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B Deriving the ionization energy of a hydrogen atom

  1. Apr 28, 2017 #1
    Hi, I've just gotten started with basic quantum physics in physics class and we've just talked about ionization energy. It is stated that the energy of a hydrogen atom is -13.60eV (or -2.179aJ). I assume this is the potential energy (and that this is the reason the atom has a lower mass than the sum of the mass of its components on their own). I'm curious as to how this can be derived.

    My book briefly states that the radius of the hydrogen atom (i.e. the distance between electron an proton) is 5.29*10-11 m. My thinking is the following: The electrostatic force on the electron is F=k Q1 Q2 / r2 = - k e2 / r2 in this case. The potential energy is thus Ep = F r = -k e2 / r = -8.988*109 (1.60218*10-19)2 / (5.29*10-11) ≈ -4.36*10-18 J ≈ -27.22eV ≈ 2*(-13.60eV).

    So I'm off by a factor of two. But I feel like I'm on the right track since my calculated energy is so close to the double of what I want to get. I'm thinking that the electrons wave function shouldn't have any effects on this, but I'm not very familiar with this. What am I missing? Thanks for any input!
     
    Last edited by a moderator: Apr 28, 2017
  2. jcsd
  3. Apr 28, 2017 #2
    You forget to add the kinetic energy of the electron :)
     
  4. Apr 28, 2017 #3
    I tried this out and got it working! And it made a lot of sense, Thanks!
    Assuming the electron is obeying bohr's model of the atom its centripetal acceleration a = F / m, and by a = v2 / s, v = √(a s). The kinetic energy Ek = ½ m v2 = ½ e2/s 8.988*109 ≈ 2.18*10-18 J. This makes the extra energy gained of the assembly of the atom E = Ep + Ek ≈ -2.18*10-18 J ≈ -13.60eV which is what I was after!
     
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