A small particle of mass m moves in such a way that the potential energy `U=(1)/(2) m^2 omega^2 r^2,` where `omega` is a constant and r is the distance of the particle from the origin. Assuming Bohr model of quantization of angular momentum and circular orbits, show that radius of the nth allowed orbit is proportional to `sqrtn.`

Explanation

`F=(delU)/(delr)={-(1)/(2)mb^(2)r}=mb^(2)r`....(i)
Quantization of angular momentum
`L=(nh)/(2pi)`……(ii)
Centipetal force `=(mv^(2))/(r )`
`F` will provide centripetal force, So, from (i) `F`
So, `mb^(2)r=(mv^(2))/(r )`
`b^(2)r^(2)=v^(2)`
`v=br`.....(iii)
`L=mvr=mbr^(2)`
From (ii)
`mbr^(2)=(nh)/(2pi)`
`r^(2)=(n)/(mb)(h)/(2pi)`
So, `r propsqrt(n)` Hence proved.