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 formula for the orbital angular momentum of an electron in an atom is \( L = \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{l(l+1)} \hbar \). Here, \( l \) represents the azimuthal quantum number (also called the orbital angular momentum quantum number), and \( h \) is Planck's constant, with \( \hbar = \frac{h}{2\pi} \) denoting the reduced Planck constant.

For a '2s' orbital, \( n = 2 \) and \( l = 0 \). Applying \( l = 0 \) to the orbital angular momentum formula yields \( L_{2s} = \sqrt{0(0+1)} \frac{h}{2\pi} = 0 \). Therefore, an electron in a 2s orbital has zero orbital angular momentum. For a '2p' orbital, \( n = 2 \) and \( l = 1 \). Substituting \( l = 1 \) into the formula gives \( L_{2p} = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi} \). Consequently, an electron in a 2p orbital possesses an orbital angular momentum of \( \sqrt{2} \frac{h}{2\pi} \). The orbital angular momenta for electrons in '2s' and '2p' orbitals are 0 and \( \sqrt{2} \frac{h}{2\pi} \), respectively.

This aligns with option (4).