This link describes a method for determining the most probable radius of an electron for a Hydrogen atom in the ground state.
It states that :
The radial probability density for the hydrogen ground state is obtained by multiplying the square of the wavefunction by a spherical shell volume element.
When I went to solve this problem myself I multiplied the square of the given wavefunction by the volume of a sphere, which gave me the wrong answer as I know it should be the Bohr radius.
When I thought about this problem it seemed reasonable to multiply it by the volume of a sphere rather than the surface area of a sphere (my gut feeling). The link doesn't really explain to me why it uses a surface area and not a volume of a sphere and I would like some help in getting an intuition for why the surface area is used and gives the right answer and the volume of a sphere does not.
Best Answer
Note: ChocoPouce's answer is the same as mine but is more mathematical.
You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this distribution is in at the $0.1\mathrm{nm}$ shell? We won't exactly $0.1\mathrm{nm}$, so we take a thin shell: how much is between $0.1$ and $0.1+ε\mathrm{nm}$? Amount $=$ (average prob. density in shell)*(volume of shell). Since the shell is so thin ($\varepsilon$ is very small), the density will be almost constant and the shell's volume is given by $\mathrm{area}\times\mathrm{thickness} = 4\pi*(0.1nm^2)*\varepsilon nm$.
Thus the probability is $4\pi\rho(r)r^2\varepsilon$, where $\rho$ only depends on $r$ since it is spherically symmetric. But $\varepsilon$ is an arbitrary "small" thickness we defined, and it is best to divide by $\varepsilon$ to get the probability per unit radius, which is called the (radial) probability density $4\pi\rho(r)r^2$. The most likely radius (which is different from the average radius) maximizes this function.