Question
The interaction energy between two atoms is given by where r is the inter-atomic separation. Show that for the particles to be in equilibrium,
and show that in stable equilibrium the energy of attraction is seven times the energy of repulsion.
Answer :
Word Count : 422
We are given the interaction energy between two atoms: \[ E(r) = -\frac{A}{r} + \frac{B}{r^7} \] where \( r \) is the inter-atomic separation, and \( A \) and \( B \) are constants. ### 1. To find the equilibrium distance \( r_e \), we need to minimize the total energy \( E(r) \). At equilibrium, the derivative of the potential energy \( E(r) \) with respect to \( r \) should be zero (since this corresponds to a minimum or maximum of the ___ ________ _____ _________ _____ ____ _________.
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We are given the interaction energy between two atoms: \[ E(r) = -\frac{A}{r} + \frac{B}{r^7} \] where \( r \) is the inter-atomic separation, and \( A \) and \( B \) are constants. ### 1. To find the equilibrium distance \( r_e \), we need to minimize the total energy \( E(r) \). At equilibrium, the derivative of the potential energy \( E(r) \) with respect to \( r \) should be zero (since this corresponds to a minimum or maximum of the ___ ________ _____ _________ _____ ____ _________.
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