MCQ
For an electron, with $n = 3$ has only one radial node. The orbital angular momentum of the electron will be
- A$0$
- B$\sqrt 6 \frac{h}{{2\pi }}$
- ✓$\sqrt 2 \frac{h}{{2\pi }}$
- D$3\left( {\frac{h}{{2\pi }}} \right)$
$3 - \ell - 1 = 1$
$\therefore \,\ell = 1$
Orbital angular momentum $ =\sqrt{\ell(\ell+1) \frac{\mathrm{h}}{2 \pi}} $
$=\sqrt{2} \frac{\mathrm{h}}{2 \pi}$
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