The orbital angular momentum of an electron in 2s orbital is: (a) h / 4π (b) zero (c) h / 2π (d) √2 * h / 2π

To find the orbital angular momentum of an electron in a 2s orbital, we need to understand the concept of orbital angular momentum in quantum mechanics.

The orbital angular momentum \( L \) of an electron in an orbital is given by the formula:
\[ L = \sqrt{l(l+1)} \hbar \]

where \( l \) is the azimuthal quantum number (also known as the angular momentum quantum number), and \( \hbar \) is the reduced Planck's constant:
\[ \hbar = \frac{h}{2\pi} \]

For an electron in a 2s orbital:
- The principal quantum number \( n \) is 2.
- The azimuthal quantum number \( l \) for an s orbital is 0.

Therefore:
\[ L = \sqrt{0(0+1)} \hbar = \sqrt{0} \hbar = 0 \]

So, the orbital angular momentum of an electron in the 2s orbital is zero.

Thus, the correct answer is (b) zero.