Question

What is the sum of radial and angular nodes in the following orbitals of H-atom?
$\left(I\right){\Psi }_{2p_x}$
$\left(II\right){\Psi }_2$
$\left(III\right){\Psi }_{3d_x}$
$\left(IV\right){\Psi }_{3d_{x^2-y^2}}$

Answer 1

Answer: 6

Radial nodes $=n-l-1$
$\Rightarrow \left(I\right)n=2,l=1$,radial node $=2-1-1=0$
Angular node $=1\left(YZ\text{plane}\right)$
$\left(II\right)n=2,l=0$, Radial node $=2-0-1=1$
Angular node $=0$
$\left(III\right)n=3,l=2$, Radial node $=3-2-1=0$, and
Angular node $=2$ ($XY$ and $YZ$ planes)
$\left(IV\right)n=3,l=2$, Radial node $=0$, and Angular nodes $=2$
Total $=I+II+III+IV=1+1+2+2=6$