Saved Bookmarks
| 1. |
The potential energy between two atoms in a molecule is given by `U=ax^(2)-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 `:-`A. `0`B. `(2b)/(3a)`C. `(3a)/(2b)`D. `(3b)/(2a)` |
|
Answer» Correct Answer - B `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. |
|