JEE Advanced 2024 Paper 1 · Q05 · Bohr Model
A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r) = -kr$, corresponding to the potential energy $V(r) = kr^2/2$, where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by $L = n\hbar$, where $\hbar = h/(2\pi)$, $h$ is the Planck's constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, then which of the following expression(s) is(are) correct?
Reveal answer + step-by-step solution
Correct answer:A, B, C
Solution
Centripetal: $\dfrac{mv^2}{r} = kr \Rightarrow v = r\sqrt{k/m}$, so $\dfrac{v}{r} = \sqrt{k/m}$. Bohr quantisation: $L = mvr = n\hbar$.
(C) $\dfrac{L}{mr^2} = \dfrac{v}{r} = \sqrt{k/m}$. ✓
(A) From $L = mvr$ and $v = r\sqrt{k/m}$: $n\hbar = mr^2\sqrt{k/m} = r^2\sqrt{mk}$, hence $r^2 = \dfrac{n\hbar}{\sqrt{mk}} = n\hbar\sqrt{\dfrac{1}{mk}}$. ✓
(B) $v^2 = (k/m) r^2 = (k/m)\cdot \dfrac{n\hbar}{\sqrt{mk}} = n\hbar\sqrt{\dfrac{k}{m^3}}$. ✓
(D) $E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kr^2 = kr^2$ (since $\tfrac{1}{2}mv^2 = \tfrac{1}{2}kr^2$). So $E = k\cdot\dfrac{n\hbar}{\sqrt{mk}} = n\hbar\sqrt{\dfrac{k}{m}}$, which is twice option (D). ✗
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