INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI Department of Physics MID SEMESTER EXAMINATION Statistical Mechanics: PH704 Solution

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1 INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI Department of Physics MID SEMESTER EXAMINATION Statistical Mechanics: PH74 Solution. There are two possible point defects in the crystal structure, Schottky and Frenkel defects. In Schottky defect, the atoms are migrated to the surface of the crystal creating vacancies inside whereas in Frenkel defect, the atoms can occupy either a regular lattice site or an interstitial site. Consider an isolated monatomic crystalline solid of N atoms with n defects, either Schottky or Frenkel, with fixed total energy nǫ. Suppose there are N interstitial sites in a lattice of N sites. (a Calculate the number of possible microstates for both the defects. (b Calculate entropy S as a function of the number of defects n for both the defects. [Given that lnx! = xlnx x] Find the temperature of the systems and defect distribution for both defects when n N. [++6 = ] Solution: (a Schottky defects: N atoms are distributed over N + n lattice sites. so the number of microstates are N +n (N +n! Ω S = = N n!n! Frenkel defects: There are n out of N lattice sites are vacant, as well as n out of N interstitial sites are occupied. Hence, the number of microstates are (b Schottky defects : S S = k B ln Frenkel defects : S F = k B ln ( N N! Ω F = = n n!(n n! (N +n! n!n! = k B [(N +nln(n +n N lnn nlnn] N! n!(n n! = k B[N lnn (N nln(n n nlnn] T = S ds dn dnd ds ǫ dn, nǫ ds S N +n Schottky defects : dn = k B[ln(N +n lnn] = k B ln n T = k B N +n ǫ ln n

2 Frenkel defects : N n + = eǫ/k BT, since n N, n Ne ǫ/k BT ds F N n dn = k B[ln(N n lnn] = k B ln n T = k B N n ln ǫ n N n = eǫ/k BT, since n N, n Ne ǫ/k BT. A gas of N non-interacting diatomic molecules of type A A in a container of volume V is in thermal equilibrium at temperature T. The single molecule Hamiltonian translational and rotational parts and it is given by H = H trs +H rot, Htrs = p M ( p and Hrot = θ I + p Isin θ where M is the mass of the molecule, I is the moment of inertia of the molecule about an axis perpendicular to the line joining the two atoms and passing through the center of mass. θ and are the rotational coordinates, both are within the range [,π]. p θ and p vary from to +. (a Obtain the canonical partition function Z of the gas. (bfindtheequationofstate. DetermineC P C V forthisgasknowing thatc P C V = TVα P /κ T, where α P is the volume expansion coefficient and κ T is the isothermal compressibility. Calculate the mean energy E of the gas and find the values of both the specific heats C P and C V. [5+3+ = ] Solution: (a The canonical partition function is given by Z(N,V,T = ( Z trs N Z rot N where Z trs and Z rot are the singe particle canonical partition functions corresponding to H trs and H rot. Z trs = V h 3 4πp exp ( p dp = V Mk B T h (πmk BT 3/ 3

3 Z rot = h dθd = π d h = π πi h β + π [ exp dθ { p β θ I + p }] Isin dp θ dp ( θ βp π + exp θ dp θ dθ exp I πi β sinθ = I β ( βp I sin dp θ 3N/ N πm I Therefore, Z(N,V,T = V N β 5N/ h (b = V P κ T V = V P = k B T lnz = Nk BT V V ( Nk BT = P and α V P = V Therefore, C P C V = TVP T = PV T = Nk B Hence, C V = E T = 5 Nk B lnz β = 5 Nk BT V T = Nk B VP = T and C P = C V +Nk B = 7 Nk B 3. Consider a three dimensional ideal gas of N Fermions enclosed in a volume V in thermal equilibrium at temperature T. (a Derive the grand canonical partition function Q of the gas? Obtain the mean occupation number n ǫ of the state of energy ǫ. (b Calculate the density of states g(ǫdǫ for the Fermions between energy ǫ and ǫ+dǫ. Obtain pressure P in terms of the internal energy E at any temperature T. [3++5 = ] Solution: (a Q(µ,V,T = z N Z(N,V,T, where z = e µ/k BT N= and, Z(N,V,T = {nk} e β k= n kǫ k, with n k = N k= 3

4 Therefore, { } Q(µ,V,T = exp β n k (ǫ k µ = exp{ βn k (ǫ k µ} N= {n k } k= k=n k = Since for Fermions, n k =, only ( Q(µ,V,T = +e β(ǫ k µ k= (b N = β lnq µ = k e β(ǫ k µ g(ǫdǫ = g s V h 34πp dp = g s πv P = lnq β V = πg s β = 3 πg s = E 3V +e = β(ǫ k µ e β(ǫ k µ + = k k n ǫ = e β(ǫ µ + n ǫ 3/ m ǫdǫ, p = mǫ h 3/ m ǫ 3/ g(ǫ n ǫ ǫdǫ = Vπg s h e β(ǫ µ + dǫ [ = 3/ m πvg s ln ( ] +e β(ǫ µ ǫdǫ β V h 3/ m ln ( +e β(ǫ µ d(ǫ 3/ h 3 3/ m ǫ 3/ h e β(ǫ µ + dǫ 4. The mean occupation number n ǫ of a state of energy ǫ of an ideal Bose gas is given by n ǫ = e β(ǫ µ where µ is the chemical potential. Photon and phonon are two massless bosons. (a Obtain the average energy E of a photon gas enclosed in a volume V in terms of temperature T. (b The density of states of a phonon gas for a three dimensional solid of N atoms is given by { 9N ω g(ω = Dω for ω ω 3 D otherwise 4

5 where ω D is the Debye frequency. Calculate the average energy E of the gas and obtain the specific heat C V of the solid for temperatures T θ D and T θ D where θ D = ω D /k B is the Debye temperature. Given that : ζ(n = x n dx Γ(n e x and ζ(4 = π 4 /9 [4+6 = ] Solution: (a For photons, µ = and ǫ = pc. Thus the average energy is given by where Therefore, g(ǫ n ǫ ǫdǫ g(ǫdǫ = g p V h 34πp dp = 8πV h 3 c 3ǫ dǫ with g p =, and n ǫ = e βǫ 8πV h 3 c 3 = π4 5 8πVk 4 B h 3 c 3 T 4 ǫ 3 dǫ e βǫ = 8πV h 3 c 3(k BT 4 e x = 8πV h 3 c 3(k BT 4 Γ(4ζ(4 (b For phonons, µ = and ǫ = ω. The occupation number is given by Thus the average energy is given by ωd = 9N ω 3 D g(ω n ω ωdω ωd = 9Nk /T B T 4 3 n ω = e β ω ω 3 dω e β ω = 9N (ω D 3(k BT 4 e x ωd /k B T (i If T θ D, then x and x 3 /(e x x. Hence, 9Nk B T ( (ii If T θ D, then x. Hence, T e x, x = ω/k BT 3 = 3Nk B T and C V = 3Nk B 9Nk B T 4 3 e x = 9Nk B θ 3 D T 4π4 5 = 3π4 Nk B 5θ 3 D T 4 and C V = π4 Nk B 5 ( T θ D 3 5