Chemistry Q52 – JEE-MAIN 2026

Home Chemistry Structure of AtomQuestion #603

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:

Clue 1 of 3: Bohr's Model for Hydrogen-like Species

Recall the fundamental principles of Bohr's model, specifically how it describes the electron's orbit and energy levels for single-electron systems.

🧠 Full Solution Path

Step 1: Identify the Formula for Bohr Radius✦ Active

The radius of the $n$ -th orbit for a hydrogen-like species (single-electron species) is given by Bohr's formula:

r_{n} = 0.529 \frac{n^{2}}{Z} Å

For two species to have identical radii, their $\frac{n^{2}}{Z}$ values must be equal.

Step 2: Calculate

n^{2} / Z

for each given species○ Expand

We calculate the value of $\frac{n^{2}}{Z}$ for each species and orbit combination:

A. H (first orbit): $Z = 1, n = 1 ⟹ \frac{1^{2}}{1} = 1$

B. He $^{+}$ (first orbit): $Z = 2, n = 1 ⟹ \frac{1^{2}}{2} = 0.5$

C. He $^{+}$ (Second orbit): $Z = 2, n = 2 ⟹ \frac{2^{2}}{2} = \frac{4}{2} = 2$

D. Li $^{2 +}$ (first orbit): $Z = 3, n = 1 ⟹ \frac{1^{2}}{3} \approx 0.333$

E. Be $^{3 +}$ (Second orbit): $Z = 4, n = 2 ⟹ \frac{2^{2}}{4} = \frac{4}{4} = 1$

💡 Teacher's Secret Hint

Remember that Z is the atomic number, and n is the principal quantum number (orbit number).

Step 3: Compare values and select the correct option○ Expand

Comparing the calculated $\frac{n^{2}}{Z}$ values, we find that H (first orbit) and Be $^{3 +}$ (Second orbit) both have a value of 1. Therefore, species A and E have identical radii. This corresponds to option 2.

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Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Bohr's Model for Hydrogen-like Species

🧠 Full Solution Path