Question

In the given wave function, $\psi = \frac{1}{81} (\frac{2}{\pi})^{\frac{1}{2}} (\frac{1}{a_0})^{\frac{3}{2}} (6 - \frac{r}{a_0} - \frac{r^2}{a_0^2}) e^{-\frac{r}{3a_0}} \cos \theta$

Identify the correct choice(s).

• A. The given orbital is $3p_z$.
• B. The given orbital is $3d_{z^2}$.
• C. The nodal surface is at a radical distance of $6a_0$ from the nucleus.
• D. The number of angular nodes in the given orbital are 2.

Answer 1

Answer: (C)

The nodal surface is at a radical distance of $6a_0$ from the nucleus.

Answer 2

Solution:

The given wave function is

$$\psi = \frac{1}{81}\left(\frac{2}{\pi}\right)^{1/2}\left(\frac{1}{a_0}\right)^{3/2}\left(6 - \frac{r}{a_0} - \frac{r^2}{a_0^2}\right)e^{-\frac{r}{3a_0}}\cos\theta.$$

1. Angular Part:
   The factor $\cos\theta$ is the angular part. For hydrogen-like orbitals, $\cos\theta$ corresponds to the spherical harmonic $Y_1^0(\theta,\phi)$ which represents a $p_z$ orbital (with $l=1$). In contrast, a $3d_{z^2}$ orbital would have an angular part proportional to $(3\cos^2\theta - 1)$.

2. Radial Part and Radial Node:
   The radial part is given by

$$R(r) \propto \left(6 - \frac{r}{a_0} - \frac{r^2}{a_0^2}\right)e^{-\frac{r}{3a_0}}.$$

Setting the polynomial equal to zero to find the radial node:

$$6 - \frac{r}{a_0} - \frac{r^2}{a_0^2} = 0.$$

Multiply by $a_0^2$:

$$r^2 + r\,a_0 - 6a_0^2 = 0.$$

Let $x = \frac{r}{a_0}$; then

$$x^2 + x - 6 = 0.$$

Solving, we get:

$$x = \frac{-1 \pm \sqrt{1+24}}{2} = \frac{-1 \pm 5}{2}.$$

The positive root is:

$$x = 2 \quad\Rightarrow\quad r = 2a_0.$$

So the radial node is at $2a_0$ and not at $6a_0$.

3. Angular Nodes:
   Since $\cos\theta$ vanishes at $\theta = \pi/2$, there is 1 angular node. (For $l=1$ there is exactly one angular node.)