Chemistry Q52 – JEE-MAIN 2026

Home Chemistry Structure of AtomQuestion #607

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)

Clue 1 of 3: Atomic Structure

Recall the Bohr model for hydrogen-like species and how the radius of an orbit depends on the principal quantum number and atomic number.

🧠 Full Solution Path

Step 1: Identify the relevant formula✦ Active

The Bohr radius for a hydrogen-like species is given by the formula:

r_{n} = r_{0} \frac{n^{2}}{Z}

Where $r_{n}$ is the radius of the $n$ -th orbit, $r_{0}$ is the Bohr radius of hydrogen ( $52.9$ pm), $n$ is the principal quantum number, and $Z$ is the atomic number.

Step 2: Substitute values from options and calculate

r_{n}

○ Expand

We will test each option by substituting the atomic number ( $Z$ ) and principal quantum number ( $n$ ) into the formula:

1. For ${Li}^{2 +}$ ( $Z = 3$ ), $n = 3$ : $r_{3} = 52.9 pm \times \frac{3^{2}}{3} = 52.9 pm \times 3 = 158.7 pm$

2. For ${He}^{+}$ ( $Z = 2$ ), $n = 3$ : $r_{3} = 52.9 pm \times \frac{3^{2}}{2} = 52.9 pm \times 4.5 = 238.05 pm$

3. For ${He}^{+}$ ( $Z = 2$ ), $n = 2$ : $r_{2} = 52.9 pm \times \frac{2^{2}}{2} = 52.9 pm \times 2 = 105.8 pm$

4. For ${Li}^{2 +}$ ( $Z = 3$ ), $n = 2$ : $r_{2} = 52.9 pm \times \frac{2^{2}}{3} = 52.9 pm \times \frac{4}{3} \approx 70.53 pm$

💡 Teacher's Secret Hint

Remember that $Z$ is the atomic number, which is 3 for Lithium and 2 for Helium.

Step 3: Compare calculated radius with given value○ Expand

The calculated radius for ${Li}^{2 +}$ in the $n = 2$ state ( $70.53$ pm) matches the given Bohr radius of the hydrogen-like species ( $70.53$ pm). Therefore, option 4 is the correct answer.

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Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Atomic Structure

🧠 Full Solution Path