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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? $\endgroup$ Dec 15, 2020 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. $\endgroup$ Dec 15, 2020 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. $\endgroup$
    – AChem
    Dec 15, 2020 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. $\endgroup$ Dec 15, 2020 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, 2020 at 9:13

1 Answer 1

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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: $$\pu{8.79E-25}\kappa^{2}\Phi_{D}n^{-4}J(\lambda)[mol^{-3}M^{-1}cm^{-1}nm^{4}]$$

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  • $\begingroup$ Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    Jun 14, 2022 at 6:44
  • $\begingroup$ Thanks for the insight! I used the 9000 form due to the fact that that was what was given in my primary source (Principles of Fluorescence Spectroscopy, 3rd Ed. by J.R. Jakowicz) - In section 13.2 the formula is shown (equation 13.4), and in the text it's indicated that $N$ is just "Avogadro's Number"... Why would he have used the $\text{mmol}^{-1}$ version over the normal one? $\endgroup$ Jun 15, 2022 at 18:01
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    $\begingroup$ To be honest I'm not sure, I'll have to go back and re-read Lakowicz. $\endgroup$
    – Mchiribo
    Jun 15, 2022 at 19:21

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