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The wave function of 3s and 3p(z) orbita...

The wave function of 3s and `3p_(z)` orbitals are given by :
`Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6=6sigma+sigma)e^(-sigma//2)`
`Psi_(3s_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0,`
`sigma=(2Zr)/(nalpha_(0))`
where`alpha_(0)=1st` Bohr radius , Z= charge number of nucleus, r= distance from nucleus.
From this we can conclude:

A

Number of nodal surface for `3p_(z)` & `3s` orbitals is equal.

B

The angualr nodal surface of `3p_(z)` orbital has the equation `theta=(pi)/(2)`

C

The radial nodal surfaces of 3 orbital and `3p_(z)` orbital are at equal distance from the nucleus.

D

3s electron have greater penetrating power intot he nucleus in comparison to 3p electrons.

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
(A) Total nodal surfaces in `3s=2`
Total nodal surfaces in `3p=2`
(B) Angular node in `3p,costheta=0`
`theta=(pi)/(2)`
(C) let `(zr)/(a_(0))=x`
Radial nodes in 3s. `(6-4x+(4)/(9)x^(2))=0`
`implies(3-2x+(2)/(9)x^(2))=0implies27-18x+2x^(2)=0`
`impliesx=(18+-sqrt(324-216))/(2xx2)=(18+-sqrt(108))/(4)`
`x=(9+3sqrt(3))/(2),x=(9-3sqrt(3))/(2)`
Radial nodes in `3p` `4-(2)/(3)x=0impliesx=6`
Hence their radial nodes are not same
(D) it is a fact
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