JEE Main 2026 — Chemistry Physical Chemistry

2026mcqeasy

The species having identical radii according to the Bohr's theory are:

A. $H$ (first orbit)

B. $He^+$ (first orbit)

C. $He^+$ (Second orbit)

D. $Li^{2+}$ (first orbit)

E. $Be^{3+}$ (Second orbit)

Choose the correct answer from the options given below:

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
A and C Only
B
A and E Only
C
B and E Only
D
C and D Only

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Solution

According to Bohr's theory, the radius of the $n$ -th orbit of a hydrogen-like species is given by $r_n = 0.529 \times \dfrac{n^2}{Z} \text{ \AA}$ .

For $H$ (first orbit), $n = 1$ , $Z = 1 \Rightarrow r = 0.529 \times \dfrac{1^2}{1} = 0.529 \text{ \AA}$ .

For $He^+$ (first orbit), $n = 1$ , $Z = 2 \Rightarrow r = 0.529 \times \dfrac{1^2}{2} = 0.2645 \text{ \AA}$ .

For $He^+$ (second orbit), $n = 2$ , $Z = 2 \Rightarrow r = 0.529 \times \dfrac{2^2}{2} = 1.058 \text{ \AA}$ .

For $Li^{2+}$ (first orbit), $n = 1$ , $Z = 3 \Rightarrow r = 0.529 \times \dfrac{1^2}{3} = 0.176 \text{ \AA}$ .

For $Be^{3+}$ (second orbit), $n = 2$ , $Z = 4 \Rightarrow r = 0.529 \times \dfrac{2^2}{4} = 0.529 \text{ \AA}$ .

The radii of $H$ (first orbit) and $Be^{3+}$ (second orbit) are identical. Therefore, species A and E have identical radii.

Answer: A and E Only

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Solution

According to Bohr's theory, the radius of the $n$ -th orbit of a hydrogen-like species is given by $r_n = 0.529 \times \dfrac{n^2}{Z} \text{ \AA}$ .

For $H$ (first orbit), $n = 1$ , $Z = 1 \Rightarrow r = 0.529 \times \dfrac{1^2}{1} = 0.529 \text{ \AA}$ .

For $He^+$ (first orbit), $n = 1$ , $Z = 2 \Rightarrow r = 0.529 \times \dfrac{1^2}{2} = 0.2645 \text{ \AA}$ .

For $He^+$ (second orbit), $n = 2$ , $Z = 2 \Rightarrow r = 0.529 \times \dfrac{2^2}{2} = 1.058 \text{ \AA}$ .

For $Li^{2+}$ (first orbit), $n = 1$ , $Z = 3 \Rightarrow r = 0.529 \times \dfrac{1^2}{3} = 0.176 \text{ \AA}$ .

For $Be^{3+}$ (second orbit), $n = 2$ , $Z = 4 \Rightarrow r = 0.529 \times \dfrac{2^2}{4} = 0.529 \text{ \AA}$ .

The radii of $H$ (first orbit) and $Be^{3+}$ (second orbit) are identical. Therefore, species A and E have identical radii.

Answer: A and E Only

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2026 — 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