For a 3s - orbitalΨ(3s)=1a√3(1a0)32(6−6σ+σ2)ϵ−σ2,Where σ=2 – InterviewSolution

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1.	For a 3s - orbitalΨ(3s)=1a√3(1a0)32(6−6σ+σ2)ϵ−σ2,Where σ=2rZ3a0What is the maximum radial distance of node from nucleus is
Answer» For a 3s - orbital $Ψ (3 s) = \frac{1}{a \sqrt{3}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (6 - 6 σ + σ^{2}) ϵ^{\frac{- σ}{2}},$ Where $σ = \frac{2 r Z}{3 a_{0}}$ What is the maximum radial distance of node from nucleus is

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