent of ϵ.(c) Explain how the chemical potential of this system should behave as a function of temperature, both when. kT ≪ ϵF and when T is much higher. (d) Because g(ϵ) is a constant for this system, it is possible to carry out the integral for the number of particles analytically. Do so, and solve for µ as a function of N. Show that the resulting formula has the expected qualitative behavior. (e) Show that in the high-temperature limit, kT ≫ ϵF, the chemical potential of this system is the same as that of an ordinary ideal gas.Equation:... Get solution

29. Carry out the Sommerfield expansion for the energy integral 1 to obtain equation 2. Then plug in the expansion for µ to obtain the final answer, equation 3.Equation:1...Equation:2...Equation:3... Get solution

30. The Sommerfield expansion is an expansion in powers of KT/ϵF, which is assumed to be small. In this section I kept all terms through order (KT/ϵF)2, omitting higher-older terms. Show at each relevant step that the term proportional to T3 is zero, so that the next nonvanishing terms in the expansions for µ and U are proportional to T4. (If you enjoy such things, you might try evaluating the terms, possibly with the aid of a computer algebra program.) Get solution

31. In Problem you 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 kT ≪ ϵF. Also show that the heat capacity has the expected behavior when kT ≫ ϵF. Sketch the heat capacity as a function of temperature.Problem:Consider a free Fermi gas in two dimensions, confined to a square area A = L2.(a) Find the Fermi energy (in terms of N and A), and show that the average energy of the particles is ϵF/2(b) Derive a formula for the density of states. You should find that it is a constant, independent of ϵ.(c) Explain how the chemical potential of this system should behave as a function of temperature, both when. kT ≪ ϵF and when T is much higher. (d) Because g(ϵ) is a constant for this system, it is possible to carry out the integral for the number of particles analytically. Do so, and solve for µ as a function of N. Show that the resulting formula has the expected qualitative behavior. (e) Show that in the high-temperature limit, kT ≫ ϵF, the chemical potential of this system is the same as that of an ordinary ideal gas. Get solution

32. Although the integrals and for N and U cannot be carried out analytically for all T, it’s not difficult to evaluate them numerically using a computer. This calculation has little relevance for electrons in metals (for which the limit kT ≪ ϵF is always sufficient), but it is needed for liquid 3He and for astrophysical systems like the electrons at the center of the sun.(a) As a warm-up exercise, evaluate the N integral for the case KT = ϵF and µ = 0, and check that your answer is consistent with the graph shown above. (Hint: As always when solving a problem on a computer, it’s best to first put everything in terms of dimensionless variables. So let t = KT/ϵF, c = µ/ϵF, and x = ϵ/ϵF. Rewrite everything in terms of these variables, and then put it on the computer.) (b). The next step is to vary µ, holding T fixed, until the integral works out to the desired value, N. Do this for values of KT/ϵF ranging from 0.1 up to 2, and plot the results to reproduce Figure. (It’s probably not a good idea to try to use numerical methods when KT/ϵF is much smaller than 0.1, since you can start getting overflow errors from exponentiating large numbers. But this is the region where we’ve already solved the problem analytically.) (c) Plug your calculated values of µ into the energy integral, and evaluate that integral numerically to obtain the energy as a function of temperature for KT up to 2ϵF. Plot the results, and evaluate the slope to obtain the heat capacity. Check that the heat capacity has the expected behavior at both low and high temperatures.Equation 1:...Equation 2:... Get solution

