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 4.03mum m but I keep getting 8.06mum can I really just be forgetting to divide by two somewhere?

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 .