For a 3S-orbital value of $ψ$ is given by following relation $ψ_{3 s} = \frac{1}{9 \sqrt{3}} {(\frac{1}{a_{0}})}^{3 / 2} (6 - 6 σ + σ^{2}) e^{- σ / 2}$ where $σ = \frac{2 r . Z}{3 a_{0}}$ .
What is the maximum Radial distance of node from nucleus

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$\frac{a_{0}}{Z}$

$\frac{2 a_{0}}{Z}$

$\frac{3}{2} \frac{(3 + \sqrt{3}) a_{0}}{Z}$

$\frac{(3 + \sqrt{3}) a_{0}}{Z}$

answer is C.

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Detailed Solution

$∴ ψ^{2} = 0 (o r) ψ = 0 6 - 6 σ + σ^{2} = 0 σ = 3 \pm \sqrt{3}$

For maximum distance $r = \frac{3}{2} \frac{(3 + \sqrt{3}) a_{0}}{Z}$

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