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Velocity of alpha particle within nucleus

  1. Sep 10, 2013 #1
    The decay constant for alpha decay is the product of the transmission probability P and the collision frequency f of the alpha with the potential barrier, where f depends on velocity of the alpha in the nucleus.

    My question is about this Hyperphysics page, where alpha decay is modelled.

    Here in both the diagram and description, the alpha seems to have the same velocity within and outside the nucleus. Is that correct? I would have thought that attenuation of the wavefunction by the potential barrier would lead to an energy difference between alpha within and outside the nucleus.
  2. jcsd
  3. Sep 10, 2013 #2
    ah so it only applies for potential wells with their baseline at zero so that really

    Ekin, in = Ekin, out + |Ewell depth|

    where the last term is the depth of negative potential

    still I am surprised that alpha energy is not dissipated by crossing the barrier
  4. Sep 10, 2013 #3


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    Staff: Mentor

    No, it has the same total energy (potential plus kinetic) inside and outside the nucleus. The horizontal line represents the total energy. Outside the nucleus it has more potential energy (represented by the height of the shaded region), so it has less kinetic energy, and the velocity is therefore smaller.
  5. Sep 10, 2013 #4


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    Staff: Mentor

    It doesn't "cross" the barrier in the sense of physical travel.

    Barrier energy works like rolling an object up a hill and then back down again: The energy you spend pushing the object up to the top of the hill you get back when it rolls down again.

    Without tunneling, the object rolls back down on the same side that it started up, unless you can push it all the way to the top of the hill so it can go back down on either side. With tunneling, the object always a definite height but not a definite position, so it can come rolling down on either side of the hill even if it isn't at the top.

    [Warning - this is a pretty decent intuitive semi-classical model of what's going. It is, however, just a model and if you push it too far it will stop working. If you want to really understand, you'll have to get hold of a first-year QM text and try working some of the problems in it].
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