Answer to Question #106812 in Molecular Physics | Thermodynamics for MS

Question

A simple harmonic one dimensional oscillator has energy level is given by En=(n+{1÷2}h cross omega.where omega is angular frequency and h=0,1,2,3,.... .Suppose that this oscillator is in thermal contact with a heat reservior at tempararature T low enough so that kT<<hcross omega.Find the ratio of probability of oscillator being in first excited state to the probability of its being in the ground state.

Answer 1

As per the given question,

Energy of the one dimensional simple harmonic oscillator $E_n=(n+\dfrac{1}{2})\hbar\omega$

Now, we know that probability of the oscillator in nth excited state,

$P=e^{-\beta E_n}$

Now,

probability of the oscillator being in the first excited state $P_1=e^{-\beta E_1}$

probability of the oscillator being in the ground state $P_0=e^{-\beta E_0}$

Hence, the required ratio $\dfrac{P_1}{P_o}=\dfrac{e^{-\beta E_1}}{e^{-\beta E_0}}=\dfrac{e^{-\beta (1+\frac{1}{2})\hbar\omega}}{e^{-\beta (0+\frac{1}{2})\hbar\omega}}$

$=\dfrac{e^{-\beta (\frac{3}{2})\hbar\omega}}{e^{-\beta (\frac{1}{2})\hbar\omega}}$

$=e^{-\beta\hbar\omega(-\frac{3}{2}+\frac{1}{2})}$

$=e^{-\beta \hbar\omega}$