InterviewSolution
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InterviewSolution
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The potential energy funtions for the force between two along in a distance molecule is approximatily given by `U(x) = (a)/(x^(12)) - b)/(x^(6)) ` where `a` and `b` are constant and `x` is the distance between the aloms , if the discision energy of the molecale is `D = [U(x = oo) - U` atequlibrium ] , D isA. `(b^(2))/(12a)`B. `(b^(2))/(4a)`C. `(b^(2))/(6a)`D. `(b^(2))/(2a)` |
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Answer» Correct Answer - B `U_((x))-(a)/(x^(12))-(b)/(x^(6))` `F=-(dU(x))/(dx)=-(12a)/(x^(13))+(6b)/(x^(7))` At equilibrium distance between the molecules, `F=0` `:. -(12a)/(x^(13))+(6b)/(x^(7))=0` or `a=((2a)/(b))^(1//6)` From `(i)` `U_((oo))=(a)/(oo^(12))-(b)/(oo^(6))=0` Dissociation energy of molecule, `D=U_((x=oo))-U_("at equilibrium")` `=0-((a)/(x^(12))-(b)/(x^(6)))` where`x=((2a)/(b))^(1//6)` `D=(-a)/(x^(12))+(b)/(x^(6))=-(a)/((2a//b)^(12//6))+(b)/((2a//b)^(6//6))` `=-(b^(2))/(4a)+(b^(2))/(2a)=(b^(2))/(4a)` |
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