Help with Forster distance conversion for FRET

Question

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!

Answer 1

A relevant discussion of the 9000-form vs 9-form can be found in section 3.4 of Förster Resonance Energy Transfer-From theory to application edited by Medintz and Hildebrandt and discussion in Braslavsky, et. al., 2008.

I think the confusion might be stemming from how Förster used $N'$ to represent the number of particles per mole instead of $N_{A}$.

Both ($N'=\pu{6.02E20mmol^{-1}}$) and ($N_{A}=\pu{6.02E23mol^{-1}}$) represent equivalent numbers of particles per mole. But, it is improper to use $N_{A}$ with the 9000-form.

For example the first term should be $\frac{9ln(10)}{128\pi^{5}(6.02*10^{23})}$ which is $\pu{8.79E-22}$.

I think you may be inadvertently using $\frac{9000ln(10)}{128\pi^{5}(6.02*10^{23})}$ which is why you are getting $\pu{8.79E-25}$.

Otherwise I think you would need to correct your units by using: $$\pu{8.79E-25}\kappa^{2}\Phi_{D}n^{-4}J(\lambda)[mol^{-3}M^{-1}cm^{-1}nm^{4}]$$