33. When the attractive forces of the ions in a crystal are taken into account, the allowed electron energies are no longer given by the simple formula 7.36; instead, the allowed energies are grouped into bands, separated by gaps where there are no allowed energies. In a conductor the Fermi energy lies within one of the bands; in this section we have treated the electrons in this band as “free’’ particles confined to a fixed volume. In an insulator, on the other hand, the Fermi energy lies within a gap, so that at T = 0 the band below the gap is completely occupied while the band above the gap is unoccupied. Because there are no empty states close in energy to those that are occupied, the electrons are “stuck in place” and the material does not conduct electricity. A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature. Figure 7.17 shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor, and defines some terminology and notation to be used in this problem.(a) As a first approximation, let us model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas, with an appropriate zero-point: ..., where g0 is the same constant as in equation 7.51. Let us also model the density of states near the top of the valence band as a mirror image of this function. Explain why, in this approximation, the chemical potential must always lie precisely in the middle of the gap, regargless of temperature.(b) Normally the width of the gap is much greater than KT. Working in this limit, derive an expression for the number of conduction electrons per unit volume, in terms of the temperature and the width of the gap. (c) For silicon near room temperature, the gap between the valence and con­duction bands is approximately 1.11 eV. Roughly how many conduction electrons are there in a cubic centimeter of silicon at room temperature? How does this compare to the number of conduction electrons in a similar amount of copper? (d) Explain why a semiconductor conducts electricity much better at higher temperatures. Back up your explanation with some numbers. (Ordinary conductors like copper, on the other hand, conduct better at low temperatures.) (e) Very roughly, how wide would the gap between the valence and conduction bands have to be in order to consider a material an insulator rather than a semiconductor?Figure: The periodic potential of a crystal lattice results in a density-of-states function consisting of “bands” (with many states) and “gaps” (with no states). For an insulator or a semiconductor, the Fermi energy lies in the middle of a gap so that at T = 0, the “valence band” is completely full while the “conduction. band” is completely empty.... Get solution

34. In a real semiconductor; the density of states at the bottom of the conduction band will differ from the model used in the previous problem by a numerical factor, which can be small or large depending on the material. Let us therefore write for the conduction band ... g0c is a new normalization constant that differs from go by some fudge factor. Similarly, write g(ϵ) at the top of the valence band in terms of a new normalization constant gov.(a) Explain why, if g0u ≠ I g0c, the chemical potential will now vary with temperature. When will it increase, and when will it decrease? (b) Write down au expression for the number of conduction electrons, in terms of T, µ ϵc, and g0c Simplify this expression as much as possible, assuming ϵc − µ ≫ KT(c) An empty state in the valence band is called a hole. In analogy to part (b), write down an expression for the number of holes, and simplify it in the limit µ − ϵv ≫ KT. (d) Combine the results of parts (b) and (c) to find an expression for the chemical potential as a function or temperature. (e) For silicon g0c/g0 = 1.09 and g0v/g0 = 0.44* Calculate the shift in µ for silicon at room temperature. Get solution

35. The previous two problems dealt with pure semiconductors, also called intrinsic semiconductors. Useful semiconductor devices are instead made from doped semiconductors, which contain substantial numbers of impurity atoms. One example of a doped semiconductor was treated in Problem 7.5. Let us now consider that system again. (Note that in Problem we measured all energies relative to the bottom of the conduction band, ϵc. We also neglected the distinction between g0 and g0c this simplification happens to be ok for conduction electrons in silicon.)(a) Calculate and plot the chemical potential as a function of temperature, for silicon doped with 1017 phosphorus atoms per cm3 (as in Problem). Continue to assume that the conduction electrons can be treated as an ordinary ideal gas. (b) Discuss whether it is legitimate to assume for this system that the conduction electrons can be treated as an ordinary ideal gas, as opposed to a Fermi gas. Give some numerical examples. (c) Estimate the temperature at which the number of valence electrons excited to the conduction band would become comparable to the number of conduction electrons from donor impurities. Which source of conduction electrons is more important at room temperature?Problem: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 neighbouring 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 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, Nc/V. (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 Nc in terms of the number of donor atoms (Nd), eliminating µ. Solve for Nc using the quadratic formula. (Hint: It’s helpful to introduce some abbreviations for dimensionless quantities. Try x = Nc/ Nd, t = kT / I, and so on.) (d) For phosphorus in silicon, the ionization energy is 0.044 eV. Suppose that there are 1017 P atoms per cubic centimeter. Using these numbers, calculate and plot the fraction of ionized donors as a function of temperature. Discuss the results. Get solution

36. Most spin-1/2 fermions, including electrons and helium-3 atoms, have nonzero magnetic moments. A gas of such particles is therefore paramagnetic. Consider, for example, a gas of free electrons, confined inside a three-dimensional box. The z component of the magnetic moment of each electron is ±µB. In the presence of a magnetic field B pointing in the z direction, each “up” state acquires an additional energy of −µB B. while each “down” state acquires an additional energy of + µB B.(a) Explain why you would expect the magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets studied in Chapters 3 and 6, for a given number of particles at a given field strength. (b) Write down a formula for the density of states of this system in the presence of a magnetic field B, and interpret your formula graphically. (c) The magnetization of this system is µB(N↑ − N↓) Where N↑ and N↓ are the numbers of electrons with up and down magnetic moments, respectively. Find a formula for the magnetization of this system at T = 0, in terms of N µB, B and the Fermi energy. (d) Find the first temperature-dependent correction to your answer to part (c), in the limit T ≪ TF. You may assume that μB B ≪ KT; this implies that the presence of the magnetic field has negligible effect on the chemical potential µ, (To avoid confusing µB with µ, I suggest using an abbreviation such as δ for the quantity µB B.) Get solution

