Blog: 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...

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

$\displaystyle{E}_{{n}}={\left({n}+\frac{{1}}{{2}}\right)}{h}{f{}}$

where

$\displaystyle{E}_{{n}}$ is the energy level of harmonic oscillator in Joules

n is the quantum level

h is the Planck's constant $\displaystyle{\left({6.623}\times{{10}}^{{-{34}}}{J}{s}\right)}$

and f is the frequency of oscillator.

To be able to apply this formula, convert the given energy to Joules. Take note that $\displaystyle{1}{e}{V}={1.602}\times{{10}}^{{-{19}}}{J}$ .

$\displaystyle{E}_{{n}}={5.45}{e}{V}\cdot\frac{{{1.602}\times{{10}}^{{-{19}}}{J}}}{{{1}{e}{V}}}$

$\displaystyle{E}_{{n}}={8.7309}\times{{10}}^{{-{19}}}\ldots$

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

$\displaystyle{E}_{{n}}={\left({n}+\frac{{1}}{{2}}\right)}{h}{f{}}$

where

$\displaystyle{E}_{{n}}$ is the energy level of harmonic oscillator in Joules

n is the quantum level

h is the Planck's constant $\displaystyle{\left({6.623}\times{{10}}^{{-{34}}}{J}{s}\right)}$

and f is the frequency of oscillator.

To be able to apply this formula, convert the given energy to Joules. Take note that $\displaystyle{1}{e}{V}={1.602}\times{{10}}^{{-{19}}}{J}$ .

$\displaystyle{E}_{{n}}={5.45}{e}{V}\cdot\frac{{{1.602}\times{{10}}^{{-{19}}}{J}}}{{{1}{e}{V}}}$

$\displaystyle{E}_{{n}}={8.7309}\times{{10}}^{{-{19}}}{J}$

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

$\displaystyle{E}_{{n}}={\left({n}+\frac{{1}}{{2}}\right)}{h}{f{}}$

$\displaystyle{8.8309}\times{{10}}^{{-{19}}}{J}={\left({3}+\frac{{1}}{{2}}\right)}{\left({6.623}\times{{10}}^{{-{34}}}{J}{s}\right)}{f{}}$

$\displaystyle{8.7309}\times{{10}}^{{-{19}}}{J}={\left({2.31805}\times{{10}}^{{-{33}}}{J}{s}\right)}{f{}}$

Then, isolate the f.

$\displaystyle{f{=}}\frac{{{8.7309}\times{{10}}^{{-{19}}}{J}}}{{{2.31805}\times{{10}}^{{-{33}}}{J}{s}}}$

$\displaystyle{f{=}}{3.76648476}\times{{10}}^{{14}}$ $\displaystyle/{\sec{}}$

$\displaystyle{f{=}}{3.76648476}\times{{10}}^{{14}}{H}{z}$

So the frequency of the oscillator is $\displaystyle{3.76648476}\times{{10}}^{{14}}{H}{z}$ .

To determine the angular frequency, apply the formula:

$\displaystyle\omega={2}\pi{f{}}$

$\displaystyle\omega={2}\pi\cdot{\left({3.76648476}\times{{10}}^{{14}}{H}{z}\right)}$

$\displaystyle\omega={2.366552170}\times{{10}}^{{15}}$ rad/s

Rounding off to two decimal places, it becomes:

$\displaystyle\omega={2.37}\times{{10}}^{{15}}$ rad/s

Therefore, the angular frequency of harmonic oscillator is $\displaystyle{2.37}\times{{10}}^{{15}}$ radian per second.

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