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NEET CHEMISTRYEasy

The orbital angular momentum of a p-electron is given as:

A

3h2π\frac{\sqrt{3}h}{2\pi}

B

32hπ\sqrt{\frac{3}{2}}\frac{h}{\pi}

C

6h2π\frac{\sqrt{6}h}{2\pi}

D

h2π\frac{h}{\sqrt{2}\pi}

Step-by-Step Solution

The orbital angular momentum (LL) of an electron in an orbital is determined by the azimuthal quantum number (ll) according to the expression: L=l(l+1)h2πL = \sqrt{l(l+1)} \frac{h}{2\pi}

For a p-electron, the value of the azimuthal quantum number l=1l = 1 . Substituting this value into the formula: L=1(1+1)h2πL = \sqrt{1(1+1)} \frac{h}{2\pi} L=2h2πL = \sqrt{2} \frac{h}{2\pi} L=h2πL = \frac{h}{\sqrt{2}\pi}

Note: The value l=0l=0 corresponds to s-orbitals, l=2l=2 to d-orbitals, etc.

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