According to Bohr’s model of the hydrogen atom, the radius of the orbit is directly proportional to which of the following?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( n^2 \)
In the Bohr model, the angular momentum of the electron is quantised, \( L=nh/2\pi \). Bohr used this quantisation of angular momentum with the balance of the electrostatic and centripetal forces. Together they give the radius of the nth orbit of the hydrogen atom:
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)
Everything except n is a constant, so the radius is proportional to the square of n. Hence, \( r_n\propto n^2 \), and the orbits are at 1, 4, 9, ... times the radius of the innermost orbit.
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)\( \qquad (12.7) \)