Question

For electron in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are :

• A. \( \sqrt{2} \frac{h}{2\pi} \) and 0
• B. \( \frac{h}{2\pi} \) and \( \sqrt{2} \frac{h}{2\pi} \)
• C. 0 and \( \sqrt{6} \frac{h}{2\pi} \)
• D. 0 and \( \sqrt{2} \frac{h}{2\pi} \)

Hint:

The orbital angular momentum depends only on the azimuthal quantum number \( l \). Remember the values of \( l \) for different orbitals: s (\( l=0 \)), p (\( l=1 \)), d (\( l=2 \)), f (\( l=3 \)), and so on. Use the formula \( L = \sqrt{l(l+1)} \frac{h}{2\pi} \) to calculate the orbital angular momentum for each orbital.

Answer 1

The Correct Option is D

The orbital angular momentum of an electron in an atom is given by the formula: \[ L = \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{l(l+1)} \hbar \] where \( l \) is the azimuthal quantum number (also known as the orbital angular momentum quantum number), and \( h \) is Planck's constant, with \( \hbar = \frac{h}{2\pi} \) being the reduced Planck constant. 

For a '2s' orbital, the principal quantum number \( n = 2 \), and for an 's' orbital, the azimuthal quantum number \( l = 0 \). Substituting \( l = 0 \) into the formula for orbital angular momentum: \[ L_{2s} = \sqrt{0(0+1)} \frac{h}{2\pi} = \sqrt{0} \frac{h}{2\pi} = 0 \] So, the orbital angular momentum for an electron in a 2s orbital is 0. For a '2p' orbital, the principal quantum number \( n = 2 \), and for a 'p' orbital, the azimuthal quantum number \( l = 1 \). Substituting \( l = 1 \) into the formula for orbital angular momentum: \[ L_{2p} = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{1(2)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} \] So, the orbital angular momentum for an electron in a 2p orbital is \( \sqrt{2} \frac{h}{2\pi} \). The orbital angular momentum values for electrons in '2s' and '2p' orbitals are 0 and \( \sqrt{2} \frac{h}{2\pi} \) respectively. 

This corresponds to option (4).