JEE Main 2023 — Structure of Atom Question with Solution

The wave function ($\mathrm{\Psi }$) of 2s is given by,

${\mathrm{\Psi }}_{2\mathrm{s}}=\frac{1}{2\sqrt{2\mathrm{\pi }}}{\left(\frac{1}{{\mathrm{a}}_0}\right)}^{1/2}\left(2-\frac{\mathrm{r}}{{\mathrm{a}}_0}\right){\mathrm{e}}^{-\mathrm{r}/2{\mathrm{a}}_0}$

At $\mathrm{r} = {\mathrm{r}}_0$, radial node is formed.

${{\mathrm{\Psi }}_{2\mathrm{s}}}^2$= Probability density of finding an electron

At node, ${{\mathrm{\Psi }}_{2\mathrm{s}}}^2 = 0$ ($\because$Probability of finding an electron is zero at the node)

${{\mathrm{\Psi }}_{2\mathrm{s}}}^2 ={\left[\frac{1}{2\sqrt{2\mathrm{\pi }}}{\left(\frac{1}{{\mathrm{a}}_0}\right)}^{1/2}\left(2-\frac{\mathrm{r}}{{\mathrm{a}}_0}\right){\mathrm{e}}^{-\mathrm{r}/2{\mathrm{a}}_0}\right]}^2$

$\Rightarrow 0 =\frac{1}{8\mathrm{\pi }}\left(\frac{1}{{\mathrm{a}}_0}\right) {\left(2-\frac{{\mathrm{r}}_0}{{\mathrm{a}}_0}\right)}^2{\mathrm{e}}^{-2{\mathrm{r}}_0/{\mathrm{a}}_0}$

$\Rightarrow \left[2 - \frac{{\mathrm{r}}_0}{{\mathrm{a}}_0}\right] = 0$

$\Rightarrow 2 = \frac{{\mathrm{r}}_0}{{\mathrm{a}}_0}$

$\Rightarrow 2{\mathrm{a}}_0 = {\mathrm{r}}_0$

So, the correct option is D.