Calculating Probability of Excited State of Harmonic Oscillator

[quickclick330]

Harmonic Oscillator

9. An ideal harmonic oscillator has energy levels separated by a constant 0.05 eV. That
is, the difference between the ground state and the first excited state of this oscillator is
an energy of 0.05 eV. This single oscillator is brought into contact with a large solid
composed of an enormous number of oscillators, and characterized by a temperature of
T=150 K. What is the probability that the single oscillator will be found in its first excited
state?
1. 0.021
2. 0.055
3. 0.098
4. 0.144
5. 0.234
6. 1.000


What relationship should I even start with? Thanks for the help!

 

[uber]

This is a simple Boltzmann factor problem.

Recall that the probability of an individual energy level is:

exp(-e/(kT))/sum(exp(-ne/(kT))

where

e is the separation between energy levels
k is the Stefan-Boltzmann constant (make sure you get its units right!)
T is the absolute temperature

in this case, we can neglect all terms n > 1 as they decay rapidly to zero. Also, recal that the energy level for the ground state is 0.

That should help you complete the problem.

 

[ogb p]

I would approach this problem by first understanding the concept of thermal equilibrium and the Boltzmann distribution, which describes the probability of a system being in a particular energy state at a given temperature. In this case, the single oscillator is in contact with a large solid at a temperature of 150 K, which means that it will reach thermal equilibrium with the solid and follow the Boltzmann distribution.

Based on this understanding, the probability of the single oscillator being in its first excited state can be calculated using the following formula:

P(E1) = (g1/g0) * e^(-E1/kT)

where P(E1) is the probability of the oscillator being in its first excited state, g1 is the degeneracy (number of energy states) of the first excited state, g0 is the degeneracy of the ground state, E1 is the energy difference between the first excited state and the ground state, k is the Boltzmann constant, and T is the temperature.

In this case, g1 = 1 (since there is only one energy state at the first excited state), g0 = 1 (since there is only one energy state at the ground state), E1 = 0.05 eV, k = 8.617 x 10^-5 eV/K, and T = 150 K.

Plugging these values into the formula, we get:

P(E1) = (1/1) * e^(-0.05/8.617 x 10^-5 * 150)

= 0.021

Therefore, the probability of the single oscillator being in its first excited state is 0.021 or approximately 2.1%.

It is important to note that this calculation assumes that the oscillator is in thermal equilibrium with the solid and follows the Boltzmann distribution. Additionally, this calculation only considers the single oscillator and does not take into account any interactions or effects from the surrounding oscillators in the solid.