The Schrodinger wave equation for H-atom is
`nabla^(2) Psi = (8pi^(2)m)/(h^(2)) (E-V) Psi = 0`
Where `nabla^(2) = (del^(2))/(delx^(2)) +(del^(2))/(dely^(2)) +(del^(2))/(delz^(2))`
E = Total energy and V=potential energy wave function `Psi_(((r, theta,phi)))R_((r))Theta_((theta))Phi_((phi))`
R is radial wave function which is function of ''r'' only, where r is the distance from nucleus. `Theta` and `Phi` are angular wave function. `R^(2)` is known as radial probability density and `4pir^(2)R^(2)dr` is known as radial probability function i.e., the probability of finding the electron is spherical shell of thickness dr.
Number of radial node =n -l - 1
Number of angular node = l
For hydrogen atom, wave function for 1s and 2s-orbitals are:
`Psi_(1s) = sqrt((1)/(pia_(0)^(a)))e^(-z_(r)//a_(0))`
`Psi_(2s) = ((Z)/(2a_(0)))^(½) (1-(Zr)/(a_(0)))e^(-(Zr)/(a_(0)))`
The plot of radial probability function `4pir^(2)R^(2)` aganist r will be:

Answer the following questions:
What will be number of angular nodes and spherical nodes for 4f atomic orbitals respectively.

Options

A. 0,0
B. 1,3
C. 3,0
D. 0,3

Answer

C

Explanation

no of angular nodes = 1