Homework 9 Solution Physics Spring a) We can calculate the chemical potential using eq(6.48): µ = τ (log(n/n Q ) log Z int ) (1) Z int =

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1 Homework 9 Solutions uestion 1 K+K Chater 6, Problem 9 a We can calculate the chemical otential using eq6.48: µ = τ logn/n log Z int 1 Z int = e εint/τ = 1 + e /τ int. states µ = τ log n/n 1 + e /τ b The artition function in the resence of internal degrees of freedom is Z = Z IG Z int. As a result the free energy, which is roortional to the logarithm of Z is additive: F = F IG + F int. F IG is the known ideal gas result. Let us calculate the contribution from the internal degrees of freedom. F int = Nτ log Z int 4 F int = Nτ log 1 + e /τ 5 [ ] n/n F = F IG + F int = Nτ log Nτ e /τ c The entroy can be easily calculated using the free energy obtained in art b using the formula: σ int = F int 7 = Nτ log 1 + e /τ 8 { = N log 1 + e /τ + /τ } 9 Just like the free energy, entroy is additive as well: σ = σ IG + σ int 1 { = N log n /n + 5/ + log 1 + e /τ + /τ } 11 { n 1 + e /τ = N log + 5 n + /τ } 1 3 1

2 d Pressure is not effected by the existence of the internal degrees of freedom: = F 13 τ,n = { { n 1 + e /τ N log + 5 n + /τ }} 14 = Nτ 15 e Heat caacity at constant ressure C = τ σ = U + 16 U = F + τσ = = 3 Nτ + N 17 U = 3 N + N /τ e /τ e /τ C = 3 N + N /τ e /τ e /τ N 19 where in the last term of the last equation we used the ideal gas law to obtain. Hence the final answer reads C = 5 N + N /τ e /τ e /τ + 1 uestion K+K Chater 6, Problem 1 a In K+K Chater 7, Problem 1, art b we calculate the density of states for an electron in dimensions. We can borrow that result here but we need to divide it by because electrons have sin 1/ and degenerate states associated with every energy. For the ideal gas we will assume zero sin hence no degeneracy factor. Hence from Eq.39 below: Dε = Am π h 1 where A is the area of the container. Using this we can calculate the one article artition function Z 1 = dεdεe ε/τ = maτ π h nd A

3 where the last equality defines the two dimensional quantum density. Chemical otential is found by enforcing the number constraint: N = λz 1 = λ n D A 3 λ = N/A n D = nd n D µ = τ log n D n D b What is the energy of the gas? Find the free energy first. This can be done in two ways. One can integrate µ = F N τ,. I will calculate F directly from the artition function for N articles. Since we are considering classical indistinguishable articles Z = Z1 N /N! using Stirlings formula for large N we get [ ] Z N F = τ log Z = τ log 1 n D = τn log N! n D once we have the free energy the internal energy is obtained by artial differentiation wrt temerature: U = τ F 7 τ,n = τ { N log n D /n D } N 8 = Nτ 9 c Entroy is also simly obtained from the free energy σ = F 3 N, [ ] n D σ = N log n D + 31 uestion 3 K+K Chater 7, Problem 1 a In 1D the energy levels are E n = n x π h ml, n x = 1,, 3,. The Fermi energy is defined as ε F = n F π h ml 3 3

4 The energy levels are equally saced in n x sace. The total number of orbitals filled u to ε F are given by ml ε F n F = π h 33 where the factor of in front of the square root comes from the sin degeneracy. Electrons, having sin 1/, can have two sin states having the same energy. So and the density of states is Nε = ml π h ε1/ 34 Dε = dn dε = L m π h ε 1/ 35 b Free electrons in D. The energy levels are given by E n = n x +n y π h ml, n x, n y = 1,,. The energy levels are equally saced in the n x, n y sace. The total number of states u to energy ε F by definition add u to the number of articles at zero temerature: N = 1 4 πn F = πn F / 36 so n F = N 1/. π Above the factor again originates from sin degeneracy. The factor of 1/4 is necessary because we want to add states with ositive n x, n y airs. The negative values do not reresent different states. Another way of saying this is that we only consider the first quadrant in the n x, n y lane. The Fermi energy can be calculated using n F : ε F = h π n F m L 37 = Nπ h ml 38 Again thinking of N as the number of states having energies u to ε F we can solve it for N as a function of ε F. This gives Nε = ml ε. The density of π h states is obtained by taking derivative wrt energy Dε = dn dε = ml π h = ma π h 39 which is a constant. In the last line A = L is the area of the box. uestion 4 K+K Chater 7, Problem 5 4

5 a 3 He has sin 1/ and is a Fermion. ρ =.81gm/cm 3. Mass er atom is 3amu gm. Using these we can calculate the concentration of atoms: n = N gm =.81 cm 3 1 atoms = gm/atoms cm 3 4 So the Fermi energy is ε F = h 3π N /3 = = J 41 m = 49µe 4 T F = ε F = 4.98K 43 k B εf v F = = m/s 44 m b Use the exression for degenerate free electrons eq7.38 C = Nπ k B T T F 4.98K Nk BT 45 =.99Nk B T 46 The observed heat caacity is C =.89Nk B T. The enhancement comes from the fact that the liquid 3 He is not an ideal gas. When one 3 He atom moves, others have to get out of its way making the effective mass of the 3 He atom larger. It aears to be about 3 times larger than the bare mass of the 3 He atom. uestion 5 a The heat caacity C v = u 3 Dɛ F τ N ɛ F τ 48 N τ τ F 49 where ɛ F τ F, from Eq36 in Chater 7. b The heat caacity ca be written as C = τ δ 5

6 Hence, δ = c The Holmholtz free energy: = τ τ π C τ dτ 5 N ε F N τ τ 1 dτ 51 τ F τ dτ 5 N ε F τ 53 F = u τδ 54 = u + π 6 Dε F τ π N τ ε F 55 = u Aτ 56 where A = Dε F [ π 3 π 6 ] 57 6 Dε F 58 = Nπ 4ε F 59 uestion 6 a The chemical otential can be found from the Helmholtz free energy as µ = F N Solve for the free energy: F = = N N τ, µdn 6 εf 1 3 π τ ε F dn 61 6

7 Note that ε F = h F = N [ h m 3π N /3, hence m 3π /3 N /3 1 3 π 1 h = 3 h 5 m 3π /3 N /3 N 1 3 π 1 h m N 3π /3 m 3π N /3 ] dn 6 /3 3N 63 = 3 5 Nε F π N ε F τ = u Aτ where u = 3 5 Nε F and A = π N ε F. 64 b The ressure can be obtained from Helmholtz free energy as P = F 65 τ, = 3 5 N ε F π N τ, ε τ ε F F 66 τ, And ε F = τ, h N m 3π /3 67 Substitute it into the above equation: = ε F 3 68 P = Nε F π N ε F τ 69 When τ, the ressure: P = Nε F 5 7 c From the above equation, the ressure will reach a finite value when temerature aroaches. This is because the Pauli Princile force the fermions into higher energy states as they are added into the box, creating an effective ressure even at τ =. This degeneracy ressure is what revents white dwarfs and neutron stars from collasing under gravitational attraction. 7