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Physics Question on mechanical properties of solids

If the potential energy between two molecules is given by U=−Ar6+Br12,U = - \frac{ A }{ r ^{6}}+\frac{ B }{ r ^{12}}, then at equilibrium, separation between molecules, and the potential energy are :

A

(BA)1/6,0\left(\frac{ B }{ A }\right)^{1 / 6}, 0

B

(B2A)1/6,−A22B\left(\frac{ B }{2 A }\right)^{1 / 6},-\frac{ A ^{2}}{2 B }

C

(2BA)1/6,−A24B\left(\frac{2 B}{A}\right)^{1 / 6},-\frac{A^{2}}{4 B}

D

(2BA)1/6,−A22B\left(\frac{2 B }{ A }\right)^{1 / 6},-\frac{ A ^{2}}{2 B }

Answer

(2BA)1/6,−A24B\left(\frac{2 B}{A}\right)^{1 / 6},-\frac{A^{2}}{4 B}

Explanation

Solution

U=−Ar6+Br12U =\frac{- A }{ r ^{6}}+\frac{ B }{ r ^{12}}
F=−dUdr=−(A(−6r−7))+B(−12r−13)F =-\frac{ d U }{ dr }=-\left( A \left(-6 r ^{-7}\right)\right)+ B \left(-12 r ^{-13}\right)
0=6Ar7−12Br130=\frac{6 A }{ r ^{7}}-\frac{12 B }{ r ^{13}}
6A12B=1r6⇒r=(2BA)1/6\frac{6 A }{12 B }=\frac{1}{ r ^{6}} \Rightarrow r =\left(\frac{2 B }{ A }\right)^{1 / 6}
U(r=(2BA)1/6)=−A2B/A+B4B2/A2U \left( r =\left(\frac{2 B }{ A }\right)^{1 / 6}\right)=-\frac{ A }{2 B / A }+\frac{ B }{4 B ^{2} / A ^{2}}
=−A22B+A24B=−A24B=\frac{- A ^{2}}{2 B }+\frac{ A ^{2}}{4 B }=\frac{- A ^{2}}{4 B }