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

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

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

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

answer is C.

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Detailed Solution

Radial node occurs where probability of finding $e^{-}$ is zero
$∴ ψ^{2} = 0 or ψ = 0 ∴ 6 - 6 σ + σ^{2} = 0; σ = 3 \pm \sqrt{3}$ .
For max. distance $r = \frac{3}{2} \frac{(3 + \sqrt{3}) a_{0}}{Z}$

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