The electron in a hydrogen atom revolves in an orbit of radius \(r\). Its de Broglie wavelength in this orbit is \(\lambda\). The angular momentum of the electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(hr/\lambda\)
For an electron moving in a circle, the angular momentum is \( L=mvr \), where \( mv \) is the linear momentum \( p \). De Broglie’s relation gives this momentum in terms of the wavelength, \( p=h/\lambda \).
\( \begin{aligned} L &= pr \\ &= \frac{hr}{\lambda} \end{aligned} \)
From the above, \( L=hr/\lambda \). With \( 2\pi r=n\lambda \) for an allowed orbit, this becomes \( L=nh/2\pi \), which is Bohr’s quantum condition.
NCERT (2026-27):
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… From Chapter 11, we have λ = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mvn. Thus, λ = h/ mvn.”
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… we have”
\( \displaystyle 2\pi r_n=\frac{nh}{mv_n}\quad\text{or}\quad \)\( \displaystyle mv_nr_n=\frac{nh}{2\pi} \)