The angular momentum of the electron in an orbit of a hydrogen atom is \(8.4\times10^{-34}\ \mathrm{J\,s}\). How many de Broglie wavelengths of the electron fit into the circumference of this orbit? (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
Bohr’s second postulate gives the angular momentum as \( L=nh/2\pi \). De Broglie showed that the same whole number \( n \) counts the wavelengths that fit round the orbit, because \( 2\pi r_n=n\lambda \) leads to \( mv_nr_n=nh/2\pi \).
\( \begin{aligned} n &= \frac{2\pi L}{h} \\ &= \frac{2\times3.14\times8.4\times10^{-34}}{6.6\times10^{-34}} \\ &= 8 \end{aligned} \)
From the above, \( n=8 \). Hence, eight de Broglie wavelengths fit into the circumference of this orbit.
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”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… From Eq. (12.12), we have”
\( \displaystyle 2\pi r_n=\frac{nh}{mv_n}\quad\text{or}\quad \)\( \displaystyle mv_nr_n=\frac{nh}{2\pi} \)