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Electric potential, etc.

  1. Feb 16, 2005 #1
    Hi,

    I finished the following problem and have gotten all the right answers, but I want to make sure my reasoning is correct.

    In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus) in a circular orbit of radius 0.53 x 10^-10 m.

    (a) What is the electric potential at the electron's orbit due to the proton?
    27 volts using v = k (q/r) -- this one is simple, you just use the radius, and the charge, and k. OK.

    (b) What is the kinetic energy of the electron?
    Change in potential energy is the opposite of change in kinetic energy. Potential energy is QV, or -2.2 x 10^-18.

    (c) What is the total energy of the electron in orbit?
    -2.2 x 10^-18. Same as above because when it's in the orbit all of its energy is kinetic energy.

    (d) What is the ionization energy -- that is, the energy required to remove the electron from the atom and take it to r = infinity, at rest?
    2.2 x 10^-18. Same but opposite because you are pulling the electron away from a charge to which it is attracted. This requires a force equal and opposite to its kinetic energy in that position.

    Do these answers seem complete?
     
  2. jcsd
  3. Feb 16, 2005 #2
    a) This seems right

    b) This is not right. Try setting the electric force equal to the centripetal force. Solve for the electrons velocity and then find its kinetic energy. The number is right but you reasoning seems off. Also it is not possible to have negative kinetic energy.

    c) This is not right either. Total energy should be the sum of kinetic and potential. The number is right however. The kinetic energy should be positive and the potential energy should be negative. Use the suggestion in part b or use kepler's laws.

    d) This has the right idea and the right number.
     
  4. Feb 16, 2005 #3
    Hi Davorak,

    Thanks for your help. Right, sorry about that, the number for B was supposed to be positive. I was under the impression, though, that applying conservation you would have potential energy having to equal kinetic energy. Otherwise, I don't really understand how the numbers could be the same except opposite (I checked the back of the book and the answers are all right).
     
  5. Feb 16, 2005 #4
    The numbers are right, but potential energy does not equal kinetic energy for a stable orbit. This is part of orbital theroy or kepler's laws.

    The easy way to find this for yourself is to:
    Setting the electric force equal to the centripetal force. Solve for the electrons velocity and then find its kinetic energy.

    If you do this you will see that in fact the magitude of the kinetic energy does not equal the magnitude of the potential energy.

    Edit:
    Total energy = kinetic + potential
    if kinetic was equal but oppsite of the potentail energy then total energy would equal zero

    I forgot to mention:
    Potential energy QV[tex]\neq [/tex] -2.2*10^-18.
     
    Last edited: Feb 16, 2005
  6. Feb 16, 2005 #5
    Right, becuase i get 4.32 * 10^-18 when i multiply the charge of electron by the Voltage.
    But if you divide it by 2, you will get 2.2*10^-18

    However, i have no idea why should it be divided by two....
     
  7. Feb 17, 2005 #6
    Yes, I have the same problem -- why is it divided by 2?????
     
  8. Feb 17, 2005 #7
    Ok.
    You know.
    Electric potential:
    [tex]
    \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r} = k \frac{q}{r} = 4.32 * 10^{-18}
    [/tex]

    The problem assumes the electron takes a circular orbit around the proton.
    For a stable circular orbit the electric force must equal the centripetal force.
    [tex]
    \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} = m \frac{v^{2}}{r}
    [/tex]
    [tex]
    \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r} = m v^{2}
    [/tex]
    [tex]
    \frac{1}{4 m \pi \epsilon_{0}} \frac{q}{r} = v^{2}
    [/tex]
    [tex]
    v = sqrt{\frac{1}{4 m \pi \epsilon_{0}} \frac{q}{r}}
    [/tex]
    kinetic energy equals[tex]\frac{1}{2}mv^{2}[/tex]. So the kinetic energy of the electron is:
    [tex]
    E_{Ke} = \frac{1}{2} \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}
    [/tex]
    Look familiar?

    Google search on kepler's laws good history and orbital theory
    http://www.google.com/search?hl=en&q=kepler's+laws&spell=1
     
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