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

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 $= I + II + III+ IV= 1 + 1 +2 + 2 = 6$