Probability density versus radial distribution function For a 1s orbital the radial probability density is maximum at the nucleus while the radial distribution function is zero at the nucleus, while the maximum radial distribution function is maximum at a particular distance from the nucleus. so my question is where is the maximum probability of finding an electron?.Is it at the nucleus or where the radial distribution function is maximum?
 A: You are confused between the radial part of the eigenfunctions and the radial probability density.
For the 1s level of a hydrogen atom, the eigenfunction is
$$\psi(r,\theta, \phi) = \frac{1}{2\pi}a_0^{-3/2} \exp(-r/a_0) $$
and there is no angular dependence.
But when you want to work out a probability density $P(r)$ for the electron to be found between $r$ and $r +dr$, then you need to consider an integral of the square of the modulus of the eigenfunction over the volume enclosed by the spherical shell between $r$ and $r+dr$ and this volume is $4\pi r^2\ dr$.
In other words, the (unnormalised) probability density as a function of radius
$$ P(r) = 4\pi r^2 \psi(r) \psi^{*}(r)$$.
So whilst $\psi(r)$ peaks at the origin, $P(r)$ is zero at the origin. To work out at what radius the electron is most likely to be you look for a maximum in $P(r)$.
A: The orbitals are just probability distributions in the spherical coordinates. You need three coordinates to situate the electron, two angles $\theta$ and $\phi$ and a radius $r$. 
The probability of finding the electron at a given angle (both $\theta$ and $\phi$) is given by the so called "orbital". For the 1s this "orbital" is a sphere. This means that no matter the angle you choose, the probability for finding the electron is the same. It is just given by the length of the vector that goes from the center of the coordinate system to the surface of the "orbital" with angles $\theta$ and $\phi$ . Now that we have the angles we need to know the radius. To do so we go to the radial distribution function. As you say it has a maximum at some point $r_0$. Therefore:


*

*For the 1s orbital the probability of finding the electron at a position $(\theta,\phi,r)$ is the same for all angles and maximum for $r=r_0$.


Note: To be more rigorous, what you need to find is the value of the integral of the angular and radial distributions between the values of $(\theta\pm d\theta,\phi\pm d\phi,r\pm dr)$ you want to find the electron in.
