What are the radial probability densities, and what is its relation with $R_{nl}(r)$? I have read the pages suggested as similar and not found the answer to my question, or if it's there I didn't understand it. I have seen radial probability density described on various different webpages as:
$$r^2 R_{nl}^2(r)$$
$$4 \pi r^2 R_{nl}^2(r)$$
$$R_{nl}^2(r)$$
Where $R_{nl}(r)$ is the radial wavefunction. I've seen that the $4\pi$ factor difference between the first two definitions doesn't affect a calculation of, for example, most probable radius but will make a difference to my calculation of the probability of finding an electron inside the nucleus of a ground state hydrogen atom. I've seen suggestions that the $4\pi$ factor is related to normalisation. I'm using a normalised wavefunction so which expression for probability density is appropriate? Can someone explain why there are different versions? 
 A: Those factors usually come out when separating the wave-function for a 3-dimensional problem (like the Hydrogen atom) in its radial and angular parts:
$$ \psi(r, \theta, \phi) = R(r) Y(\theta, \phi). $$
If you are then interested in the probability of finding the particle (or whatever you are studying) at a distance from the origin between $r$ and $r+dr$ around the solid angle characterized by the angles $\theta$ and $\phi$, you have to integrate $\lvert\psi\rvert^2$ between $r$ and $r+dr$.
In spherical coordinates this reads
$$ \lvert\psi(r,\theta,\phi)\rvert^2 \hspace{-8pt}\underbrace{d^3r}_{r^2\sin\theta drd\theta d\phi} \hspace{-10pt}
= r^2R(r)^2 \lvert Y(\theta,\phi) \rvert^2 \,\,\sin(\theta) \,\,dr d\theta d\phi $$
hence the term $r^2 R^2(r)$.
If the problem at hand is spherical, it can be more interesting to ask what is the probability of the particle being found in the spherical shell between the radii $r$ and $r+dr$, regardless of the solid angle.
This amounts to integrate the above in the whole solid angle, which if there is no angular variation of $Y$ (that is, if $\psi(r,\theta,\phi)=\psi(r)$), gives $4\pi$:
$$ \operatorname{P}(r_0 \le r \le r_0 + dr) = 4\pi r_0^2 R^2(r_0) dr $$
A: The radial part of the Schroedinger equation is solved using $u(r) \equiv R(r) r$. Here, the $R(r)$ is the actual radial part of the wavefunction. Hence the $r^2R_{nl}^2$ comes from that. The substitution is made so that it looks analogous to the classical energy equation with a central potential(effective). The $4\pi$ is most probably the result of the integration of the angular part of spherical polar co-ordinates. So when you are required to use radial probability density for your calculations, you should use $R_{nl}^2$
