In `psi_(321)`, the sum of angular momentum, spherical nodes and angular node is

Options

A. `(sqrt(6)h + 4pi)/(2 pi)`
B. `(sqrt(6)h)/(2pi) + 3`
C. `(sqrt(6h) + 2pi)/(2pi)`
D. `(sqrt(6h) + 8pi)/(2pi)`

Answer

To solve the question regarding the sum of angular momentum, spherical nodes, and angular nodes for the orbital represented by \( \psi_{321} \), we will follow these steps:

### Step 1: Identify the quantum numbers
The notation \( \psi_{321} \) corresponds to the quantum numbers:
- \( n = 3 \) (principal quantum number)
- \( l = 2 \) (azimuthal quantum number)
- \( m = 1 \) (magnetic quantum number)

### Step 2: Determine the type of orbital
Given \( l = 2 \), this indicates that we are dealing with a d-orbital. The specific orbital is identified by the magnetic quantum number \( m = 1 \), which corresponds to the \( d_{yz} \) or \( d_{xz} \) orbital. However, for this question, we will focus on the calculations rather than the specific orbital shape.

### Step 3: Calculate angular momentum
The formula for angular momentum \( L \) is given by:
\[ L = \sqrt{l(l + 1)} \frac{h}{2\pi} \]
Substituting \( l = 2 \):
\[ L = \sqrt{2(2 + 1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi} \]

### Step 4: Calculate spherical nodes
The formula for the number of spherical nodes is:
\[ \text{spherical nodes} = n - l - 1 \]
Substituting \( n = 3 \) and \( l = 2 \):
\[ \text{spherical nodes} = 3 - 2 - 1 = 0 \]

### Step 5: Calculate angular nodes
The number of angular nodes is equal to the azimuthal quantum number \( l \):
\[ \text{angular nodes} = l = 2 \]

### Step 6: Sum the values
Now we will sum the values calculated:
- Angular momentum: \( \sqrt{6} \frac{h}{2\pi} \)
- Spherical nodes: \( 0 \)
- Angular nodes: \( 2 \)

Thus, the total sum is:
\[ \text{Total} = \sqrt{6} \frac{h}{2\pi} + 0 + 2 = \sqrt{6} \frac{h}{2\pi} + 2 \]

### Final Answer
The sum of angular momentum, spherical nodes, and angular nodes for \( \psi_{321} \) is:
\[ \sqrt{6} \frac{h}{2\pi} + 2 \]