In a hydrogen atom, the radius of an electron orbit is 9 times the radius of the ground-state orbit. How many de Broglie wavelengths of the electron fit into the circumference of this orbit?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
In Bohr’s model the orbit radius is proportional to \( n^2 \). A radius 9 times the ground-state radius belongs to the \( n=3 \) orbit.
\( \begin{aligned} \frac{r_n}{r_1} &= n^2 \\ &= 9 \end{aligned} \)
\( n=3 \)
In de Broglie’s picture, the electron wave forms a standing wave round the orbit. The circumference then holds a whole number of wavelengths, \( 2\pi r_n=n\lambda \). Therefore, three de Broglie wavelengths fit into the circumference of this orbit.
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)\( \qquad (12.7) \)
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”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)