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For a 3s-orbital Phi(3s)=(1)/(asqrt(3)...

For a 3s-orbital
`Phi(3s)=(1)/(asqrt(3))((1)/(a_(0)))^(3//2)(6-6sigma+sigma^(2))in^(-sigma//2)`
where `sigma=(2rZ)/(3a_(sigma))`
What is the maximum radial distance of node from nucleus?

A

`r=(1)/(2)(3+sqrt(3))a_(0)`

B

`r=(1)/(2)(3-sqrt(3))a_(0)`

C

`r=(3)/(2)(3-sqrt(6))a_(0)`

D

`r=(3)/(2)(3+sqrt(3))a_(0)`

Text Solution

Verified by Experts

The correct Answer is:
D
For radial node,
`psi(r )=0`
`27-18sigma+2sigma^(2)=0`
`sigma=(18pmsqrt(18^(2)-4xx2xx27))/(2xx2)=(18pmsqrt(108))/(4)=(18pmsqrt(36xx3))/(4)`
`sigma=(18pm6sqrt(3))/(4)`
`=(3)/(2)(3pmsqrt(3))`
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