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In $ψ_{321}$ the sum of angular momentum, spherical nodes and angular nodes is

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$\frac{\sqrt{6} h + 2 π}{2 π}$

$\frac{\sqrt{6} h + 8 π}{2 π}$

$\frac{\sqrt{6} h + 4 π}{2 π}$

$\frac{\sqrt{6} h}{2 π} + 3$

answer is A.

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Detailed Solution

$n = 3 l = 2 m = 1$
Angular momentum $= \frac{h}{2 π} \sqrt{l (l + 1)} = \frac{\sqrt{6} h}{2 π}$
Spherical nodes $= 3 - 2 - 1 = 0$
Angular nodes $= 2$
Sum of all $= \frac{\sqrt{6} h + 4 π}{2 π}$

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