In psi(321) the sum of angular momentum,...

In `psi_(321)` the sum of angular momentum, spherical nodes and angular node is:

`(sqrt(6) h + 4 pi)/(2pi)`

`(sqrt(6) h)/(2pi) +3`

`(sqrt(6) h + 2pi)/(2pi)`

`(sqrt(6) h + 8pi)/(2pi)`

Text Solution

Verified by Experts

The correct Answer is:

`Psi_(321): n=3,l=2,m=1`
Angular momentum `= (h)/(2pi) sqrt(l(l+1))= (sqrt(6)h)/(2pi)`
Spherical nodes `= 3-2 -1 =0`,
Angular node `=2`
Sum of all the above `= (sqrt(6) h)/(2pi) +2 = (sqrt(6)h + 4pi)/(2pi)`

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`h/(sqrt2n), 0,1`

`h/(2pi),5,2`

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`h/(4pi),1,1`

The number of angular nodes and radical nodes in 3s orbital are

0 and 2, respectively

1 and 0, respectively

3 and 0, respectively

0 and 1, respectively

The number of angular nodes and radial nodes is 3s orbital are