37. Prove that the peak of the Planck spectrum is at x = 2.82. Get solution

38. It’s not obvious from Figure how the Planck spectrum changes as a function of temperature. To examine the temperature dependence, make a quantitative plot of the function u(ϵ) for T = 3000 K and T = 6000 K (both on the same graph). Label the horizontal axis in electron-volts.Figure: The Planck spectrum, plotted in terms of the dimensionless variable x = ϵ/KT = hf/kT. The area under any portion of that graph, multiplied by 8π(kT)4/(hc)3, equals the energy density of electromagnetic radiation with in the corresponding frequency (or photon energy) range; see equation....Equation:... Get solution

39. 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).Equation:... Get solution

40. Starting from equation, derive a formula for the density of states of a photon gas (or any other gas of ultrarelativistic particles having two polarization states). Sketch this function.Equation:... Get solution

41. Consider any two internal states, s1 and s2, of an atom. Let s2 be the higher-energy state, so that E(s2) − E(s1) =ϵ for some positive constant ϵ. 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 ϵ. 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 a 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) at f = ϵ/h The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coefficient:...Finally, it is also possible for the atom to make a stimulated transition from s2 down 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:...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, N1 and N2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related byB′ = Β and ... Get solution

42. Consider the electromagnetic radiation inside a kiln, with a volume of 1 m3 and a temperature of 1500 K.(a) What is the total energy of this radiation? (b) Sketch the spectrum of the radiation as a function of photon energy. (c). What fraction of all the energy is in the visible portion of the spectrum, with with wavelengths between 400 nm and 700 nm? Get solution

43. At the surface of the sun, the temperature is approximately 5800 K.(a) How much energy is contained in the electromagnetic radiation filling a cubic meter of space at the sun’s surface? (b) Sketch the spectrum of this radiation as a function of photon energy. Mark the region of the spectrum that corresponds to visible wavelengths, between 400 nm and 700 nm. (c)What fraction of the energy is in the visible portion of the spectrum? (Hint: Do the integral numerically.) Get solution

44. 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...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). Get solution

45. Use the formula ... to show that the pressure of a photon gas is 1/3 times the energy density (U/V). Compute the pressure exerted by the radiation inside a kiln at 1500 K, and compare to the ordinary gas pressure exerted by the air. Then compute the pressure of the radiation at the center of the sun, where the temperature is 15 million K. Compare to the gas pressure of the ionized hydrogen, whose density is approximately 105 kg/m3. Get solution

46. Sometimes it is useful to know the free energy of a photon gas.(a) Calculate the (Helmholtz) free energy directly from the definition F = U − TS. (Express the answer in terms of T and V.) (b) Check the formula ... for this system. (c) Differentiate F with respect to V to obtain the pressure of a photon gas. Check that your result agrees with that of the previous problem. (d) A more interesting way to calculate F is to apply the formula F = –kT In Z separately to each mode (that is, each effective oscillator), then sum over all modes Carry out this calculation, to obtain...Integrate by parts, and check that your answer agrees with part (a). Get solution

47. In the text I claimed that the universe was filled with ionized gas until its temperature cooled to about 3000 K. To see why, assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 109 photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionized as a function of temperature, for temperatures between 0 and 6000 K. How does the result change if the ratio of photons to atoms is 108, 1010? (Hint: Write everything in terms of dimensionless variables such as t = kT/I, where I is the ionization energy of hydrogen.) Get solution

48. In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and anti neutrinos ..., currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.(a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so that their chemical potentials are equal: ... Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction...(where γ is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that µ = 0 for both the neutrinos and the antineutrinos. (b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: There are very few differences between this “neutrino gas” and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To . evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a Lable or consult Appendix B. (c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K. (d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn’t have affected the production of neutrinos in the early universe, when mc2 would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m What would mc2 have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the to the mass of ordinary matter? Get solution

