I have a homework question where I should calculate $R_0$ of an energy transfer between two chromophores. However, I don't understand how my professor arrived at the equation that we should apply (which is important to understand for the exam).

In a radiationless energy transfer experiment (RET) it was observed that the efficiency of transfer was $0.0185$ at $\pu{5 nm}$ distance between the two chromophores. Estimate $R_0$ for the two chromophores, assuming the energy transfer efficiency is $0.7$ at short distances.

In our lecture we are given the formula:
$$[E_\mathrm{eff}] = \frac{R_0^6}{r^6 + R_0^6}$$

I understand that $E_\mathrm{eff}$ is the efficiency of the transfer, and $r$ is the non-variable distance between the two chromophores. However, I am not able to solve for $R_0$. I tried to ask my professor but he thinks faster than I do and skips steps. So I was not able to follow completely, but he somehow mathematically solved the equation to (where $Eff_\mathrm{max}$ is maximum efficiency of energy transfer): $$r = R_0 \sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}$$

When I apply his formula I "plug in" the given numbers and solve for $R_0$ as following: \begin{align} R_0 &= \frac{r}{\sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}}\\ R_0 &= \frac{\pu{5 nm}}{\sqrt[6]{\frac{0.0185}{1}}}\\ R_0 &= \pu{9.72 nm} \end{align}

I would like to understand how he arrived at the final formula that I apply. Is it correct to solve for $R_0$ like I have shown above?


2 Answers 2


I think I found the answer to my own question through some help from some classmates. What the professor did, was to solve the equation mathematically by converting $R_0^6$ to $1$.

So first, the efficiency of the transfer $E_\mathrm{eff}$ is actually a ratio of the actual transfer efficiency divided by the maximum efficient (which we should assume is "one", unless we are told otherwise.

The next step is a bit hard to explain, as I am not that good in math, but I will try my best. What I understand is that you can "gather" the two $R_0^6$ by multiplying both the numerator and denominator by what we should think of as the number one, but we actually divide by $R_0^6$ and in this way we don’t have to include $R_0^6$ that we divide by on both sides of the equal sign - as it is technically the number one (mathematically allowed), as following: $$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{\frac{R_0^6}{R_0^6}}{\frac{R_0^6}{R_0^6}+ \frac{r^6}{R_0^6}}$$

We then get as following: $$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$

So by applying some math and assumptions we can solve the equation like this, and what I have explained in my question should therefore be correct.

  • $\begingroup$ As you have r = 5nm and the efficiency at the same distance using your first equation only $R_0$ is unknown. The 0.7 bit seems to be a red herring! $\endgroup$
    – porphyrin
    Commented Jun 29, 2016 at 13:54

This is a very late reply.

The transfer rate constant is defined as


where $\chi$ is an orientation parameter, (assume $\chi^2=2/3$ in solution) and $k_f$ is the fluorescence decay rate constant of the donor molecule. The transfer efficiency (or quantum yield) is


from which with some rearranging becomes


The reason for the form of the efficiency equation is that only two processes are assumed from the excited state, fluorescence and energy transfer and the yield is the fraction going to energy transfer relative to the total of all possible events, i.e. fluorescence and energy transfer.


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