JEE MAIN - Physics (2010 - No. 25)
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$$ where $$a$$ and $$b$$ are constants and $$x$$ is the distance between the atoms. If the dissociation energy of the molecule is $$D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$$ is
$${{{b^2}} \over {2a}}$$
$${{{b^2}} \over {12a}}$$
$${{{b^2}} \over {4a}}$$
$${{{b^2}} \over {6a}}$$
Explanation
Given $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$$
$${U\left( {x = \infty } \right)}$$ = 0
We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$
At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$
$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$
$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$
$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$
$$ = - {{{b^2}} \over {4a}}$$
$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$
$${U\left( {x = \infty } \right)}$$ = 0
We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$
At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$
$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$
$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$
$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$
$$ = - {{{b^2}} \over {4a}}$$
$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$
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