What will be the orbital angular momentum of an electron in 2s-orbital?

Zero

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One

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Solution

The correct option is C Zero
Orbital angular momentum $μ_{L} = \sqrt{l (l + 1)} \frac{h}{2 π}$
For 2s-orbital n=2, l=s=0
upon substitution we get:
$μ_{L} = \sqrt{0 (0 + 1)} \frac{h}{2 π} = 0$

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