For an electron with $n = 3$,there is only one radial node. The orbital angular momentum of the electron will be

  • A
    $0$
  • B
    $\sqrt{6} \frac{h}{2\pi}$
  • C
    $\sqrt{2} \frac{h}{2\pi}$
  • D
    $3 \left(\frac{h}{2\pi}\right)$

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Similar Questions

$A$: No two electrons in an atom can have the same set of all four quantum numbers.
$R$: Two electrons in an atom can exist in the same shell,subshell,or orbital only if they have opposite spins.

The principle which states that electron occupies the available orbitals singly before pairing in any one orbital occurs is known as:

The shape of $2p$ orbital is

The quantum numbers $n$ and $l$ for four electrons are given as: $(1) \, n = 4, l = 1; \, (2) \, n = 4, l = 0; \, (3) \, n = 3, l = 2; \, (4) \, n = 3, l = 1$. Arrange them in increasing order of their energy.

Any $p-$ orbital can accommodate up to:

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