Question

2. System of particles obey M-B statistics in thermal equilibrium with its enviroment at temprature T. the percent of occupation of energy levels is given at the table. Energy in ev Energy level propapility percent 0.002 60% 0.011 20% 0.020 8.0% 0.030 1.9% (i) Calculate the temperature of the system. If the vibrationally excited states of nitrogen molecule is given by &, =(n +) @ where n;= 0, 1, 2,..., the differences between anytwo levels is given by ) w=0.5 ev,If the gas is in thermal equilibrium at T= 1500 k (ii) Calculate the ratio between the occupation probability of second exited state and the ground state.

Answer 1

Considering defferent occupation index,.. Eiz Eenergy of ith level. We have average Energy, Pia occupation (€[ Ei pa : 0.002 x 60 + 0.011x hot + 0.020* probablity. + 0.030 x 1.9 10.12 + 0:22 +0.16 + 0.057 . 100 0.557 100 ev = 0.00557 ev

We have, w = 0.5 ev = 0.5x1.6x10 s at T = 1500k ret, W KT or & 0.5x1.64 10 18 1500 - 5.33 x 10-23 1/4 let us consider, Energy equivalene with temperature relation, So from ean (i), let the Temperature be To : So, Tis: 0.00557 x 7.6x10-19 5.33 x 10-23 - 16.71 K . (1) and Ratio between Occupation probablity of selona Cexeited ground state, 0.020 KT. e -0.002 . .. e KT Here Kis Boltzmano const, Tis Absolute temp. 0.020 -0.002) ., Р е. с кт 1.0,018 x 1.6x10-19 L e 138 x10-23 x 16:41 - e - 12.489 33.76 x 10-6
do comment for any difficulty.