The potential energy function for the force between two atoms in … - Vidyadip

MCQ

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x)=$ $\;\frac{a}{{{x^{12}}}} - \frac{b}{{{x^6}}}$ where a and b are constants and $x $ is the distance between the atoms. If the dissociation energy of the molecule is $D = [ U ( x = \infty) - U_{ at\ equilibrium}], D$ is

A
$\frac{{{b^2}}}{{2a}}$
B
$\;\frac{{{b^2}}}{{6a}}$
✓
$\;\frac{{{b^2}}}{{4a}}$
D
$\;\frac{{{b^2}}}{{12a}}$

✓

Answer

Correct option: C.

$\;\frac{{{b^2}}}{{4a}}$

c
At equilibrium : $\frac{{dU(x)}}{{dx}} = 0$

$ \Rightarrow \frac{{ - 12a}}{{{x^{11}}}} = \frac{{ - 6a}}{{{x^5}}} \Rightarrow x = {\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}$

$\begin{gathered}
{U_{at}}\,equilibrium = \frac{a}{{{{\left( {\frac{{2a}}{b}} \right)}^2}}} - \frac{b}{{\left( {\frac{{2a}}{b}} \right)}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{b^2}}}{{a4}}\,and\,{U_{\left( {x = \infty } \right)}} = 0 \hfill \\
\end{gathered} $

$\therefore $ $D = 0$$ - \left( { - \frac{{{b^2}}}{{4a}}} \right) = \frac{{{b^2}}}{{4a}}$

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