Part D Using the Boltzmann distribution, calculate nj/no for 1H3Cfor 20 at T-1050 K
Solution
Using Boltzmann distribution,
rotational constant for H 1 Cl 35 B 0 = 10.5934 cm -1
h = planck\'s constant = 6.626 x 10 -34 joule. sec
c = 3 x 10 10 cm/sec
B = hc B 0 = 6.626 x 10 -34 x 3 x 10 10 x 10.5934 = 210.576 x 10 -24 joules
n j /n o = (2J+1)e -BJ(J+1)/K B T
K B = 1.38 X 10 -23 m 2 kg s -2 K -1
J = 20 and T = 1050 K
(2J+1) = 2X20+1 = 41
BJ(J+1) = 210.576 x 10 -24 joules(20+1) =21X 210.576 x 10 -24 = 4422.096 X 10 -24
K B T = 1.38 X 10 -23 m 2 kg s -2 K -1 X 1050 K = 1449 X 10 -23 m 2 kg s -2
n j /n o = (2J+1)e -BJ(J+1)/K B T
n j /n o = 41 e -4422.096 X 10-24/1449 X 10-23
n j /n o = 41 e -0.3052 = 41 x 0.737 = 30.216
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Part D Using the Boltzmann distribution- calculate nj-no for 1H3Cfor 2.docx
1. Part D Using the Boltzmann distribution, calculate nj/no for 1H3Cfor 20 at T-1050 K
Solution
Using Boltzmann distribution,
rotational constant for H 1
Cl 35
B 0 = 10.5934 cm -1
h = planck's constant = 6.626 x 10 -34
joule. sec
c = 3 x 10 10
cm/sec
B = hc B 0 = 6.626 x 10 -34
x 3 x 10 10
x 10.5934 = 210.576 x 10 -24
joules
n j /n o = (2J+1)e -BJ(J+1)/K
B
T
K B = 1.38 X 10 -23
m 2
kg s -2
K -1
J = 20 and T = 1050 K
(2J+1) = 2X20+1 = 41
BJ(J+1) = 210.576 x 10 -24
joules(20+1) =21X 210.576 x 10 -24
= 4422.096 X 10 -24
K B T = 1.38 X 10 -23
m 2
kg s -2
K -1
X 1050 K = 1449 X 10 -23
m 2
kg s -2
n j /n o = (2J+1)e -BJ(J+1)/K
B
T
n j /n o = 41 e -4422.096 X 10-24/1449 X 10-23
n j /n o = 41 e -0.3052
= 41 x 0.737 = 30.216
