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Kinetic energy of a nucleus

  1. Jun 25, 2013 #1
    Hello everybody, I am a first year student and I have a question that I am not sure if I can answer.

    A radioactive isotope of Tin(50), with 85 neutrons in its nucleus and without any electrons bound to it, undergoes a beta minus decay. The electron pops into existence, 10-15m from the center of the nucleus (consider that the nucleus is uniformly charged), and it leaves with a speed of 0.999c.

    (We consider Mn=Mp=1.6x10-27)

    1. What is the final speed of, both the electron and the nucleus?

    2. Will the the new nucleus, be able to bind the electron, into one of its orbits?

    Tips: The new nucleus will have 51 protons and 84 neutrons.

    The atom is a hydrogen like one, so we can find the probable bound energies through the Bohr model. (We don't care about its angular distribution)

    The kinetic energy of the electron has to be calculated through the K = (γ-1)mc2
    and the potential energy from the Coulomb law U = -Ze2/(4πε0r).

    I assume that the energy levels will be 51x13.6/n2eV.

    So the hard part, is the equation of potential-kinetic energy with two varying speeds, that can give us the result.
     
  2. jcsd
  3. Jun 25, 2013 #2

    mfb

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    "two varying speeds"?
    Did you compare the kinetic energy of the electron to its potential energy?

    That could indicate that you have to take the finite size of the nucleus into account. Do you know models about the size of nuclei?
     
  4. Jun 27, 2013 #3
    No. I am just saying that, as the electron is travelling, it is 'pulling' the nucleus with it. A positive and a negative charge, that are getting separated (cause of the moving electron and the stationary, at the beginning nucleus.)

    Personally I do not. However I am thinking of it as : It is of a finite size and quite small. 10-15m, is its whole length.
     
  5. Jun 27, 2013 #4

    mfb

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    Well that would be interesting!

    In that case, the electron would "appear" (directly) outside the nucleus, and the given uniform charge distribution in the nucleus would not matter.
     
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