49. For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of “background radiation,” in addition to the photons and neutrinos (see Figure). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (each with the same mass), and each has two independent polarization states. During the time period of interest the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in the previous problem. Recall from special relativity that the energy of a massive particle is ...(a) Show that the energy density of electrons and positrons at temperature T is given by...(b) Show that u(T) goes to zero when kT ≪ mc2 and explain why this is a reasonable result. (c) Evaluate u(T) in the limit kT ≫ mc2, and compare to the result of the previous problem for the neutrino radiation. (d) Use a computer to calculate and plot u(T) at intermediate temperatures. (e) Use the method of Problem, part (d), to show that the free energy density of the electron-positrou radiation is...Evaluate f(T) in both limits, and use a computer to calculate and plot f(T) at intermediate temperatures. (f) Write the entropy of, the electron-positron radiation in terms of the functions u(T) and f(T). Evaluate the entropy explicitly in the high-T limit.Figure: When the temperature was greater than the electron mass times c2/k the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutrinos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles....Problem:Sometimes it is useful to know the free energy of a photon gas.d) A more interesting way to calculate F is to apply the formula F = −KT In Z separately to each mode (that is, each effective oscillator), then sum over all modes Carry out this calculation, to obtain...Integrate by parts, and check that your answer agrees with part (a). Get solution

50. The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where KT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they “heated” the photon radiation but not the neutrino radiation.(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have been conserved in the early universe, assuming that no other species of particles interacted with these.(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by...as the universe expands and cools. Evaluate the constant by assuming that T = Tv, when the temperatures are very high. (c) Calculate the ratio T/Tv in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K. (d) Use a computer to plot the ratio T/Tv as a function of T, for kT/mc2 ranging from 0 to 3Problem:In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and anti neutrinos ..., currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.(a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so t hat their chemical potentials are equal: ... Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction...(where γ is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that µ = 0 for both the neutrinos and the antineutrinos. (b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: There are very few differences between this “neutrino gas” and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a Lable or consult Appendix B.(c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K. (d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn’t have affected the production of neutrinos in the early universe, when mc2 would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m. What would mc2 have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the to the mass of ordinary matter? Get solution

51. The tungsten filament of an incandescent light bulb has a tem­perature of approximately 3000 K. The emissivity of tungsten is approximately 1/3, and you may assume that it is independent of wavelength.(a) If the bulb gives off a total of 100 watts, what is the surface area of its filament in square millimeters? (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 wave-lengths, between 400 and 700 nm.(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 3695 K. Get solution

52. (a) Estimate (roughly) the total power radiated by your body, neglecting any energy that is returned by your clothes and environment. (Whatever the color of your skin, its emissivity at infrared wavelengths is quite close to 1; almost any nonmetal is a near-perfect blackbody at these wavelengths.) (b) Compare the total energy radiated by your body in one clay (expressed in kilocalories) to the energy in the food you eat. Why is there such a large discrepancy? (c) The sun has a mass of 2 × 1030 kg and radiates energy at a rate of 3.9 × 1026 watts. Which puts out more power per units mass—the Sum of your body? Get solution

53. A black hole is a blackbody if ever there was one, so it should emit blackbody radiation, called Hawking radiation. A black hole of mass M has a total energy of Mc2, a surface area of 16πG2M2/c4, and a temperature of hc3/16π2kGM (as shown in Problem 1).(a) Estimate the typical wavelength or the Hawking radiation emitted by a one-solar-mass (2 × 1030 kg) black hole. Compare your answer to the size of the black hole. (b) Calculate the total power radiated by a one-solar-mass black hole. (c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing. As it loses energy, its mass must, decrease; one could say it “evaporates.” Derive a differential equation for the mass as a function of time, and solve this equation to obtain an expression for the lifetime of a black hole in terms of its initial mass.(d) Calculate the lifetime of a one-solar-mass black hole, and compare to the estimated age of the known universe (1010 years) . (e) Suppose that a black hole that was created early in the history of the universe finishes evaporating today. What was its initial mass? In what part of the electromagnetic spectrum would most or its radiation have been emitted?Problem 1:Use the result of below Problem 2 to calculate the temperature of a black hole, in terms of its mass M. (The energy is Mc2.) Evaluate the resulting expression for a one-solar-mass black hole. Also sketch the entropy as a function of energy, and discuss the implications of the shape of the graph.Problem 2:A black hole is a region of space where gravity is so strong that nothing, not even light, can escape. Throwing something into a black hole is therefore an irreversible process, at least in the everyday sense of the word. In fact, it is irreversible in the thermodynamic sense as well: Adding mass to a black hole increases the black hole’s entropy. It turns out that there’s no way to tell (at least from outside) what kind of matter has gone into making a black hole. Therefore, the entropy of a black hole must be greater than the entropy of any conceivable type of matter that could have been used to create it. Knowing this, it’s not hard to estimate the entropy of a black hole.a) Use dimensional analysis to show that a black hole of mass M should have a radius of order GM/c2, where G is Newton’s gravitational constant and c is the speed of light. Calculate the approximate radius of a one-solar-mass black hole (M = 2 × 1030 kg).b) In the spirit of below Problem 3, explain why the entropy of a black hole, in fundamental units, should be of the order of the maximum number of particles that could have been used to make it.c) To make a black hole out of the maximum possible number of particles, you should use particles with the lowest possible energy: long-wavelength photons (or other mass less particles). But the wavelength can’t be any longer than the size of the black hole. By setting the total energy of the photons equal to Mc2, estimate the maximum number of photons that could be used to make a black hole of mass M. Aside from a factor of 8π2, your result should agree with the exact formula for the entropy of a black hole, obtained through a much more difficult calculation....d) Calculate the entropy of a one-solar-mass black hole, and comment on the result.Problem 3:For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by Nk times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say S ~ Nk. That is, tire entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose (400 kg of water); the sun (2 × 1030 kg of ionized hydrogen). Get solution

