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Repulsion energy of two atoms

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The dissociation energy of a hypothetical molecule AB (composed of A+ and B- ions) is 5.24 eV, and the equilibrium separation of AB is 0.216 nm. The electron affinity of a B atom is 3.36 eV and the ionization energy of element A is 4.00 eV. Determine the core-repulsion energy of AB.

    2. Relevant equations

    U(rep)=-[E(diss)+deltaE+Ue], where delta E is the dissociation energy minus the electron affinity

    Ue=-(ke^2)/r

    3. The attempt at a solution

    First I found delta E. Delta E= 4.00ev-3.36ev=.64eV.

    The dissociation energy, E(diss), is given as 5.24eV.

    I calculated the electrostatic potential energy and converted it to electron volts and got -6.6472eV.

    Therefore, U(rep)=-[5.24eV+.64eV-6.6472eV]=.767eV

    However, my assignment is telling me this is incorrect. Can anyone find where I went wrong? Any help is greatly appreciated.
     
  2. jcsd
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