The energy of a particle in the n = 3 excited state of a harmonic oscillator. Potential is 5.45 eV. What is the classical angular frequency of oscillation of this particle?

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To solve, use the formula of energy level of harmonic oscillator.

`E_n = (n + 1/2)hf`


`E_n` is the energy level of harmonic oscillator in Joules

n is the quantum level

h is the Planck's constant `(6.623 xx 10^(-34) Js)`

and f is the frequency of oscillator.

To be able to apply this formula, convert the given energy to Joules. Take note that `1eV = 1.602xx10^(-19)J` .

`E_n=5.45 eV * (1.602xx10^(-19)J)/(1eV)`

`E_n = 8.7309xx10^(-19) J`

Plug-in this value of En to the formula of energy level of harmonic oscillator.

`E_n = (n+1/2)hf`

`8.8309 xx10^(-19) J= (3+1/2)(6.623 xx 10^(-34) Js)f`

`8.7309 xx10^(-19) J=(2.31805 xx10^(-33) Js) f`

Then, isolate the f.

`f = (8.7309 xx 10^(-19)J)/(2.31805xx10^(-33)Js)`

`f=3.76648476xx10^14``/ sec`

`f=3.76648476 xx 10^14Hz`

So the frequency of the oscillator is `3.76648476xx10^14Hz` .

To determine the angular frequency, apply the formula:

`omega = 2pif`

`omega =2pi * (3.76648476xx10^14 Hz)`

`omega=2.366552170 xx10^15` rad/s

Rounding off to two decimal places, it becomes:

`omega =2.37 xx10^15` rad/s


Therefore, the angular frequency of harmonic oscillator is `2.37xx10^15` radian per second.

Approved by eNotes Editorial Team

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