54. The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan’s law.(a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of 2.4 eV, while Sirius A is approximately 24 times as luminous as the sun. How does the radius of Sirius A compare to the sun’s radius? (b) Sirius B, the companion of Sirius A (see Figure) , is only 3% as luminous as the sun. is spectrum, plotted as a function of energy, peaks at about 7 eV. How does its radius compare to that. of the sun(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of 0.8 eV, while Betelgeuse is approximately 10,000 times as luminous as the sun. How does the radius of Betelgeuse compare to the sun’s radius? Why is Betelgeuse called a “red supergiant”?Figure:The double star system Sirius A and B. Sirius A (greatly overexposed in the photo) is the brightest star in our night sky. Its companion, Sirius B, is hotter but very faint, indicating that it must be extremely small—a white dwarf. From the orbital motion of the pair we know that Sirius B has about the same mass as our sun. (UCO/Lick Observatory photo.)... Get solution

55. 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.) Get solution

56. 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 70 times that of earth’s atmosphere. You can therefore model the atmosphere of Venus as 70 successive “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.) Get solution

57. Fill in the steps to derive equations 7.112 and 7.117.Equation 7.117:... Get solution

58. The speed of sound in copper is 3560 m/s. Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure, and compare. Get solution

59. Explain in some detail why the three graphs in Figure all intercept the vertical axis in about the same place, whereas their slopes differ considerably. Get solution

60. Sketch the heat capacity of copper as a function of temperature from 0 to 5 K, showing the contributions of lattice vibrations and conduction electrons separately. At what temperature are these two contributions equal? Get solution

61. The heat capacity of liquid 4He below 0.6 K is proportional to T3, with the measured value CV/NK = (T/4.67 K)3. This behavior suggests that the dominant excitations at low temperature are long-wavelength phonons. The only important difference between phonons in a liquid and phonons in a solid is that a liquid cannot transmit transversely polarized waves—sound waves must be longitudinal. The speed of sound in liquid 4He is 238 m/s, and the density is 0.145 g/cm3. From these numbers, calculate the phonon contribution to the heat capacity of 4He in the low-temperature limit, and compare to the measured value. Get solution

62. Evaluate the integrand in equation 7.112 as a power series in x, keeping terms through x4 Then carry out the integral to find a more accurate expression for the energy in the high-temperature limit. Differentiate this expression to obtain the heat capacity, and use the result to estimate the percent deviation of CV from 3Nk at T = TD and T = 2TD. Get solution

63. Consider a two-dimensional solid, such as a stretched drumhead or a layer of mica or graphite. Find an expression (in terms of an integral) for the thermal energy of a square chunk of this material of area A = L 2 , and evaluate the result approximately for very low and very high temperatures. Also find an expression for the heat capacity, and use a computer or a calculator to plot the heat capacity as a function of temperature. Assume that the material can only vibrate perpendicular to its own plane, i.e., that there is only one “polarization.” Get solution

