The potential energy between two atoms in a molecule is given by
`U=ax^(3)-bx^(2)`
where `a` and `b` are positive constants and `x` is the distance between the atoms. The atom is in stable equilibrium when `x` is equal to `:-`

Options

A. `0`
B. `(2b)/(3a)`
C. `(3a)/(2b)`
D. `(3b)/(2a)`

Answer

B

Explanation

`U=ax^(3)-bx^(2)rArr(dU)/(dx)=3ax^(2)-2bx` and
`(d^(2)U)/(dx^(2))=6ax-2b`
For equilibrium `(dU)/(dx)=0rArr3ax^(2)-2bx=0`
`rArrx=0,(2b)/(3a)`
For `x=(2b)/(3a),(d^(2)U)/(dx^(2))=6axx(2b)/(3a)-2b=2bgt0`
Therefore at `x=(2b)/(3a)`, potential energy is minimum. This is a condition of stable equilibrium.