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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= \left [ U\left ( x=\infty \right )-U_{at \: equilibrium} \right ],     D\: is

 

  • Option 1)

    \frac{b^{2}}{6a}

  • Option 2)

    \frac{b^{2}}{2a}

  • Option 3)

    \frac{b^{2}}{12a}

  • Option 4)

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

 

Answers (2)

best_answer

As we have learned

Conservative Force -

F=\frac{-dU}{dr}

- wherein

Negative of the rate of change of potential energy with respect to position

 

 U = \frac{q}{x^{12}}- \frac{b }{x^6}

At equilibrium F = - dv/dx = 0 

\Rightarrow \frac{d}{dx }(\frac{a}{x^{12}-\frac{6}{x^6}})= 0 \Rightarrow a \frac{-12}{x^{13}}- b \frac{-b}{x^7 }= 0

or 

 x^6 = \frac{12a }{6b }\: \: \: or \: \: \: x = \left ( \frac{2a}{b} \right )^{1/6}

U_{equilibrium } = \frac{a}{\left ( \frac{2a}{b} \right )^2}-\frac{b}{\left ( \frac{2a}{b} \right )}= \frac{ab^2}{4a^2}-\frac{b^2}{4a}

D = U (x = \infty ) - U_{equilibrium } = 0 - (-\frac{b^2}{4a})= + \frac{b^2}{4a}

 

 

 

 

 


Option 1)

\frac{b^{2}}{6a}

Option 2)

\frac{b^{2}}{2a}

Option 3)

\frac{b^{2}}{12a}

Option 4)

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

Posted by

SudhirSol

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