Q 12-12-115JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
A particle of mass $m$ moves in a circular orbit in a central potential field $U(r) = \dfrac12kr^2$. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as:
Answer: (B) $r_n \propto \sqrt n$, $E_n \propto n$
Force $F = kr$ provides the centripetal force: $\dfrac{mv^2}{r} = kr \Rightarrow v = r\sqrt{\dfrac km}$.
Quantisation $mvr = \dfrac{nh}{2\pi}$: $mr^2\sqrt{\dfrac km} = \dfrac{nh}{2\pi}$, so $r^2 \propto n$ and $r_n \propto \sqrt n$.
Energy $E = \dfrac12mv^2 + \dfrac12kr^2 = kr^2 \propto n$.
Solution by Sreeraj P, M.Sc Physics