JEE Main 2023 — Chemistry Physical Chemistry

2023mcqmedium

The wave function $(\Psi )$ of $2s$ is given by

${\Psi }_{2s}=\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}$

At $r={r}_{0}$ , radial node is formed. Thus, ${r}_{0}$ in terms of ${a}_{0}$

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
${r}_{0}={a}_{0}$
B
${r}_{0}=4{a}_{0}$
C
${r}_{0}=\frac{{a}_{0}}{2}$
D
${r}_{0}=2{a}_{0}$

Did you get this right?

Solution

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

${\Psi }_{2s}=\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}$

At $r={r}_{0}$ , radial node is formed.

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

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

${{\Psi }_{2s}}^{2}={[\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}]}^{2}$

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

$\Rightarrow [2-\frac{{r}_{0}}{{a}_{0}}]=0$

$\Rightarrow 2=\frac{{r}_{0}}{{a}_{0}}$

$\Rightarrow 2{a}_{0}={r}_{0}$

So, the correct option is D.

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

2023mcqmedium

The wave function $(\Psi )$ of $2s$ is given by

${\Psi }_{2s}=\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}$

At $r={r}_{0}$ , radial node is formed. Thus, ${r}_{0}$ in terms of ${a}_{0}$

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
${r}_{0}={a}_{0}$
B
${r}_{0}=4{a}_{0}$
C
${r}_{0}=\frac{{a}_{0}}{2}$
D
${r}_{0}=2{a}_{0}$

Did you get this right?

Solution

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

${\Psi }_{2s}=\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}$

At $r={r}_{0}$ , radial node is formed.

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

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

${{\Psi }_{2s}}^{2}={[\frac{1}{2\sqrt{2\pi }}{(\frac{1}{{a}_{0}})}^{1/2}(2-\frac{r}{{a}_{0}}){e}^{-r/2{a}_{0}}]}^{2}$

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

$\Rightarrow [2-\frac{{r}_{0}}{{a}_{0}}]=0$

$\Rightarrow 2=\frac{{r}_{0}}{{a}_{0}}$

$\Rightarrow 2{a}_{0}={r}_{0}$

So, the correct option is D.

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2023 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry