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H2+ molcule

  1. May 20, 2006 #1
    hi I need help in solving this

    The protons the H2+ molculer ion are 0.106 nm apart and the binding energy of H2+ is 2.65 ev , what negetive charge must be placed halfway between two protons this distance apart to give the same binding energy?

    I calculated it for the electrons , but here i didn know for the protons ?
    I tryed to use the graph I had but I dont think its the right solution
    so if someone could help me or give an eqation I could apply

    Last edited by a moderator: May 20, 2006
  2. jcsd
  3. May 20, 2006 #2
    Please correct me if I'm wrong...

    The two protons repel each other so left to itself, the two proton system is electrostatically unstable. You need a negative charge somewhere between them so that they are both attracted to the -ve charge and effectively stay in place. The binding energy is now defined in terms of the 3 charge system: 2 protons + 1 negative charge. [No other nuclear force is being considered for the mathematical calculations]

    Does this help?

    PS--Show your work
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