The wave function of 3s and 3p_z orbitals are given by:

${\mathrm{\Psi }}_{3\mathrm{s}}=\frac{1}{9\sqrt{3}}{\left(\frac{1}{4\mathrm{\pi }}\right)}^{1/2}{\left(\frac{\mathrm{Z}}{{\mathrm{a}}_0}\right)}^{3/2}\left(6-6\mathrm{\sigma }+{\mathrm{\sigma }}^2\right){\mathrm{e}}^{-\mathrm{\sigma }/2}$

${\mathrm{\Psi }}_{3{\mathrm{p}}_{\mathrm{z}}}=\frac{1}{9\sqrt{6}}{\left(\frac{3}{4\mathrm{\pi }}\right)}^{1/2}{\left(\frac{\mathrm{Z}}{{\mathrm{a}}_0}\right)}^{3/2}(4-\mathrm{\sigma }){\mathrm{\sigma e}}^{-\mathrm{\sigma }/2}\cos \mathrm{\theta }$

$\sigma =\frac{2\mathrm{Zr}}{{\mathrm{na}}_0}$

where a_{0 =} 1st Bohr radius, Z = charge number of nucleus, r = distance from conclude:

 From this, we can conclude :

Total nodal surface $=\mathrm{n}-1$

Total angular nodal surface + radial nodal surface = 2 for both orbital.
There is a finite probability of finding electron in nucleus in case of 3s orbital. Hence, 3s have more penetrating power than 3p_z.
At radial node $\psi =0$; the radial nodes in 3s and 3p_z are not at equal distance.