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:
- \( n = 3 \) (principal quantum number)
- \( l = 2 \) (azimuthal quantum number)
- \( m = 1 \) (magnetic quantum number)

### Step 2: Determine the angular momentum
The formula for the angular momentum \( L \) of an orbital is given by:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
Substituting \( l = 2 \):
\[ L = \sqrt{2(2 + 1)} \cdot \frac{h}{2\pi} = \sqrt{6} \cdot \frac{h}{2\pi} \]

### Step 3: Calculate the 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 4: Determine the angular nodes
The number of angular nodes is equal to \( l \):
\[ \text{angular nodes} = l = 2 \]

### Step 5: Sum the values
Now we can sum the angular momentum, spherical nodes, and angular nodes:
\[ \text{Sum} = L + \text{spherical nodes} + \text{angular nodes} \]
Substituting the values we have:
\[ \text{Sum} = \sqrt{6} \cdot \frac{h}{2\pi} + 0 + 2 \]

### Final Answer
Thus, the final expression for the sum of angular momentum, spherical nodes, and angular nodes is:
\[ \text{Sum} = \sqrt{6} \cdot \frac{h}{2\pi} + 2 \]