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Why do heavier atoms experience relavistic effects(contraction)?

  1. Oct 28, 2008 #1
    It is said that the electrons in heavier atoms have speed close to that of light and therefore undergo relativistic effects?

    May I ask why is this so? What is it in heavier elements that caused such a change to the speed of the electron?

    Is it related to the energy-mass equivalence formula?
     
  2. jcsd
  3. Oct 28, 2008 #2

    Jonathan Scott

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    Heavier atoms have more positive charge in the nucleus and hence a deeper electrostatic potential well, so the deepest electrons have much higher kinetic energy and hence higher momentum and higher velocity.
     
  4. Oct 28, 2008 #3
    sorry, what's electrostatic potential well?
     
  5. Oct 28, 2008 #4

    Jonathan Scott

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    The word "well" in this context simply means a region where something becomes deeper. That is, if you plotted a picture of the electrostatic potential in a plane using the height to represent the potential energy of a negative electron at various locations, it would dip down more steeply around a heavy nucleus because of the increased positive charge.
     
  6. Oct 28, 2008 #5
    That sounds a bit like gravity. The heaver a star is, the faster planets in a given orbit move.
     
  7. Oct 28, 2008 #6

    Dale

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    Do bound electrons have a high velocity relative to their nucleus? I thought they were essentially stationary which is why they don't radiate. Their orbital is a smeared out cloud that doesn't move, so I don't know how you could say it has a velocity, but my QM knowledge is very sketchy.
     
  8. Oct 28, 2008 #7

    atyy

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    I think it's an ok "quick and dirty" heuristic? The first link uses it, the second points to a more quantum mechanical formulation.
    http://www.fourmilab.ch/documents/golden_glow/
    http://www.math.ucr.edu/home/baez/physics/Relativity/SR/gold_color.html

    This is the most common explanation for the colour of gold. As far as I can tell, the calculation involves adding a small relativistic "correction", which looks reasonable to me, but I don't know how it is actually justified from a fully relativistic framework.
     
  9. Oct 28, 2008 #8

    Jonathan Scott

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    The simplified Bohr model of an orbiting charge with a orbit that has quantized angular moment gives a fairly good idea of kinetic energy, momentum and velocity values for the lowest energy orbits around different central charges. See the Wikipedia entry on the "Bohr model" for more details.
     
  10. Oct 29, 2008 #9
    Hmmm, I thought nuclear charge is constant throughout all the the entire atom? from n= 1 to n=infinity.

    Isn't the screening effect and the penetrating power of higher energy atomic orbitals the factors that reason why the potential well is created?
     
  11. Oct 29, 2008 #10

    Jonathan Scott

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    The electric charge on the nucleus as a whole is determined by the number of protons and is equal to the atomic number Z multiplied by the elementary charge unit.

    The potential in the absence of electrons would be proportional to Z/r. For purposes of the electrons closest to the nucleus, the other electrons are sufficiently spread out that this can be used to determine the approximate energy in the same way as for a hydrogen atom.
     
  12. Oct 29, 2008 #11

    I'm sorry Scott, I really don't understand. Can you put in a layman term for me?
    and a side note, is it relavistic or is it relativistic?
     
  13. Oct 29, 2008 #12

    Jonathan Scott

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    That was definitely meant to be in layman's terms!

    The word is "relativistic". When relating to a speed, it means that the speed is a sufficiently large fraction of the speed of light c that accurate calculations need to take relativity into account.

    Basically, the Bohr simplified model of quantum theory for a hydrogen-like atom says that electrons effectively "whiz round" the nucleus with a fixed amount of angular momentum. For heavier atoms where the electric charge of the atomic nucleus is higher, the electron is attracted more strongly, so it has to whiz round faster to "stay up". Whizzing round faster increases momentum, so the orbit has to be smaller to have the same angular momentum, and the smaller orbit causes a further increase in the force on the electron. The energy of the electron therefore increases proportionally to the square of the atomic number. The kinetic energy of the electron determines the speed via the conventional (mv^2)/2 formula for non-relativistic speeds, but when this speed starts to be significant compared with the speed of light it must be calculated instead using a relativistic formula.

    (Of course, for a real atom with many electrons there are other corrections required as well in order to obtain exact energy levels).

    I suggest you look at the "Bohr Model" page in Wikipedia, which explains how the energy of an electron is calculated for a hydrogen-like model of an atom.
     
