Question

(TOTAL MARKS: 25) QUESTION 4 (15 marks] Q4(a) Assume 4 fermionic particles (N=N,+NA+N, -4) populate 3 degenerate energy levels E <E, <E, with 8, = 4,8, = 3.8, = 2 and N, 2N, 2N, What are the possible macrostates of this system ? (3 marks) (l) For each macrostate found at (), count the number of possible microstates using sketches showing the quantum state occupation number in each energy level. (7 marks) (H) Retrieve your results at (ii) if the number of possible microstates for a given macrostate is given by 02(N.)-II-.&. (5 marks) N! Q4(b) (10 marks] It can be shown that the entropy s of the above three-level system can be given by the Boltzmann formula: S =k, In MIN.IN! We where ke=1.38 x 10-2 JK. The total population is N = 4x10”. Calculate the entropy of this system when ON, = N, = 0 and (ii) N=0.1N, N, =0.02N. Stirling formula: InN!= N INN-N [End of Question 4

Answer 1

4a) No of fermions, N=4. Jermions & Fermi Dirac statistics each state can particles - (9) degeneracy states available for each energy level. 9, = 4; 9 = 3; 93=2 manimum no. of fermions allowed in an energy lowel degenadace of that leold.

acrostates I No. of particles in Energy level Е, Е3 Total Energy È lu YEI I 3 E, & { M N N 26, +2E, 3 E, tez 28, + 2 & 3 128, + E + Ez E, + & 2 + 2 & 3 N t ett E, + 2 E + Ez IE, + 3 E,

al E. (g =4) T ar E Cg = 3) 1 No. of Ernstul - Eniti microstate UE, I 1 (y=2) - Slate 36,+ E₂ | 12 126, +2E₂ 18

mauwstate E, G =4) I E Cg = 3) E39=2) EM EN & Total no. of microstates 36 + E3 2E,+ 2 63 6 22222222222LLLLLLLL12 19111111111 2 Et Ent &

Total no. of E39=2) Estul Eroil microstate Macw state &lg=4) 19-3) &, + E2 +263 12 222222222222 & + 2E + E3 24 Eit 3 E TX

ali n i) = . n gi! Ni! (gi-Ni)! for macro state! 31 In 6) - 31 31 1 1 21 — 2111 0121 31 211 0 31 2101 sa is 31 ) 3)-36 (2121) 1121 1111

4 a) rii No. of meowstate for Moow state 21 12 1131 2lol so = ( 13 ) ( 112 ) (-21) = 12 VIIS ME G .) - 24 x 15 = 6 ) ) 4 01

s=ke lu N! NI! Ne ! N3! ke = 10 38 x 10 23 5/6 N = 4 x 1023 - total population. ů) Entropy when Nr = N3=0 Ni=N Nz = N₂ = 0 then So if & S = KB In N! N!.01.0! = ko. ln N1 = kg en s = kB en I & s=0 ili Entropy when N₂ = 0.IN Ng = 0.02 N the should haw, N = Ni to .IN+0.02 N A N, N ON=0:02N N- 0.12 N N = 0.88 N

values in so now putting all these S = KR en NI (O.IN)! (0.02N)! (0.88N)! In N! - ln (0.1N)! - ln (0.02N)! - en 10 188N)! 7. s = kB NON-N -ort on IN en O. IN & O-IN -0.02 N ln 0.02N + 0.02 N-0088 N en (0.88N). + 0,88 N _ by Stirling's formula an N o: IN IO-IN -0.02 N ln 10. 62N - 0.88 N en 60 0.88 N | 2438 x 10-23 [ 4x7093 tot 4*70 23). = Nke [ enn - 0.1 lno. IN - 0.02 ln 0.02 N - 0.88 in 0.88N] = 4X1023 x 1-38 X10-23 [ n(4 X 100) 0.1 An (0.184X 16") -0.02 en (0.02 x 4 x 1023) - 0.88 en (0.88 x 4 X 103)

= 5.52 T 23 - o.1 noil - oln 4-234 0. [ en 4 + 23 - oor en o • 02 a 0.8.8 eno.88 - 0.02 en 0.02-0.02ln 4 - 257 0. 88 x 23 28 x 23 0.88 inoo 88 - 0.88 0.88 en 4 en 4 - - 0.02 x 2 en 2 - 0.46 n2 + 23-0.1 (-20303) - 0. 12 ln2 = 5.52 [2 ln 2 + 23-0.1 (-2303) - = 2.3 - 0.02 (-3.912) - 0.02 x 2 en - 0.88 (-0.128) - 0.88 x 2 ln2 - 20-27 J 102 - 20.24 | 2 - 0.693 - 2 3 - 0.2303 - 0.2 x 5.52 2x0.693 +23+ 0.2303 - 0.2 x Oobe -0.078 - 0.04 x 0.693 - 0.46 + 0.113 - 1-7X0-692 - 20.24 ] = 5.52 [ 1.386 + 23 + 0 0 2303 -0.1386 - 203 +0.078 - 0.62772 - 0:46 +0.113 - 1.18 - 20:24 5:52 (0.46098] 5 = 2.554 J/x . S = 2.554 Ilk A positive value of entropy