According to Bohr’s model of the hydrogen atom, the ratio of the electrostatic force on the electron in the orbit \(n=1\) to that in the orbit \(n=2\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(16:1\)
The electrostatic force on the electron is \( F=e^{2}/4\pi\varepsilon_0r^{2} \), so \( F\propto1/r^{2} \). In Bohr’s model, the orbit radius grows as \( n^{2} \). So the force falls as \( 1/n^{4} \).
\( \displaystyle F\propto\frac{1}{r^{2}}\propto\frac{1}{n^{4}} \)
\( \begin{aligned} \frac{F_1}{F_2} &= \frac{2^{4}}{1^{4}} \\ &= \frac{16}{1} \end{aligned} \)
Hence, the force on the electron in the first orbit is 16 times the force in the second orbit. The ratio is \( 16:1 \).
NCERT (2026-27):
Electron orbits – “… Thus, for a dynamically stable orbit in a hydrogen atom Fe = Fc”
\( \displaystyle \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}} \)\( \displaystyle =\frac{mv^{2}}{r} \qquad (12.2) \)
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^{2}}{m}\left(\frac{h}{2\pi}\right)^{2}\frac{4\pi\varepsilon_0}{e^{2}} \)\( \qquad (12.7) \)