# 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 5 nm distance between the two chromophores. Estimate $R_o$ for the two chromophores, assuming the energy transfer efficiency is 0.7 at short distances.

In our lecture we are given the formula: $$[Eff] =\dfrac{R_0^6}{r^6 + R_0^6}$$

I understand that "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 $Eff_{max}$ is maximum efficiency of energy transfer):

$$r =R_0 \sqrt{\dfrac{Eff}{Eff_{max}}}$$

When I apply his formula I "plug in" the given numbers and solve for $R_0$ as following:

$$R_0 = \dfrac{r}{\sqrt{\dfrac{Eff}{Eff_{max}}}}$$

$$R_0 = \dfrac{5}{\sqrt{\dfrac{0.0185}{1}}}$$

$$R_0 = 9.72$$

Questions:

• 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?

So 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 "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 nominator 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:

$$\dfrac{Eff}{Eff_{max}} = \dfrac{\dfrac{R_0^6}{R_0^6}}{\dfrac{R_0^6}{R_0^6}+ \dfrac{r^6}{R_0^6}}$$

we then get as following: $$\dfrac{Eff}{Eff_{max}} = \dfrac{1}{1+ \dfrac{r^6}{R_0^6}}$$

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.

• 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