JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

The Bohr radius of a hydrogen like species is $70.53$ pm. The species and the stationary state $(n)$ are respectively (Given : Hydrogen atom Bohr radius is $52.9$ pm)

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\text{Li}^{2+}, 3$
B
$\text{He}^+, 3$
C
$\text{He}^+, 2$
D
$\text{Li}^{2+}, 2$

Did you get this right?

Solution

The radius of the $n$ -th Bohr orbit for a hydrogen-like species is given by the formula:

$r_n = a_0 \dfrac{n^2}{Z}$

where $a_0$ is the Bohr radius of the hydrogen atom ( $52.9$ pm), $n$ is the principal quantum number, and $Z$ is the atomic number.

Given $r_n = 70.53$ pm, we can write:

$70.53 = 52.9 \times \dfrac{n^2}{Z}$

$\dfrac{n^2}{Z} = \dfrac{70.53}{52.9} = 1.333 = \dfrac{4}{3}$

Checking the given options:

For $\text{Li}^{2+}$ and $n = 3$ , $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{3} = 3$

For $\text{He}^+$ and $n = 3$ , $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{2} = 4.5$

For $\text{He}^+$ and $n = 2$ , $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{2} = 2$

For $\text{Li}^{2+}$ and $n = 2$ , $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{3}$

The ratio matches for $\text{Li}^{2+}$ and $n = 2$ .

Answer: $\text{Li}^{2+}, 2$

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

The Bohr radius of a hydrogen like species is $70.53$ pm. The species and the stationary state $(n)$ are respectively (Given : Hydrogen atom Bohr radius is $52.9$ pm)

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\text{Li}^{2+}, 3$
B
$\text{He}^+, 3$
C
$\text{He}^+, 2$
D
$\text{Li}^{2+}, 2$

Did you get this right?

Solution

The radius of the $n$ -th Bohr orbit for a hydrogen-like species is given by the formula:

$r_n = a_0 \dfrac{n^2}{Z}$

where $a_0$ is the Bohr radius of the hydrogen atom ( $52.9$ pm), $n$ is the principal quantum number, and $Z$ is the atomic number.

Given $r_n = 70.53$ pm, we can write:

$70.53 = 52.9 \times \dfrac{n^2}{Z}$

$\dfrac{n^2}{Z} = \dfrac{70.53}{52.9} = 1.333 = \dfrac{4}{3}$

Checking the given options:

For $\text{Li}^{2+}$ and $n = 3$ , $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{3} = 3$

For $\text{He}^+$ and $n = 3$ , $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{2} = 4.5$

For $\text{He}^+$ and $n = 2$ , $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{2} = 2$

For $\text{Li}^{2+}$ and $n = 2$ , $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{3}$

The ratio matches for $\text{Li}^{2+}$ and $n = 2$ .

Answer: $\text{Li}^{2+}, 2$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

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