Question

The potential energy of a diatomic molecule(a two-atom system like

H_{2}

or

O_{2}

) is given by

U = \frac{A}{r^{12}} - \frac{B}{r^{6}}

, where

r

is the separation of the two atoms of the molecule and

A

and

B

are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Accepted Answer

(a) By using the relation,

$\Rightarrow F = - \frac{d u}{d r}$

$\Rightarrow F = - \frac{d}{d r} (\frac{A}{r^{12}} - \frac{B}{r^{6}})$

$\Rightarrow F = - A \frac{d}{d r} (r^{- 12}) + B \frac{d}{d r} (r^{- 6})$

$\Rightarrow F = \frac{12 A}{r^{13}} - \frac{6 B}{r^{7}}$ .

Now for equilibrium position,

$\Rightarrow F = 0$ then

$\Rightarrow \frac{12 A}{r^{13}} - \frac{6 B}{r^{7}} = 0$

$U = \frac{A}{r^{12}} - \frac{B}{r^{6}}$

$\Rightarrow r^{6} = \frac{2 A}{B}$

$\Rightarrow r = {(\frac{2 A}{B})}^{\frac{1}{6}}$ .

(b) To find the nature of force between molecules when separation gets changed, you can use the concept of maxima and minima of potential energy to find the type of equilibrium.

$\Rightarrow \frac{d^{2} u}{d r^{2}} = \frac{d}{d r} (\frac{d u}{d r}) = \frac{d}{d r} (- F) = - \frac{d F}{d r}$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = - \frac{d}{d r} (\frac{12 A}{r^{13}} - \frac{6 B}{r^{7}})$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = 12 A \frac{d}{d r} (r^{- 13}) + 6 B \frac{d}{d r} (r^{- 7})$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = + \frac{156 A}{r^{14}} - \frac{42 B}{r^{8}}$ .

Now put $r = {(\frac{2 A}{B})}^{\frac{1}{6}}$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = \frac{156 A}{{(\frac{2 A}{B})}^{\frac{14}{6}}} - \frac{42 B}{{(\frac{2 A}{B})}^{\frac{8}{6}}}$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = \frac{156 {(B)}^{\frac{7}{3}}}{{(2 A)}^{\frac{7}{3}}} - \frac{42 B (B)^{\frac{4}{3}}}{{(2 A)}^{\frac{4}{3}}}$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = \frac{156}{2} \frac{{(B)}^{\frac{7}{3}}}{{(2 A)}^{\frac{4}{3}}} - 42 \frac{(B)^{\frac{7}{3}}}{{(2 A)}^{\frac{4}{3}}}$

$\Rightarrow \frac{d^{2} u}{d r^{2}} = 36 \frac{{(B)}^{\frac{7}{3}}}{{(2 A)}^{\frac{4}{3}}}$ .

Now as $\frac{d^{2} u}{d r^{2}}$ is (+ve) then potential energy is minimum this equilibrium position, hence, type of equilibrium is stable.

(b) When the separation between molecules is smaller, then the force will be repulsive.

(c) When the separation between the molecules is larger, then the force will be attractive.

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