64. A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbours. At T = 0 the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are N atoms, the total magnetization is typically ~2µBN where µB is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h/2π and therefore reduces the magnetization by, ~2µB. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin wave is proportional to the square of 1/λ (in the limit of long wavelengths). Therefore, since ϵ = hf and p = h/λ for any “particle,” the energy of a magnon is proportional to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write ϵ = p2/2m*, where m* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24 × 10−29 kg, about 14 times the mass of an electron. Another difference between magnons and phouons is that each tnagnon (or spin wave mode) has only one possible polarization.(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by...Evaluate the integral numerically. (b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(0) − M(T))/M(0). Write your answer in the form (T /T0)3/2, and estimate the constant T0 for iron. (c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low termperature. You should find CV / Nk = (T /T1)3/2, where T1 differs from T0 only by a numerical constant. Estimate T1 for iron and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470 K.) (d) Consider a two-dimensional array or magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (three-dimensional) direct ion, so spin waves are still possible. Show that the integral for the total number of magnous diverges in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Sect ion 8.2 we will consider a different two-dimensional model in which magnetization does occur.)Figure: In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves , in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy, because the difference in direction between neighbouring dipoles is very small.... Get solution

65. Evaluate the integral in equation 7.124 numerically to confirm the value quoted in the text. Get solution

66. Consider a collection of 10,000 atoms of rubidium-87 confined inside a box of volume (10–5 m)3.(a) Calculate ϵ0, the energy of the ground state. (Express your answer in both joules and electron-volts.) (b) Calculate the condensation temperature, and compare kTc to ϵ0. (c) Suppose that T = 0.9Tc. How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states? (d) Repeat parts (b) and (c) for the case of 106 atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state. Get solution

67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, a gas of approximately 2 × 1010 atoms was trapped and cooled until its peak density was 1.8 × 1014 atoms/cm3 . Calculate the condensation temperature for this system, and compare to the measured value of 50 μK. Get solution

68. Caluculate the condensation temperature for liquid helium-4, pretending that the liquid is a gas of nonintcracting atoms. Compare to the observed temperature of the superfluid transition, 2.17 K. (The density of liquid helium-4 is 0.145 g/cm3.) Get solution

69. If You have a computer system that can do numerical integrals, It’s not particularly difficult to evaluate μ for T > Tc.(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/Tc , c = μ/kTc, and x = ϵ/kTc. Express the integral that defines μ, equation , in terms of these variables. You should obtain the equation...(b) According to Figure 7.33, the correct value of c when T = 2Tc is approximately –0.8. Plug in these values and check that the equation above is approximately satisfied. (c) Now vary μ, holding T fixed, to find the precise value of μ for T = 2Tc Repeat for values of T/Tc ranging from 1.2 up to 3.0, in increments of 0.2. Plot a graph of μ as a function of temperature. Get solution

70. Figure shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph. (a) Write down an expression for the total energy of a gas of N bosons confined to a volume V, in terms of an integral (analogous to equation 7.122). (b) For T Tc you can set µ = 0. Evaluate the integral numerically in this case, then differentiate the result with respect to T to obtain the heat capacity. Compare to Figure. (c) Explain why the heat capacity must approach ... Nk in the high-T limit. (d) For T > Tc you can evaluate the integral using the values of µ calculated in Problem. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat. capacity. Plot the heat capacity, and check that your graph agrees with Figure.Figure: Heat capacity of an ideal Bose gas in a three-dimensional box....Problem:If You have a computer system that can do numerical integrals, It’s not particularly difficult to evaluate μ for T > Tc.(a) As usual when solving a problem on a computer, it’s best to start by putting everything m terms of dimensionless variables. So define t == T /Tc , c = μ/kTc, and x = ϵ/kTc. Express the integral that defines μ, equation , in terms of these variables. You should obtain the equation...(b) According to Figure 7.33, the correct value of c when T = 2Tc is approximately –0.8. Plug in these values and check that the equation above is approximately satisfied. (c) Now vary μ, holding T fixed, to find the precise value of μ for T = 2Tc Repeat for values of T/Tc ranging from 1.2 up to 3.0, in increments of 0.2. Plot a graph of μ as a function of temperature. Get solution

71. Starting from the formula for CV derived in Problem, calculate the entropy, Helmholtz free energy, and pressure of a Bose gas for: TTc Notice that the pressure is independent of volume; how can this be the case.Problem:Figure shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.(b) For T Tc you can set µ = 0. Evaluate the integral numerically in this case, then differentiate the result with respect to T to obtain the heat capacity. Compare to Figure.Figure: Heat capacity of an ideal Bose gas in a three-dimensional box.... Get solution

