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The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt{l (l + 1)} \frac{n}{2 π}$ . This momentum for an $s$ -electron will be given by :

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+ \frac{1}{2} \cdot \frac{h}{2 π}

Zero

\frac{h}{2 π}

\sqrt{2} \cdot \frac{h}{2 π}

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detailed solution

Correct option is B

The azimuthal quantum number $l$ for an s electron, is equal to zero.

$Orbital angular momentum = \sqrt{l (l + 1)} \cdot \frac{h}{2 π}$

the s-electron is replaced with the value of $l$

$= \sqrt{0 (0 + 1)} \cdot \frac{h}{2 π} = 0$

Hence, option B is correct.

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The orbital angular momentum for an electron revolving in an orbit is given by l(l+1)n2π. This momentum for an s-electron will be given by :