If a 0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength ( λ ) of the electron present in the second orbit of hydrogen atom? [ n : any integer]

Chemistry · Structure of Atom

JEE Main 2025 — 29 January, Morning Shift — Question 12

If $\mathrm{a}_{0}$ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength ( $\lambda$ ) of the electron present in the second orbit of hydrogen atom? [ n : any integer]

Option A:
$\; 2\pi a_0$
Option B:
$\; 8\pi a_0$
Option C:
$\; 4\pi a_0$
Correct
Option D:
$\; \frac{4\pi a_0}{2}$

Answer: C

Step-by-step solution

Standing wave condition, $2\pi r_n = n\lambda$

Bohr radius relation, $r_n = n^2 a_0$

On substituting, we have

$2\pi(n^2 a_0) = n\lambda$

$\lambda = 2\pi n a_0$

For second orbit, $n = 2$

Therefore, $\lambda = 4\pi a_0$ Thus, the correct answer is $4\pi a_0$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Wave-Particle Duality of Matter - de Broglie, Heisenberg