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The orbital angular momentum for an electron resolving in an orbit is $\large\frac{h}{2\pi}$$\sqrt{l(l+1)}$.This momentum of a $s$-electron is :

$\begin{array}{1 1}(a)\;\large\frac{h}{2\pi}&(b)\;\sqrt{2}\large\frac{h}{2\pi}\\(c)\;\pm \large\frac{1}{2}\frac{h}{2\pi}&(d)\;zero\end{array}$

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$l=0$ for $s$-orbital.
Hence (d) is the correct answer.
answered Jan 13, 2014 by sreemathi.v
 

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