Find the wavelength of the photon emitted during the transition from the second excited state to the ground state in a harmonic oscillator with a...

Sunday, October 7, 2012

Find the wavelength of the photon emitted during the transition from the second excited state to the ground state in a harmonic oscillator with a...

First, compute the energy level for each state. The formula of energy level of a harmonic oscillator is:

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

where

n is the quantum number

h is the Planck's constant $\displaystyle{\left({6.625}{x}{{10}}^{{-{34}}}{J}{s}\right)}$ and

f is the frequency of the oscillator

At second excited state, the quantum number of harmonic oscillator is n=2. So its energy level at this state is:

$\displaystyle{E}_{{2}}={\left({2}+\frac{{1}}{{2}}\right)}{\left({6.625}\times{{10}}^{{-{34}}}{J}{s}\right)}{\left({3.72}\times{{10}}^{{13}}{H}{z}\right)}$

$\displaystyle{E}_{{2}}={6.126125}\times{{10}}^{{-{20}}}{J}$

At ground state, the quantum number of harmonic oscillator is n=0. So its energy level at this state is:

$\displaystyle{E}_{{0}}={\left({0}+\frac{{1}}{{2}}\right)}{\left({6.625}\times{{10}}^{{-{34}}}{J}{s}\right)}{\left({3.72}\times{{10}}^{{13}}{H}{z}\right)}$

$\displaystyle{E}_{{0}}={1.23225}\times{{10}}^{{-{20}}}{J}$

Then, determine transition energy from n=2 to n=0.

$\Delta E = E_2 - E_0$

$\Delta E = 6.126125 xx 10^(-20)J - 1.23225 xx 10^(-20)J$

$\Delta E = 4.929 xx 10^(-20) J$

So during the transition from n=2 to n=0, $\displaystyle{4.929}{x}{{10}}^{{-{20}}}{J}$ of energy is emitted. This is the energy of the photon emitted during the transition.

Energy of photon, $\displaystyle{E}={4.929}\times{{10}}^{{-{20}}}{J}$

To determine the wavelength of the photon, apply the formula of energy of photon.

$\displaystyle{E}={h}{f{}}$

where f is the frequency of light.

Since the frequency of light is $\displaystyle{f{=}}\frac{{c}}{\lambda}$ , the formula can be re-written as:

$\displaystyle{E}={h}\cdot\frac{{c}}{\lambda}$

where

c is the speed of light $\displaystyle{\left({3}\times{{10}}^{{8}}\frac{{m}}{{s}}\right)}$ and

$\displaystyle\lambda$ is the wavelength of photon

Isolating the wavelength, the formula becomes:

$\displaystyle\lambda=\frac{{{h}\cdot{c}}}{{E}}$

Plugging in the values, the wavelength will be:

$\displaystyle\lambda=\frac{{{\left({6.625}\times{{10}}^{{-{34}}}{J}{s}\right)}\cdot{\left({3}\times{{10}}^{{8}}\frac{{m}}{{s}}\right)}}}{{{4.929}\times{{10}}^{{-{20}}}{J}}}$