Chemistry · Structure of Atom

JEE Main 2026 — 6 April, Evening Shift — Question 46

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 )

Option A:
$\mathrm{Li}^{2+}, 3$
Option B:
$\mathrm{He}^{+}, 3$
Option C:
$\mathrm{He}^{+}, 2$
Option D:
$\mathrm{Li}^{2+}, 2$
Correct

Answer: D

Step-by-step solution

$\mathrm{r}_{\text {species }}=\mathrm{a}_{0} \frac{\mathrm{n}^{2}}{\mathrm{Z}}$ $70.53=52.9\left(\frac{\mathrm{n}^{2}}{\mathrm{Z}}\right)$ $\frac{\mathrm{n}^{2}}{\mathrm{Z}}=\frac{70.53}{52.9}=1.33=\frac{4}{3}$ Data will satisfy if $\mathrm{n}=2$ and $\mathrm{z}=3$ $\mathrm{Li}^{+2}, 2 \Rightarrow \frac{4}{3}=1.33$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Bohr's Model of Atom