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