JEE Mains · Chemistry · STD 11 - 2. structure of atom
The orbital angular momentum of an electron in \(3 s\) orbital is \(\frac{x h}{2 \pi}\). The value of.\(x\) is \(..........\).
- A \(1\)
- B \(2\)
- C \(3\)
- D \(0\)
Answer & Solution
Correct Answer
(D) \(0\)
Step-by-step Solution
Detailed explanation
\(l\) for \(3s\) orbital is \(0\). Orbital angular momentum \(=\) \(\sqrt{l(l+1)} \frac{h}{2\pi}\) \(= \sqrt{0(0+1)} \frac{h}{2\pi} = 0\) Given \(=\) \(\frac{x h}{2 \pi}\) \(x = 0\)
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