For an orbital,
$ψ_{300} = \frac{1}{81 \sqrt{3} π} {(\frac{z}{a_{0}})}^{3 / 2} [27 - 18 u + 2 u^{2}] \exp (\frac{- u}{3})$
where, $u = \frac{z r}{a_{0}}$
What is the maximum radial distance of node from nucleus of ${He}^{+}$ ion?

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$a_{0}$

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

$(3 + \sqrt{3}) \frac{3 a_{0}}{4}$

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

answer is B.

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Detailed Solution

Radial node occurs where probability of finding $e^{-}$ zero.

$\begin{array}{l} ∴ ψ^{2} = 0, ψ = 0 \\ ∴ 27 - 18 u + 2 u^{2} = 0 \end{array}$

$\begin{array}{l} \Rightarrow u = \frac{9 \pm 3 \sqrt{3}}{2} \\ \Rightarrow r = \frac{3 a_{0}}{2 \times z} [3 \pm \sqrt{3}] \end{array}$

$= \frac{3}{4} [3 \pm \sqrt{3}] a_{0}$

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