Question

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

• A. $\left(\frac{ B }{ A }\right)^{1 / 6}, 0$
• B. $\left(\frac{ B }{2 A }\right)^{1 / 6},-\frac{ A ^{2}}{2 B }$
• C. $\left(\frac{2 B}{A}\right)^{1 / 6},-\frac{A^{2}}{4 B}$
• D. $\left(\frac{2 B }{ A }\right)^{1 / 6},-\frac{ A ^{2}}{2 B }$

Answer 1

Answer: (C)

$U =\frac{- A }{ r ^{6}}+\frac{ B }{ r ^{12}}$
$F =-\frac{ d U }{ dr }=-\left( A \left(-6 r ^{-7}\right)\right)+ B \left(-12 r ^{-13}\right)$
$0=\frac{6 A }{ r ^{7}}-\frac{12 B }{ r ^{13}}$
$\frac{6 A }{12 B }=\frac{1}{ r ^{6}} \Rightarrow r =\left(\frac{2 B }{ A }\right)^{1 / 6}$
$U \left( r =\left(\frac{2 B }{ A }\right)^{1 / 6}\right)=-\frac{ A }{2 B / A }+\frac{ B }{4 B ^{2} / A ^{2}}$
$=\frac{- A ^{2}}{2 B }+\frac{ A ^{2}}{4 B }=\frac{- A ^{2}}{4 B }$