A small particle of mass m moves in such a way that the potential energy `U=(1)/(2)m omega^(2)r^(2)`, where `omega` is a constant and r is the distance of the particle from the origin. Assuming Bohr's model of quantisation of angular momentum and circular orbits. Find the radius of the `n^(th)` orbit.

Explanation

`U=(1)/(2)m omega^(2)r^(2)`
`F=-(dU)/(dr)=-m omega^(2)r`
-ve sign shows that force F is attractive
`F=(mv^(2))/(r )`
`(mv^(2))/(r ) implies v=r omega`
`mvr=(nh)/(2pi)`
`m omega r^(2)=(nh)/(2 pi)`
`r=r_(n)=sqrt((nh)/(2pi m omega))`