The schrodinger wave equation for hydrogen atom is
$ψ_{2 s} = \frac{1}{4 \sqrt{2 π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} (2 - \frac{r_{0}}{a_{0}}) e^{\frac{- r_{0}}{a_{0}}}$
Where $a_{0}$ is Bohr’s radius. If the radial node in 2S be at $r_{0}$ , then correct relation between $r_{0}$ and $a_{0}$

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$r_{0} = a_{0}$

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

$a_{0} = 2 r_{0}$

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

answer is D.

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Detailed Solution

At node $ψ^{2} = 0$
$0 = \frac{1}{32 π} {(\frac{1}{a_{0}})}^{3} {(2 - \frac{r_{0}}{a_{0}})}^{2} e^{\frac{- 2 r_{0}}{a_{0}}} ∴ 2 - \frac{r_{0}}{a_{0}} = 0 \Rightarrow r_{0} = 2 a_{0}$

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