Consider simple harmonic oscillator with energy quantum =10-20 J. That is, the state energies are 0,,, etc. It is in an environment with tem

Question

Consider simple harmonic oscillator with energy quantum =10-20 J. That is, the state energies are 0,,, etc. It is in an environment with temperature 320 K.

1) Over all the possible states in equilibrium, what is the most probable energy of the oscillator?

in progress 0
Kiệt Gia 2 weeks 2021-07-19T06:28:16+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-19T06:29:37+00:00

    Answer:

    Since the probability of  the energy E = 0 is maximum, this is the most probable state.

    Explanation:

    K_BT_a=1.3807*10^{-23}* 320

    = 0.442 * 10^{-20}

    \epsilon/K_BT_a =(10^{-20}) / (0.442*10^{-20})

    = 1/0.442

    = 2.26

    Probability of Energy, E = 0

    P (E = 0) = e^{-0} / [e^{-0} + e^{-\epsilon /K_BT_a} + e^{-2\epsilon /K_BT_a}]

    = 1/[1 + e^{-2.26} + e^{-4.53}]

    = 1/1.11

    = 0.9009

    =90%

    Probability of Energy, E= \epsilon

    P (E = \epsilon  ) = e^{-\epsilon /K_BT_a} / [e^{-0} + e^{-\epsilon/K_BT_a }+ e^{-2*\epsilon/K_BT_a}]= e^{-2.26}/[1 + e^{-2.26} + e^{-4.53}]= 0.104/1.11= 0.0936

    =9.36%

    Probability of Energy, E= 2*\epsilon

    P (E = 2*\epsilon ) = e^{-2\epsilon/K_BT_a }/ [e^{-0} + e^{-\epsilon/K_BT_a} + e^{-2\epsilon /K_BT_a}]

    = e^{-4.53}/[1 + e^{-2.26} + e^{-4.53}]= 0.011/1.11= 0.009

    =1%

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )