Find units of transition dipole moment

[sshrestha]

I have been trying to figure out which formula for the relationship between oscillator strength and transition dipole moment is correct.

Wiki had the following formula:

http://photonicswiki.org/index.php?title=Transition_Dipole_Moment

f = 4.703 \times 10^{29} \bar{v} \mu^{2}_{ge}\,\!

where

\bar{v}\,\! is the mean absorption frequency of the band in cm–1

\mu^{2}_{ge}\,\!, refers to the square of the transition dipole moment between the ground state and the excited state.

What I am unsure about is how they got the value of the constants and the units for it.

In another reference I had this formula:

f =[(8*pi*massofelectron)/(3*planck's const*electric)charge^2)] *vbar*u^2

When I calculate the constant using this formula I get 4.49 * 10^41. Moreover I am also confused about whether vbar is frequency or wavenumber. Since wavenumber is in cm^-1 the units don't cancel out and f is no longer dimensionless. Has anyone done this kind of calculation?

The values that I have in order to calculate the transition dipole moment is energy and f.

I use E = Planck's const*frequency to get frequency and then c= frequency*wavelenth to get the wave number and plug it in the equation. In either case the value I get is very off. Can you please help me figure out where I am missing the point?

 

[azntoon]

The correct formula for the relationship between oscillator strength and transition dipole moment is:f = [ (2*π*m*e)/(3*h*ε₀) ] *νₐᵣ*μ²ᵍᵉwhere:f is the oscillator strength,m is the mass of an electron,e is the charge of an electron,h is Planck's constant,ε₀ is the permittivity of free space,νₐᵣ is the mean absorption frequency in cm⁻¹, andμ²ᵍᵉ is the square of the transition dipole moment.