I have been looking for explanation,how to draw angular part of wave function and I have found in physicsforums.com/archive answer to my question by user Gokul43201.

He posted this:

The trick is in being able to draw polar plots roughly, from the spherical harmonics.

Consider the state, n=2, l=1, m=0

The polar part of the wavefunction is given by :

|Y_1^0|^2 = \frac{3}{4 \pi} cos^2 \theta

Ignoring the constant term, the functional behavior is cos ^2 \theta , which we want to plot against \theta

Let's draw the 2D version of this plot (or you can simply graph this using Mathematica or a calculator that can do polar plots). Draw the X (theta = 90, -90) and Z (theta = 0, 180) axes, and draw a large number of lines, all passing through the origin - like the spokes on a bicycle wheel. These lines represent the different values of theta. Now on each line, place a point at a distance (from the origin) given by cos^2 \theta . Finally, join all these points, neighbor to neighbor. Remember, the X-axis represents \theta = 90 and the Z axis represents \theta = 0

At \theta =90,~ cos^2 \theta = 0. So, the points along the horizontal spokes are at the origin. At \theta increases, or decreases, cos ^2\theta increases till it reaches maxima at \theta = 0, 180 . So, for spokes above and below the X-axis, the points move farther and farther out, reaching a maximum at the +/-Z-axis. Join these points and you'll find it looks like a vertical dumbell oriented along the Z-axis. So, clearly this is the 2p_z orbital. While this is not all of it (you must now combine the radial part with the polar part), and the best way involves using some 3D plot software, I can't really do that here.

For the s-orbitals, the total wavefunction is something like |\psi _n(r, \theta \phi)|^2 ~=~A r^{2n}e^{-2r/na_0} . Since these orbitals have no polar (\theta) or azimuthal (\phi) dependence, they are spherically symmetric, and all the "equipotential" surfaces are spheres.

If this was too confusing, I'll try and attach a picture, when I find a little more time.

PS : Not more clever...maybe a little more experienced, that's all !

And I would like to request,if somebody could post some picture to this answer,because I don t know to imagine this...

Thanks a lot