The Schrodinger wave equation for hydrogen atom of 4s-orbital is given by :
$ψ (r) = \frac{1}{16 \sqrt{4}} {(\frac{1}{a_{0}})}^{3 / 2} [(σ - 1) (σ^{2} - 8 σ + 12)] e^{- σ / 2}$
Where $a_{0} = 1^{st}$ Bohr radius and $σ = \frac{2 r}{a_{0}}$ . The distance from the nucleus where there will be no radial node will be:

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$r = 2 a_{r}$

$r = a_{0}$

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

$r = 3 a_{0}$

answer is D.

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Detailed Solution

Here, we have;

For radial nodes, put $ψ (r) = 0$

$\begin{array}{l} \Rightarrow σ = 1, σ^{2} - 8 σ + 12 = 0 \\ \Rightarrow σ = 6 and σ = 2 \\ \Rightarrow r = \frac{a_{0}}{2}, 3 a_{0}, a_{0} \end{array}$

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