The orbital angular momentum of an electron in 3s orbital is \(\frac{xh}{2π}\) The value of x is___.
Correct Answer: 0
Solution and Explanation
The orbital angular momentum is given by: \[ L = \frac{\sqrt{l(l+1)}h}{2\pi}. \] For an \(s\)-orbital, \(l = 0\). Substituting \(l = 0\): \[ L = \frac{\sqrt{0(0+1)}h}{2\pi} = 0. \]
Final Answer: \( \boxed{0} \).
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