JEE Mains · Chemistry · STD 11 - 2. structure of atom
For electron in ' 2 s ' and ' 2 p ' orbitals, the orbital angular momentum values, respectively are :
- A \(\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\) and 0
- B \(\frac{\mathrm{h}}{2 \pi}\) and \(\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\)
- C 0 and \(\sqrt{6} \frac{\mathrm{~h}}{2 \pi}\)
- D 0 and \(\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\)
Answer & Solution
Correct Answer
(D) 0 and \(\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Orbital angular momentum }=\sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi} \\ & \therefore \quad \text { For } 2 \mathrm{~s} \text { orbital }: \ell=0 \\ & \text { Orbital angular momentum }=0 \\ & \therefore \quad \text { For } 2 \mathrm{p} \text {…
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