The position and energy of an electron is specified with the help of four quantum numbers namely, principal quantum number (n), azimuthal quantum number (l), magnetic quantum number `(m_l)` and spin quantum number `(m_s)` . The permissible values of these are :
`n = 1,2.....`
`l = 0,1,.....(n-1)`
`m_l = -l,......0,......+l`
`m_s = +1/2 and -1/2` for each value of `m_l`.
The angular momentum of electron is given as `sqrt(l(l + 1)) cdot h/(2pi)`
While spin angular momentum is given as `sqrt(s(s+1)) cdot (h/(2pi))` where `s = 1/2`
The electrons having the same value of `n, l and m_l` are said to belong to the same orbital. According to Pauli's exclusion principle, an orbital can have maximum of two electrons and these two must have opposite spin.
For an electron having `n = 3 and l = 0`, the orbital angular momentum is

Options

A. `((sqrt3)h)/(pi)`
B. `sqrt(6) h/(2pi)`
C. zero
D. `2sqrt(3) h/pi`

Explanation

To determine the orbital angular momentum of an electron with the given quantum numbers \( n = 3 \) and \( l = 0 \), we can follow these steps:

### Step 1: Identify the quantum numbers
We have:
- Principal quantum number \( n = 3 \)
- Azimuthal quantum number \( l = 0 \)

### Step 2: Understand the formula for orbital angular momentum
The formula for the orbital angular momentum \( L \) of an electron is given by:
\[ L = \sqrt{l(l + 1)} \cdot \frac{h}{2\pi} \]
where \( h \) is Planck's constant.

### Step 3: Substitute the value of \( l \)
Since \( l = 0 \):
\[ L = \sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} \]

### Step 4: Calculate the value
Calculating the expression:
\[ L = \sqrt{0} \cdot \frac{h}{2\pi} = 0 \]

### Conclusion
Thus, the orbital angular momentum for an electron with \( n = 3 \) and \( l = 0 \) is:
\[ \boxed{0} \]

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