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If the potential energy of two molecules is given by \(\sf U=\frac{A}{r^{12}} - \frac{B}{r^6}\)then at equilibrium position, its potential energy is equal to
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Answer given is \(\sf \frac{-B^2}{4A}\)
try this \[ \frac{dU}{dr}=\frac{-12A}{r^{13}}+\frac{6B}{r^7} = 0\] \[r_e^6 = \frac{2A}{B}\] \[ U_e=\frac{A}{(\frac{2A}{B})^2}−\frac{B}{\frac{2A}{B}}\] etc
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