Question

Consider following Schrodinger wave equation for an orbital of hydrogen atom.
$(a_0$ is first Bohr radius)
$\Psi =\frac{\sqrt{2r}}{81a_0^2\sqrt{\pi a_0}}(6-\frac{r}{a_0})e^{\frac{-r}{3a_0}}cos\theta$

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(I\right)& \text{Which orbital is this?}& \left(P\right)& 1\\ \left(II\right)& \text{Number of total node(s)}& \left(Q\right)& 2\\ \left(III\right)& \text{Nodal plane}& \left(R\right)& 3p_z\\ \left(IV\right)& \text{Number of radial node(s)}& \left(S\right)& 3p_x\\ & & \left(T\right)& \text{XY plane}\\ & & \left(U\right)& \text{YZ plane}\end{array}$

Match the correct combination considering List-I and List-II

• A. (III),(T) and (IV),(P)
  
  
• B. (III),(U) and (IV),(P)
• C. (III),(T) and (IV),(Q)
• D. (III),(U) and (IV),(Q)

Answer 1

The correct option is A (III),(T) and (IV),(P)



(III) Since the orbital is $3p_z$, so the nodal plane is $XY$ plane.
(IV) Number of radial nodes = $n-l-1=3-1-1=1$