The wave function of the H atom have a number of nodal surfaces equal to the first quantum number. But these nodal surfaces are not always spheres, as you think. They can be planes or still different surfaces.
In general the quantum number $n$ gives the total number of nodal surfaces. The second quantum number $l$ gives the number of nodal non-spherical surfaces of the particular atomic orbital. As a consequence, $n -l$ gives the number of nodal spheres, that you are calling "minima" in your drawing.
Let's talk about the atomic orbitals with $n = 3$, as it is your desire. The $2$nd quantum number can be $0$, $1$, or $2$.
If $n = 3$ and $l = 0$, the orbital is $3s$ : there are one nodal sphere at infinite distance, and two nodal spheres near and around the nucleus. And there is only one such orbital, as the third number $m$ is equal to zero.
If $n = 3$ and $l = 1$, the orbital is called $3p$, and it is your problem. As the third number $m$ can be $-1$, $0$, or $+1$, there are three such orbitals $3p$, called $3p_x$, $3p_y$ $3p_z$. For each orbital, there is $l = 1$ nodal plane per orbital, and $n - l = 2$ spherical nodes in each orbital, one being at infinite distance, and one near the nucleus. There are three such orbitals, with the same number of spherical and planar nodes : one has a nodal plane perpendicular to the $Ox$ axis, the second perpendicular to the $Oy$ axis, and the third perpendicular to the $Oz$ axis.
If $n = 3$ and $l = 2$, the orbitals are called $3d$. As the third quantum number $m$ can have 5 values, being $-2$. $-1$. $0$, $+1$ or $+2$, it makes five $3d$ atomic orbitals. These $5$ orbitals have all $n - l = 1$ spherical node, which is at infinite distance from the nucleus. Remains two nodal surfaces per each one of these 5 orbitals. For the first three of these five orbitals, the $2$ non-spherical nodal surfaces are two mutually perpendicular planes, namely $Oxy - Oyz$, then $Oxy - Oxz$, and $Oyz - Oxz$. The two other $3d$ orbitals must have two nodal surfaces which are neither spheres, nor planes. They may be conical, but they are difficult to explain without drawing.
This long development explains why the distribution curve of $3p$ orbitals have two minima (one at ∞), so that your figures $3$ and $4$ are equally valid. To explain why Fig. 3 is to be preferred, you should use the radial wave function for $3p$ orbital, which is $$\ce{R(r) = constant · (4 - {r/a}) ({r/a}) e^{-{r/2a}}}$$ When calculating $\pu{R(r)}$ for different values of $\pu{r/a}$ going from $1$, to $2$, $3$, $4$, etc. up to $10$, you will find that $\pu{R(r)}$ takes the following values : $$ 1.82, 1.47, 0.67, 0.00, 0.41, 0.59, 0.63, 0.586, 0.50, 0.404$$ This shows clearly that the maximum value of $\pu{R(r)}$ must be near the nucleus, and before the minima observed at $\pu{r/a = 4}$. Now if you multiply this value by $\pu{r^2}$, to get your wanted distribution curve, you find that $\pu{R(r)·r^2}$ takes the following values : $$\pu{1.82, 5.89, 6.03, 0.00, 10.25, 21.24, 30.87, 37.5, 40.5, 40.4}$$ This shows clearly that your distribution curve has a maximum far away from the nucleus, and far away form the minima (observed at $\pu{r = 4a}$).
As a consequence, your curve $3$ has to be preferred.
Ans (3), source
