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 classical frequency of 3.72 x 10^13 Hz. My buddy says it equals 4.03`mum` m but I keep getting 8.06`mum` can I really just be forgetting to divide by two somewhere?  

Expert Answers

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

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

where

n is the quantum number

h is the Planck's constant `(6.625 x 10^(-34)Js)` 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:

`E_2= (2+1/2)(6.625 xx 10^(-34)Js)(3.72xx10^13 Hz)`

`E_2=6.126125 xx 10^(-20) J`

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

`E_0= (0+1/2)(6.625xx10^(-34)Js)(3.72xx10^13 Hz)`

`E_0=1.23225 xx 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, `4.929 x 10^(-20) J` of energy is emitted.  This is the energy of the photon emitted during the transition.

Energy of photon, `E = 4.929 xx 10^(-20) J`

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

`E=hf`

where f is the frequency of light.

Since the frequency of light is  `f = c/ lambda` , the formula can be re-written as:

`E = h * c/ lambda`

where 

c is the speed of light  `(3 xx10^8 m/s)` and

`lambda` is the wavelength of photon

Isolating the wavelength, the formula becomes:

`lambda = (h*c)/E`

Plugging in the values, the wavelength will be:

`lambda = ((6.625xx10^(-34)Js)*(3xx10^8 m/s))/(4.929xx10^(-20)J)`

`lambda= 4.03 xx 10^(-6) m`

`lambda =4.03 mum`

 

Therefore, the wavelength of the photon emitted is `4.03 mum` .

Approved by eNotes Editorial Team

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