Read the following passage and answer the questions.
In Bohr’s model of the hydrogen atom, the electron moves only in certain allowed orbits. Bohr’s second postulate picks out these orbits. Of all his postulates, it looked the most puzzling, because it gave no reason for this choice. The French physicist Louis de Broglie later gave a reason. He had suggested that material particles, such as electrons, also have a wave nature. He argued that the electron in its circular orbit must be seen as a particle wave. Waves travelling on a stretched string can form standing waves when the conditions are right. In the same way, the particle wave of the electron can form a standing wave along its orbit. This idea explains why only certain orbits are allowed. It links the allowed orbits of the atom to the wave nature of the electron.
For the electron in the \(n\)th orbit, its de Broglie wavelength \(\lambda_n\) is plotted against \(n\). The graph is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → a straight line passing through the origin
De Broglie’s condition gives the wavelength in the \( n \)th orbit as \( \lambda_n=2\pi r_n/n \). In Bohr’s model the radius of the orbit grows as \( n^{2} \).
\( r_n\propto n^{2} \)
\( \begin{aligned} \lambda_n &= \frac{2\pi r_n}{n} \\ &\propto \frac{n^{2}}{n} \\ &\propto n \end{aligned} \)
From the above, \( \lambda_n\propto n \). Hence, the graph of \( \lambda_n \) against \( n \) is a straight line through the origin.
Option (1) a curve that falls as \(n\) increases — Incorrect. A falling curve would need \( \lambda_n\propto 1/n \). Here the radius grows faster than \( n \), so \( \lambda_n \) increases.
Option (2) a curve that rises more and more steeply — Incorrect. That shape belongs to \( r_n\propto n^{2} \). Dividing \( 2\pi r_n \) by \( n \) leaves only the first power of \( n \).
Option (3) a straight line parallel to the \(n\)-axis — Incorrect. A line parallel to the \( n \)-axis would mean the same wavelength in every orbit. The wavelength changes from orbit to orbit.
Option (4) a straight line passing through the origin — Correct. With \( \lambda_n\propto n \), doubling \( n \) doubles the wavelength. So the plotted points lie on a straight line through the origin.
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”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)