The de Broglie wavelength of the electron revolving in the 4th orbit of a hydrogen atom is \( \lambda \). What is its de Broglie wavelength in the 2nd orbit?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \lambda/2 \)
In de Broglie’s picture, the electron in the nth orbit forms a standing wave. So the circumference of the orbit holds a whole number of wavelengths: \( 2\pi r_n=n\lambda_n \). In the Bohr model, the orbit radius grows as \( r_n\propto n^{2} \).
The de Broglie wavelength in the nth orbit is then:
\( \begin{aligned} \lambda_n &= \frac{2\pi r_n}{n} \\ &\propto \frac{n^{2}}{n} \\ &\propto n \end{aligned} \)
For the 2nd and the 4th orbits:
\( \begin{aligned} \frac{\lambda_2}{\lambda_4} &= \frac{2}{4} \\ &= \frac{1}{2} \end{aligned} \)
Since \( \lambda_4=\lambda \), the wavelength in the 2nd orbit is \( \lambda_2=\lambda/2 \).
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
\( 2\pi r_n=n\lambda \)\( ,\quad n=1,2,3\ldots \qquad (12.12) \)”
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is
\( \displaystyle r_n=\frac{n^{2}}{m}\left(\frac{h}{2\pi}\right)^{2}\frac{4\pi\varepsilon_0}{e^{2}} \)\( \qquad (12.7) \)”