Problem calculating the ground state radius of Toponium using strong force potential for short ranges If it were true that the top quark were stable we could, in theory, have a hydrogen-like bound state consisting of toponium $T \bar{T}$. Looking at the strong force, for small ranges it takes the form of a coulomb potential:
$$
V(r) = -\frac{4\alpha_s}{3r}
\tag1
$$
(Where $\alpha_s = 0.1 $)  Now I'm trying to find the radius of the ground state using the (knowingly outdated) Bohr model and so the process is not necessarily unique for toponium, however I am going wrong somewhere and I'm not sure how. So using the Bohr model we equate the position derivative of potential to the centrifugal force as followed:
$$
\frac{d}{dr}(-\frac{4\alpha_s}{3r}) = \frac{\mu v^2}{r}
\tag2
$$
Where $\mu$ is the reduced mass, which, for top-quarks of $m_t = 180GeV$, is $\mu = 90GeV$. Following the equation through we arrive at
$$
\mu v^2r = \frac{4\alpha_s}{3}
\tag3
$$
Now, the velocity of the top quark is not known so we use the knowledge that since the allowed orbits are quantised and are essentially standing waves, we can say they have a circumference  equal to an integer number of De Broglie wavelengths $\lambda$
$$
\lambda = \frac{h}{mv} 
$$
$$
 2\pi r = n \lambda 
$$
Combining gives us
$$
 mv^2r = \frac{n^2h^2}{4\pi^2 rm} = \frac{n^2 \hbar^2}{rm}
\tag4
$$
Equating this to $(3)$ thus gives:
$$
\frac{n^2 \hbar^2}{rm} = \frac{4\alpha_s}{3}
$$
Rearranging for $r$:
$$
r = \frac{3}{4\alpha_s}\frac{\hbar^2}{m}
$$
However when I plug numbers into this in SI units (i.e after converting GeV to kg), I do not get sensible results. (I.e $10^{-47}$ ) if I multiply the equation by $\frac{1}{\hbar c}$ however I get the result I am supposed to get of $1.6\times10^{-16}\mathrm m$. Why do I have to multiply by this factor? Any help would be appreciated.
 A: Interpreted as SI units, your  equation (1) is not imensionally correct. $V(r)$ has dimensions of energy ${\rm ML^2 T^{-2}}$ while  $r^{-1}$ has ${\rm L^{-1}}$ and you are taking $\alpha_s$ to be dimensionless. This is because $\alpha_s=0.1$ is true only   in natural units in which which $\hbar=c=1$. The division by   $\hbar^2 c$, expressed in SI, units presumably (I have not worked through the conversion) fixes the units.
A: If you suspect a unit conversion error, the problem-solving strategy is to check the units in every step. This is particularly important if you are working in “natural units” with $\hbar = c = 1$.  There are people who insist that natural units are great because you can ignore all of the $\hbar$ and $c$ and stick them in the right places at the very end of your computation; those people make fewer algebra mistakes than I do, because that almost never works out of me.
Your first unit error (or your first implicit use of natural units) is that your potential has units of inverse length rather than energy.
When I’m doing these kinds of computations I usually carry around factors of $mc^2$, with units of energy (“as you know from childhood,” one of my professors liked to say), and factors of $\hbar c$, with unit $\text{energy}\times\text{length}$.  Your final result should be of the form
$$
\text{length} = (\text{dimensionless})\times\frac{\hbar c}{mc^2}
\tag1$$
Other people like to cancel out the “redundant” factors of $c$, but I make more mistakes that way.
Note that the difference between your result and the correct form (1) is not the factor you claim in your question, so you aren’t done finding algebra errors. (Or perhaps SI unit-conversion errors; you could test this by also computing the Bohr radius $a = \frac1\alpha\frac{\hbar c}{m_ec^2} = 10^{-10}\rm\,m$.)
