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For my project on Fluorescence, I'm trying to simplify the equation for the Forster distance (I obtained this from the Lakowicz book). To keep track, the units are in square brackets.

$$R_0 = \left(\frac{9000\kappa^2\Phi_D\ln(10)}{128\pi^5N_An^4}J(\lambda)\right)^{1/6} \approx 0.211 (\kappa^2n^{-4}\Phi_DJ(\lambda))^{1/6}\, [Å]$$ I'm using $J(\lambda)$ in units of $M^{-1}cm^{-1}nm^4$ but for some reason I can't seem to obtain the $0.211$ no matter how I do the conversion. I always seem to be off by a factor of $10^{-3}$, which I can't seem to locate.

For reference, $J(\lambda)$ is the overlap integral: $$J(\lambda) = \frac{\int_0^{\infty} F_D(\lambda)\varepsilon_A(\lambda)\lambda^4 d\lambda}{\int_0^{\infty} F_D(\lambda) d\lambda}$$

For example, I simplified $R^6_0$, attempting to convert to Angstrom:

$$8.79\times10^{-25}\kappa^2\Phi_Dn^{-4} J(\lambda) [mol\,M^{-1}\,cm^{-1}\,nm^4]$$ $$=8.79\times10^{-25}\kappa^2\Phi_Dn^{-4} J(\lambda) [0.1m^210^{-36}m^4]$$ $$=R^6_0 = 8.79\times10^{-62}\kappa^2\Phi_Dn^{-4}J(\lambda) [m^6] = 8.79\times10^{-2} \kappa^2\Phi_Dn^{-4}J(\lambda)\,[Å]^6$$

Which, upon taking the 6th-root, I don't get the number $0.211$. If the power is $10^{-5}$ I do get the number.

What am I missing? Any help appreciated. I'm not sure if I'm overlooking something painfully obvious, but I can't see it.

Cheers!

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    $\begingroup$ Might be of use - in this publication, there is a 9 where you have a 9000. Haven't worked through the rest and not sure if your expression is equivalent to their eq (2), p. 5, but might be a clue? $\endgroup$ – Todd Minehardt Dec 15 '20 at 20:06
  • $\begingroup$ I did indeed wonder about that. I'm not 100% sure where the prefactor is derived, and most sources (including the lakowicz) appear to use 9000, and then have the 0.211 as well. $\endgroup$ – Frankie S. Palmer Dec 15 '20 at 20:17
  • $\begingroup$ @FrankieS.Palmer, Your numbers seem to be right, a back of the envelope calculation also results in 0.0879, which is same as yours. However, other publications also use 9000. It has have a power 10^-5 in order to get 0.211. It means there is an extra hidden factor of 1000 in the numerator. $\endgroup$ – M. Farooq Dec 15 '20 at 20:58
  • $\begingroup$ @M.Farooq Yeah, as mentioned before by Todd, if the numerator is simply a 9 then I get the right result- the issue is now more a case of why that's the case- as several sources (the Lakowicz book included) appear to give the 9000 as I have, but with the correct 0.211 result, too. $\endgroup$ – Frankie S. Palmer Dec 15 '20 at 21:43
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    $\begingroup$ what I meant was that as $\epsilon$ is in $dm^3/mol/cm$ but we work in SI then there is a factor of $10^{-5}$ in J . $\endgroup$ – porphyrin Dec 16 '20 at 9:13

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