Question

Orbital angular momentum of electron in \(p\) orbital is equal to

• A. \(2\sqrt{3}\hbar\)
• B. zero
• C. \(\sqrt{6}\hbar\)
• D. \(\sqrt{2}\hbar\)

Hint:

Orbital angular momentum depends on azimuthal quantum number \(l\): \[ L = \sqrt{l(l+1)}\,\hbar \]
\(s\)-orbital (\(l=0\)) : \(L = 0\)
\(p\)-orbital (\(l=1\)) : \(L = \sqrt{2}\hbar\)
\(d\)-orbital (\(l=2\)) : \(L = \sqrt{6}\hbar\)

Answer 1

The Correct Option is D

Step 1: Orbital angular momentum of an electron is given by: \[ L = \sqrt{l(l+1)}\,\hbar \] where \(l\) is the azimuthal quantum number.
Step 2: For a \(p\)-orbital: \[ l = 1 \]
Step 3: Substitute the value of \(l\): \[ L = \sqrt{1(1+1)}\,\hbar = \sqrt{2}\,\hbar \]
Step 4: Hence, the orbital angular momentum of an electron in a \(p\)-orbital is: \[ \boxed{\sqrt{2}\hbar} \]