The wave function of 3s and `3p_(z)` orbitals are given by :
`Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6=6sigma+sigma)e^(-sigma//2)`
`Psi_(3s_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0,`
`sigma=(2Zr)/(nalpha_(0))`
where`alpha_(0)=1st` Bohr radius , Z= charge number of nucleus, r= distance from nucleus.
From this we can conclude:

Options

A. Total number of nodal surface is same for 3s and `3p_(x)` orbitals
B. The angular nodal surface of `3p_(z)` orbital occur at `0=(pi)/2`
C. The radial nodal surface of 3s and `3p_(z)` orbitals are at equal distance from nucleus.
D. 3s electrons have greater penetrating power into the nucleus compared to `3p_(z)` electron.

Answer

a b d