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A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{\text{th}}$ orbit and the particle's speed $v$ in the orbit depend on $n$ as:
A
$r \propto n^{1/3}, v \propto n^{1/3}$
B
$r \propto n^{1/3}, v \propto n^{2/3}$
C
$r \propto n^{2/3}, v \propto n^{1/3}$
D
$r \propto n^{4/3}, v \propto n^{-1/3}$
Explanation
Centripetal balance $\frac{mv^2}{r} = F \implies v \propto r^{1/2}$. Angular momentum quantization $mvr = \frac{nh}{2\pi} \implies r^{3/2} \propto n \implies r \propto n^{2/3}$ and $v \propto n^{1/3}$.
Detailed Solution
For circular motion under a constant centripetal force $F$: $\frac{m v^2}{r} = F = \text{constant} \implies v = \sqrt{\frac{F r}{m}} \implies v \propto r^{1/2}$. According to the Bohr quantization postulate: $L = m v r = \frac{n h}{2\pi}$. Substituting $v \propto r^{1/2}$: $m (r^{1/2}) r \propto n \implies r^{3/2} \propto n \implies r \propto n^{2/3}$. Consequently, the orbital speed scales as: $v \propto r^{1/2} \propto (n^{2/3})^{1/2} = n^{1/3}$. Therefore, $r \propto n^{2/3}$ and $v \propto n^{1/3}$.