Respuesta :
Answer:
Since the probability of the energy E = 0 is maximum, this is the most probable state.
Explanation:
[tex]K_BT_a=1.3807*10^{-23}* 320[/tex]
[tex]= 0.442 * 10^{-20}[/tex]
[tex]\epsilon/K_BT_a =(10^{-20}) / (0.442*10^{-20})[/tex]
[tex]= 1/0.442[/tex]
[tex]= 2.26[/tex]
Probability of Energy, E = 0
[tex]P (E = 0) = e^{-0} / [e^{-0} + e^{-\epsilon /K_BT_a} + e^{-2\epsilon /K_BT_a}][/tex]
[tex]= 1/[1 + e^{-2.26} + e^{-4.53}][/tex]
[tex]= 1/1.11[/tex]
[tex]= 0.9009[/tex]
=90%
Probability of Energy, E= [tex]\epsilon[/tex]
[tex]P (E = \epsilon ) = e^{-\epsilon /K_BT_a} / [e^{-0} + e^{-\epsilon/K_BT_a }+ e^{-2*\epsilon/K_BT_a}][/tex][tex]= e^{-2.26}/[1 + e^{-2.26} + e^{-4.53}]= 0.104/1.11= 0.0936[/tex]
=9.36%
Probability of Energy, E= [tex]2*\epsilon[/tex]
[tex]P (E = 2*\epsilon ) = e^{-2\epsilon/K_BT_a }/ [e^{-0} + e^{-\epsilon/K_BT_a} + e^{-2\epsilon /K_BT_a}][/tex]
[tex]= e^{-4.53}/[1 + e^{-2.26} + e^{-4.53}]= 0.011/1.11= 0.009[/tex]
=1%
