In a hydrogen atom, the circumference of an electron orbit is \(1.2\times10^{-8}\ \mathrm{m}\). The de Broglie wavelength of the electron in this orbit is \(2.0\times10^{-9}\ \mathrm{m}\). The angular momentum of the electron is (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(6.30\times10^{-34}\ \mathrm{J\,s}\)
An allowed orbit holds a whole number \( n \) of de Broglie wavelengths in its circumference. This number is also the \( n \) in Bohr’s rule \( L=nh/2\pi \).
\( \begin{aligned} n &= \frac{2\pi r_n}{\lambda} \\ &= \frac{1.2\times10^{-8}}{2.0\times10^{-9}} \\ &= 6 \end{aligned} \)
\( \begin{aligned} L &= \frac{nh}{2\pi} \\ &= \frac{6\times6.6\times10^{-34}}{2\pi} \\ &= 6.30\times10^{-34}\ \mathrm{J\,s} \end{aligned} \)
Therefore, the angular momentum of the electron is \( 6.30\times10^{-34}\ \mathrm{J\,s} \).
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”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)
Bohr Model of the Hydrogen Atom – “… This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is”
\( \displaystyle L=\frac{nh}{2\pi} \)\( \qquad (12.5) \)