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For a 3s - orbital, value of Phi is give...

For a 3s - orbital, value of `Phi` is given by following realation:
`Psi(3s)=(1)/(9sqrt(3))((1)/(a_(0)))^(3//2)(6-6sigma+sigma^(2))e^(-sigma//2)," where " sigma=(2r.Z)/(3a_(0))`
What is the maximum radial distance of node from nucleus?

A

`((3+sqrt(3)))/(Z)a_(0)`

B

`(a_(0))/(Z)`

C

`(3)/(2)((3+sqrt(3)))/(Z)a_(0)`

D

`(2a_(0))/(Z)`

Text Solution

Verified by Experts

The correct Answer is:
C
At node `Psi = 0` & `6 - 6 sigma +sigma^(2) = 0`
`:. sigma = 3 +- sqrt(3)`
for max. `r = (3)/(2) ((3+sqrt(3))a_(0))/(z)`
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