Question

the interaction energy between two atoms is given by e(r)=(-A/r +B/r7) where r is the inter atomic separation.show that for the particles to be in equilibrium.r=re=(7B/A) and show that for the equilibrium the energy of attraction is
seven times the energy of repulsion

Answer 1

Answer on Question #73584-Physics-Other

The interaction energy between two atoms is given by $e(r) = (-A / r + B / r7)$ where $r$ is the inter atomic separation. Show that for the particles to be in equilibrium $r = re = (7B / A)$ and show that for the equilibrium the energy of attraction is seven times the energy of repulsion.

Solution

For the equilibrium:

$$\frac {d}{d r} e (r) = 0 = - A \left(- \frac {1}{r ^ {2}}\right) + B \left(- 7 \frac {1}{r ^ {8}}\right)$$ $$\frac {A}{r ^ {2}} = \frac {7 B}{r ^ {8}}$$ $$r ^ {6} = \frac {7 B}{A}$$ $$r _ {e q} = \sqrt [ 6 ]{\frac {7 B}{A}}$$

The ratio of energies:

$$\frac {e _ {a t}}{e _ {r e}} = \frac {\frac {A}{\tau_ {e q}}}{\frac {B}{\tau_ {e q} ^ {7}}} = \frac {A}{B} r _ {e q} ^ {6} = \frac {A}{B} \frac {7 B}{A} = 7.$$

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