I Confused about unit conversion involving natural units

AI Thread Summary
The discussion revolves around converting units for an atomic transition expression, specifically from natural units (Joules) to wavenumber (cm^-1) and the implications of using different constants. The user is confused about converting a term from 1/A (angstrom) to the required units, noting that multiplying by 10^8 would convert to 1/cm. There is a debate on whether to use hc instead of ħc for the conversion, with clarification that spectroscopists define wavenumber as 1/λ rather than k = 2π/λ. The conversation emphasizes the importance of making implicit conversion factors explicit to avoid errors in calculations.
kelly0303
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Hello! I have an expression whose natural units are Joules, but all the terms are expressed in terms of cm##^{-1}## (it is for an atomic transition). I have a term in the expression whose units are 1/A (angstrom) and I am not sure how to convert it to what I need. On one hand, if I were to go from 1/A to 1/cm, I would need to multiply that term by ##10^8##. On the other hand, if I want to convert to J, I need to multiply by ##\hbar## c, then convert from J to cm##^{-1}##, which gives ##15927759.569##. The difference between the 2 approaches is ##2\pi##, but I am not sure why. Should I actually use hc instead of ##\hbar##c? The issue is that the formula is defined using ##\hbar## and I am not sure what I am doing wrong. For reference I am talking about equation 1 in this paper (##\alpha_5##, ##A_1## and ##A_2## are unitless). Thank you!
 
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You have implicit conversion factors somewhere. You need to make them explicit. (Well, you don't need to, but the chances of screwing up are smaller if you do)
 
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kelly0303 said:
On one hand, if I were to go from 1/A to 1/cm, I would need to multiply that term by ##10^8##.
Correct.
kelly0303 said:
Should I actually use hc instead of ##\hbar##c?
Yes. What spectroscopists refer to as wave number is ## 1/\lambda ##, not ## k = 2 \pi / \lambda ##, as theorists often do. In terms of energy (## E=hc/\lambda ##), ##\rm 1 ~eV = 8065.5~cm^{-1} = 1.6022 \times 10^{-19}~J##.
 
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