In a hydrogen atom, the de Broglie wavelength of the electron in a certain orbit is \(\pi\) times the radius of that orbit. The principal quantum number \(n\) of this orbit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
De Broglie pictured the orbiting electron as a standing wave. In the \( n^{\text{th}} \) orbit, \( n \) whole wavelengths fit into the circumference. Here the wavelength is \( \lambda=\pi r_n \).
\( \begin{aligned} 2\pi r_n &= n\lambda \\ &= n\pi r_n \end{aligned} \)
\( n=2 \)
Therefore, the orbit is the one with \( n=2 \). In the ground state the wavelength would be the whole circumference, \( 2\pi r_1 \).
NCERT (2026-27):
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… For an electron moving in nth circular orbit of radius rn, the total distance is the circumference of the orbit, 2πrn. Thus”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)