A) \[\frac{\sqrt{6}h+4\pi }{2\pi }\]
B) \[\frac{\sqrt{6}h}{2\pi }+3\]
C) \[\frac{\sqrt{6}h+2\pi }{2\pi }\]
D) \[\frac{\sqrt{6}h+8\pi }{2\pi }\]
Correct Answer: A
Solution :
[a] \[{{\Psi }_{321}}:n=3,\ell =2,m=1\] Angular momentum \[=\frac{h}{2\pi }\sqrt{\ell (\ell +1)}=\frac{\sqrt{6h}}{2\pi }\] Spherical nodes \[=3-2-1=0;\] Angular node = 2 Sum of all the above \[=\frac{\sqrt{6}h}{2\pi }+2=\frac{\sqrt{6}h+4\pi }{2\pi }\]
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