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Question

In Ψ321 the sum of angular momentum, spherical nodes and angular mode is:

A
6h+4π2π
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B
6h2π+3
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C
6h+2π2π
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D
6h+8π2π
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Solution

The correct option is D 6h+4π2π
Ψ321Ψl,m,n
n=3,l=2,m=1
Angular Momentum=l(l+1)h2π=2(2+1)h2π=6h2π
Radial node or spherical node =nl1=321=0
Angular node =l=2
Thus, sum=6h2π+2=6h+4π2π

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