72. For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of ϵ (see Problem). Investigate the behaviour of a gas of non interacting bosons in a two-dimensional box. You should find that the chemical potential remains significantly less than zero as long as T is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must g(ϵ) have in order for there to be an abrupt Bose-Einstein condensation?Problem:Consider a free Fermi gas in two dimensions, confined to a square area A = L2.(a) Find the Fermi energy (in terms of N and A), and show that the average energy of the particles is ϵF/2(b) Derive a formula for the density of states. You should find that it is a constant, independent of ϵ.(c) Explain how the chemical potential of this system should behave as a function of temperature, both when. kT ≪ ϵF and when T is much higher. (d) Because g(ϵ) is a constant for this system, it is possible to carry out the integral for the number of particles analytically. Do so, and solve for µ as a function of N. Show that the resulting formula has the expected qualitative behavior. (e) Show that in the high-temperature limit, kT ≫ϵF, the chemical potential of this system is the same as that of an ordinary ideal gas.Equation:... Get solution

73. Consider a gas of N identical 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 n is any nonnegative integer and f is the classical oscillation frequency. The degeneracy of level n is (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 ausume 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 π and 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. Get solution

74. 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 numerically, without approximating it as an integral.(a) Write equation for this system as a sum over energy levels, taking degeneracy into account. Replace T and μ with the dimensionless variables t = kT/ hf and c = µ/hf. (b) Program a computer to calculate this sum for any given values of t and c. Show that, for N = 2000, equation is satisfied at t = 15 provided 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 t to 14, and adjust the value of c until the sum again equals 2000. Plot the number of particles as a function of energy. (e) Repeat part (d) for t = 13, 12, 11, and 10. You should find that the required value of c increases toward zero but never quite reaches it. Discuss the results in some detail.Equation:... Get solution

75. Consider a gas of non interacting spin-0 bosons at high temperatures, when T ≫ Tc. (Note that “high” in this sense can still mean below 1 K.)(a) Show that, in this limit, the Bose-Einstein distribution function can be written approximately as...(b) Keeping only the terms shown above, plug this result into equation to derive the first quantum correction to the chemical potential for a gas of bosons.(c) Use the properties of the grand free energy (Problems 1 and 2) to show that the pressure of any system is given by P = (kT/V) In Z, where Z is the grand partition function. Argue that, for a gas of non interacting particles, In z can be computed as the sum over all modes (or single-particle states) of In zi, where Zi is the grand partition function for the ith mode. (d) Continuing with the result of part (c), write the sum over modes as an integral over energy, using the density of states. Evaluate this integral explicitly for a gas of noninteracting bosons in the hightemperature limit, using the result of part (b) for the chemical potential and expanding the logarithm as appropriate. When the smoke clears, you should find...again neglecting higher-order terms. Thus, quantam statistics results in a lowering of the pressure of a boson gas, as one might expect. (e) Write the result of part (d) in the form of the virial expansion introduced in Problem 3, and read off the second virial coefficient, B(T). Plot the predicted B(T) for a hypothetical gas of noninteracting helium-4 atoms. (f) Repeat this entire problem for a gas of spin-1/2 ferrnions. (Very few modifications are necessary.) Discuss the results, and plot the predicted virial coefficient for a hypothetical gas of noninteracting helium-3 atoms.Problem 1:By subtracting μN from U, H, F, or G, one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),Φ ≡ U − TS − μN.(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives of Φ with respect to T, V, and μ.(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φ tends to decrease.(c) Prove that Φ = – PV.(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in below Problem 4, the prediction for such a small system is only a probabilistic one.)Problem 4:The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)Problem 2:In Section 6.5, I derived the useful relation F = − kT ln Z between the Helmholtz free energy and the ordinary partition function. Use an analogous argument to prove thatϕ = − kTlnZwhere z is the grand partition function and ϕ is The grand free energy introduced in Problem 5.Problem 5:By subtracting μN from U, H, F, or G, one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),Φ ≡ U − TS − μN.(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives of Φ with respect to T, V, and μ.(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φ tends to decrease.(c) Prove that Φ = – PV.(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in below Problem 6, the prediction for such a small system is only a probabilistic one.)Problem 6:The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)Problem 3:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive. Get solution