  14. Oct 29, 2008 #13
    alright, thanks but a few things to clarify.
    Whizzing round faster means the orbit has to be smaller to have same angular momentum? How do you determine that?
    and what is the relavistic formula for speeds close to speed of light?

    actually I'm trying to relate speed to energy but I can't seem to find the formula(for relativistic speeds.)
     
  15. Oct 29, 2008 #14

    Jonathan Scott

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    For a (simplified) circular orbit, the angular momentum is mvr, so if v is faster, r must be smaller to have the same angular momentum.

    For relativistic speeds, the mass term in the momentum should be replaced with the total energy divided by c^2 (sometimes confusingly called the "relativistic mass"), which increases with speed, so for a relativistic speed the radius decreases even further.

    Total energy = [tex]\frac{mc^2}{\sqrt{1-v^2/c^2}}[/tex]

    Momentum = [tex]\frac{mv}{\sqrt{1-v^2/c^2}}[/tex]

    In relativity, the kinetic energy is the difference between total energy and the rest energy:

    Kinetic energy = [tex]mc^2 \left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)[/tex]

    When v is much smaller than c, that expression is approximately [itex]\frac{1}{2} mv^2[/itex] as in Newtonian mechanics.

    You can determine the speed from the kinetic energy via that messy expression, if you really need to.

    However, this whole calculation only provides a rough basic idea as to why electrons move at relativistic speeds for heavier atoms, so there is little point in trying to use it to calculate accurate results.
     
  16. Oct 30, 2008 #15
    hmm, the angular momentum must always be maintained?
     
  17. Oct 30, 2008 #16

    Jonathan Scott

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    That's quantum theory for you; it says that angular momentum is quantized in multiples of [itex]\hbar[/itex].

    Bohr didn't know at the time that this overall angular momentum is actually made up of the electron's intrinsic spin of [itex]\hbar/2[/itex] combined with orbital angular momentum, which among other things also allows a zero-spin ground state. However, if you want to know more about that, you need to look at QM text books and ask any questions in the QM forum.
     
  18. Oct 30, 2008 #17

    atyy

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    Here's a cheating way to think about it. The sun is pulling us towards it. The reason we don't fall into the sun is we have a tangential velocity, so we fall round the sun, rather than into the sun. The closer you are to the sun, the greater the pull of gravity, and the greater your tangential velocity has to be to orbit the sun instead of falling into it. Similarly electrons that are closer to the nucleus go around faster.
     
  19. Nov 1, 2008 #18
    cool analogy... now i'm stilling trying to understand how angular momentum must always be specific integers of h/2pi
     
    Last edited: Nov 1, 2008
  20. Nov 1, 2008 #19

    atyy

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    Yes, you are making sense. The usual "justification" for integers is to use the de Broglie hypothesis that particles are waves with "wavelength=(Planck constant)/(mass*velocity)", and require that the circumference of the electron orbit (2*pi*radius) be an integer number of wavelengths otherwise the wave will not close up on itself. This is the famous Bohr model which had experimental successes, and was one of the things that made people think particles are waves.

    However, to deal with waves properly one needs a wave equation. The correct equation was discovered by Schroedinger, and together with ideas of Heisenberg and Dirac, forms the "real" start of non-relativistic quantum mechanics. Relativistic quantum mechanics uses the Dirac and Klein-Gordon equations, which in turn are believed to be only approximations to the equations of quantum field theory. So the Bohr model is a useful heuristic for seeing why we might need relativistic quantum mechanics, but it is not sufficient to do the actual calculations.
     
  21. Nov 2, 2008 #20
    Actually, I'm trying to explain quantatively why heavier atoms like gold have to be explained using relativistic contraction.

    I fabricated my explanation but there's this contradicting part which I need help on.

    From the probability density function of the atomic orbitals, it can be seen that electrons in higher n quantum number orbitals have certain probability to be close to the nucleus, which is what I learnt as the penetrating power of higher n quantum number orbitals and is it because of this penetration that the electrons in higher n quantum number are drawn close to the nucleus and that's why experience its strong nuclear attraction? and thus increases its Kinetic energy to compensate to the strong coulombic forces. With increase in K.E, it translates to an increase in mass since E=mc2. Following with an increase in mass of the electron, radius of the orbital the electron is in has to decrease as L=mvr=n(h/2pi) as L(angular momentum) has to be conserved and to be always an integer of (h/2pi).
    Thus, that orbital contracts and brought further in proximity to the nucleus which results in stablizing of the orbital and thus, creating an inertness of that electron(s) in that orbital to partake in any bonding due to a great energy barrier.

    Contradicting part is, I seemed to mention the orbital is drawn close to the nucleus TWICE.(bolded)
     
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