Question

Q.7) Consider a systems of N>>1 identical, distinguishable and independent particles that can be placed in three energy levels of energies 0, E and 2€, respectively. Only the level of energy sis degenerate, of degeneracy g=2. This system is in equilibrium with a heat reservoir at temperature T. a) Obtain the partition function of the system. b) What is the probability of finding each particle in each energy level? c) Calculate the average energy <B>, the specific heat at constant volume. Cv, and the entropy S, of the system. d) Define the high and low temperature limits. Give the mean energy and the entropy of the system in these limits. Justify qualitatively these results.

Answer 1

Solution. - 26 -- €19=2) - 9 Partition function, ?=? P= {gie BG E + ze Be + e-2BE 2BE = 1+2 6+B6 + 17 = (1+e+B6)2) particle in E=0 6 Cite aja Probability of finding = P = n = e Ñ Cite-At)? In tlg=2) = P₂ = 20 .-ße (lte-Beja In 2€ -2BE > P3= e Clt e-BE) © 287 mm enz (1++28€+2+) = 2€ -28€ +26e ße KEY= 26e ße (teßt) ews agentes e 2e [ 348* + cfr ] n

levite [ 28 29€ + eipts] S = ? as, F = -K gT luz fru energy, F = U-TS = -keł en Citė 2BE the BE) 'N, -2A E => SE (1) May (I len (14 é 4€ + 2 été] Szkolen (lte*)*+ I ( The - (lté 2BE +268f) S = 1 [ln (17 & Bey² + 2 BE Leapete Be) ? 1. (ite2BE +26-BE) ] @ High Temperature limit, K&T EXE e BE Mean energy E7 = 2 € x2 z 4€ .at Intropy, s = 2 kq ln 2 to = 2 kg lu2 Cs Scanned with CamScanner Low

tune limit High temperature limit, kot the eßt a 1 - 2 ke7= 26(1--) (2) 167 = (e - Pete) (- te - ha de tenia LEY <te- en nadere 5= [(2-poj* + 2 (1 + 1 - 28€ + 2 - 2B6) 4-48E ] S = kp [2 ln (2–pe) + 24€ (2-386)7 |= ks [ 2 ln(2-BE) + pe (2-3627) 2(1-BE) so in value i how temperature limit (KT4€) Mean energy e constant = 4€ & Entropy is also constant = 2kB lu 2. limit, both depends temperature high ton 1 - on 'T' CS Scanned with CamScanner