First slide

Classical atomic models

Question

Orbital angular momentum of an electron is $\sqrt{3} \frac{h}{π}$ then the number of possible orientations of this orbital in space is:

Moderate

Solution

Orbital angular momentum of electron
$= \sqrt{∣ (1 + 1)} \cdot \frac{h}{2 π} = \sqrt{3} \cdot \frac{h}{π}$
I = 3
$∴ number of orientation = 2 l + 1 = 2 \times 3 + 1 = 7$

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