Why is the radial distribution at the nucleus 0 but not the radial wave function? The modulus squared of the radial wave function gives the probability of finding an electron in an infinitesimal volume dv. On the other hand, the radial distribution gives the probability of finding an electron in a shell of radius r and thickness dr, or in other words, the probability of finding an electron at a radius r.
I know how the radial distribution is derived and it being 0 at the nucleus makes sense mathematically but not physically. The radial distribution being 0 at the nucleus means that an electron cannot be found in a shell of r=0 &amp; thickness dr ( a sphere of radius dr), but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus. These results appear to contradict each other.
Is there something wrong in my physical interpretation of the 2 functions?
You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.'
To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised
$$\int_0^\infty R_{n\ell}^*(r)R_{n\ell}(R)r^2 dr=\frac{4}{a_0^3}\int_0^\infty e^{-2r/a_0}r^2 dr=1$$
where the $r^2$ comes from the volume element needed to integrate over all angles which is $r^2\sin(\theta)drd\theta d\phi$. The probability from $r\to r+dr$ is then for the 1s orbital $\displaystyle p=\frac{4}{a_0^3}r^2e^{-2r/a_0}dr$ whereas the 1s wavefunction itself has the form $\sim e^{-r/a_0}$. For other orbitals, e.g $2p_{\pm 1}$ the wavefunction is a function with complex numbers so cannot represent anything physical.