A particle of mass m moves in a circular path in such a way that the potential energy is $U=a{r}^2$ where ‘a’ is a constant and r is the distance of the particle from the origin. Assuming Bohr's model to be valid, find the radius of $n^{th}$ allowed orbit.

Potential energy function is given as $U=a{r}^2$

$\therefore Force, F = \frac{-dU}{dr} = -2ar$

$F = -2ar$ is force which provides centripetal acceleration to the particle moving along circular path.

$\begin{array}{l}\therefore F=\frac{m{v}^2}{r}\\ \Rightarrow 2ar=\frac{m{v}^2}{r} ...........\left(1\right)\end{array}$

And it is given that angular momentum is quantised according to Bohr's model.

  $i.e., mvr=\frac{nh}{2\pi } ..........\left(2\right)$

Solving (1) and (2)

We get  $r={\left\{\frac{n^2{h}^2}{8{\pi }^{2}am}\right\}}^{1/4}$