Question

Interaction between atoms of a diatomic molecule is described by the relation P.E. (x) = $\frac { \alpha } { x ^ { 12 } } - \frac { \beta } { x ^ { 6 } }$, where P.E. is potential energy. If the two atoms enjoy stable equilibrium, the distance between them would be

• A. $\left( \frac { \alpha } { \beta } \right) ^ { 1 / 6 }$
• B. $\left( \frac { 2 \alpha } { \beta } \right) ^ { 1 / 6 }$
• C. $\left( \frac { 2 \beta } { \alpha } \right) ^ { 1 / 6 }$
• D. $\left( \frac { \beta } { \alpha } \right) ^ { 1 / 6 }$

Answer 1

Answer: (B)

$\left( \frac { 2 \alpha } { \beta } \right) ^ { 1 / 6 }$

Answer 2

Potential energy is minimum if equilibrium is stable.

$\frac { \mathrm { d } } { \mathrm { dx } } [$ P.E. $( \mathrm { x } ) ] = 0$

i.e.,

i.e., $\frac { - 12 \alpha } { x ^ { 13 } } + \frac { 6 \beta } { x ^ { 7 } }$ = 0

i.e., [-2α + βx⁶] $\frac { 6 } { x ^ { 13 } }$ = 0

i.e., (-2α + βx⁶) = 0

i.e., x = $\left( \frac { 2 \alpha } { \beta } \right) ^ { 1 / 6 }$