The potential energy function of a particle is given by `U(r)=A/(2r^(2))-B/(3r)`, where `A` and `B` are constant and `r` is the radial distance from the centre of the force. Choose the correct option (s)

Options

A. The equilibrium distance will be `r_(0)=(2A)/B`
B. The equilibrium distance will be `r_(0)=(3A)/B`
C. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vaishh ate `(r_(0))/3`, where `r_(0)` is the equilibrium distance
D. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vanish at `(r_(0))/2`,where `r_(0)` is the equilibrium distance.

Answer

B, C

Explanation

`F=-(dU)/(dr)=A/(r^(3))-B/(3r^(2))`
For equilibrium `F=0, r=(3A)/B`
`(B^(2))/(6A)=A/(2r_(0)^(2))-B/(3r_(2))` (for `r_(0)=r/3`)