Q22 / 75 · JEE Main 2025 (3 Apr Shift 2) · Chemistry

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 $\sqrt{6} \frac{\mathrm{~h}}{2 \pi}$
  • Correct answer
    D
    0 and $\sqrt{2} \frac{\mathrm{~h}}{2 \pi}$

Solution

$\begin{aligned} & \text { Orbital angular momentum }=\sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi} \\ & \therefore \quad \text { For } 2 \mathrm{~s} \text { orbital }: \ell=0 \\ & \text { Orbital angular momentum }=0 \\ & \therefore \quad \text { For } 2 \mathrm{p} \text { orbital }: \ell=1 \\ & \quad \begin{aligned} \text { Orbital angular momentum } & =\sqrt{1(1+2)} \frac{\mathrm{h}}{2 \pi} \\ & =\sqrt{2} \frac{\mathrm{~h}}{2 \pi}\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'D'.

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