If the potential energy between two molecules is given by U =-A r… - Vidyadip

MCQ

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

✓

Answer

Correct option: C.

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

c
$U =\frac{- A }{ r ^{6}}+\frac{ B }{ r ^{12}}$

$F =-\frac{ dU }{ dr }=-\left( A \left(-6 r ^{-7}\right)\right)+ B \left(-12 r ^{-13}\right)$

$0=\frac{6 A }{ r ^{7}}-\frac{12 B }{ r ^{13}}$

$\frac{6 A }{12 B }=\frac{1}{ r ^{6}} \Rightarrow r =\left(\frac{2 B }{ A }\right)^{1 / 6}$

$U \left( r =\left(\frac{2 B }{ A }\right)^{1 / 6}\right)=-\frac{ A }{2 B / A }+\frac{ B }{4 B ^{2} / A ^{2}}$

$=\frac{- A ^{2}}{2 B }+\frac{ A ^{2}}{4 B }=\frac{- A ^{2}}{4 B }$

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