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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      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 ?       AIEEE  Solved  Paper-2010

    A) \[\frac{{{b}^{2}}}{6a}\] 

    B)                        \[\frac{{{b}^{2}}}{2a}\]                 

    C)        \[\frac{{{b}^{2}}}{12a}\]               

    D)        \[\frac{{{b}^{2}}}{4a}\]

    Correct Answer: D

    Solution :

     \[U=\frac{a}{{{x}^{12}}}-\frac{b}{{{x}^{6}}}\]  \[{{U}_{x=\infty }}=0\] At equilibrium \[F=0=-\frac{dU}{dx}=-12a{{x}^{-13}}+6b{{x}^{-7}}=0\] \[\Rightarrow \] \[\frac{1}{{{x}^{6}}}=\frac{b}{2a}\]                         So \[x={{\left( \frac{2a}{b} \right)}^{\frac{1}{6}}}\] \[{{U}_{eq}}=-\frac{{{b}^{2}}}{4a}\]        \[D=\left[ 0-\left( -\frac{{{b}^{2}}}{4a} \right) \right]=\frac{{{b}^{2}}}{4a}\]


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