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Homework Help: Why electrons can't remain in the nucleus

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Estimate the momentum of an electron confined to an atomic nucleus of radius a, and deduce a formula for its speed. By comparing the speed to the escape velocity explain why electrons cannot remain in the nucleus.

    2. Relevant equations
    Uncertainty principle

    3. The attempt at a solution
    Uncertainty Principle: ∆x∆p≈h (1)

    If the electron were confined to the nucleus, the uncertainty in its position ∆x would be the radius of the nucleus. This means that the uncertainty in its momentum is given by rearranging (1) to:

    ∆p= ∆(mv)≈h/∆x => ∆v≈h/(m ∆x) (2)

    The uncertainty in the velocity ∆v is all we know about the velocity so this is the velocity that the electron may have. (Is this explained right? How may I word this better – I get confused between uncertainty and actual velocity.)

    ∆v≈(6.626×10^(-34))/(9.11×10^(-31)×10^(-15) )=(6.626×10^12)/9.11

    = 7.3×10^11 (this is faster than speed of light so must be wrong?)

    Potential Energy of an electron in the nucleus of a Bohr atom is PE= e^2/(4πε_0 a) where the radius of the nucleus is a~10^(-15) metres.

    Putting in values => PE=2.3×10^(-13) Joules

    Escape velocity can be calculated from KE = PE.

    KE=(m_e v^2)/2=PE
    => v = √(2PE/m_e ) =√((2 ×2.3×10^(-13))/(9.11×10^(-31) ))=707 106 781
    => v ≈ 7×10^8 ms^(-1)

    So Escape Velocity ≈7×10^8 ms^(-1)

    This is faster than the speed of light so I must have done something wrong as this would imply that the electron cannot escape unless its velocity is greater than the speed of light, therefore the electron would be bound???
  2. jcsd
  3. May 19, 2009 #2
    Hi there,

    From a first glance at your calculus, you do not take special relativity into it. Therefore, into classical physics, many problems can lead to solution where the speed of the object is greater than the speed of light.

  4. May 19, 2009 #3
    Hi, many thanks for your comment, however I don't think the question intends special relativity to be used - it's in part of a course on QM. Not sure if that's the right attitude to take?! However, can you see any QM related things I have missed or done wrong?
  5. May 19, 2009 #4
    Hi there,

    Not being an expert in QM, and having very bad memories from my classes in University, I don't see anything wrong with your equation, from QM point of view.

    Eventhough, since you are talking about speed that are fractions of the speed of light, you anyways need to uses special relativity.

    The problem that I see with you equation is that you suppose the mass of the electron constant, which is not true for speed that great.
    is not right from that point of view. It should be something like:
    [tex]\Delta p = \Delta (mv) \approx \frac{h}{\Delta x}[/tex]
    Since the mass cannot be considered constant, this becomes:
    [tex]\Delta v \approx \frac{h}{\Delta m \cdot \Delta x} [/tex]

  6. May 21, 2009 #5
    Many thanks for your suggestions - I am not sure how your suggestion would lead to a different answer - in other words I'm still confused!
  7. May 25, 2009 #6
    Hi there,

    I am just pointing out that electrons relevant speed in atoms must be viewed as relativistic particles. Therefore, the electron's mass is not it's rest mass anymore.

  8. May 26, 2009 #7


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    Homework Helper

    In relativity, momentum is defined as
    [tex]p = \frac{mv}{\sqrt{1 - v^2/c^2}}[/tex]
    and kinetic energy is defined as
    [tex]K = \left(\frac{1}{\sqrt{1 - v^2/c^2}} - 1\right)mc^2[/tex]
    where in each case m is what used to be called the "rest mass" (which is constant) not the "relativistic mass" (which changes with velocity).

    As for explaining how to get the velocity: you could say that an electron stuck in the nucleus, over time, has an average velocity of 0 (otherwise it wouldn't be stuck). But if the uncertainty in that velocity is ∆v, it means the electron would actually have a velocity in some range whose characteristic width is ∆v - in some sense, the "average" magnitude of velocity is ∆v (for some definition of average).
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