# Radiationless energy transfer: how to arrive at the equation that give R0 between two chromophores?

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): $$r = R_0 \sqrt{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}$$

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{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}}\\ R_0 &= \frac{\pu{5 nm}}{\sqrt{\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?

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: $$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$
• 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! – porphyrin Jun 29 '16 at 13:54