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K - shell capture

  1. May 28, 2009 #1
    To my understanding this is an electron in the K-Shell being absorbed by the the nucleus and along with a proton converted into a neutron with the emission of a neutrino. My question is what force is acting here? Nuclear? If so, how can nuclear forces act on an electron, when the range of nuclear forces barely extends beyond the nucleus?
     
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  3. May 28, 2009 #2

    alxm

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    Sometimes electrons enter the nucleus. A K-shell is an s-orbital, i.e. exp(-r) so the nucleus is actually the most probable point in space for a K-shell electron to be located at.
     
  4. May 28, 2009 #3
    I thought the size on the nucleus was so small compared to an orbital that there is a node between the nucleus and the first shell electron?
     
  5. May 28, 2009 #4

    clem

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    There is no node for s shell orbitals. The weak force is responsible for the e+p-->n+neutrino interaction. The rate is proportional to |\ps|^2 at the origin, which is nonzero.
     
  6. May 28, 2009 #5
    So can you please differentiate b/w weak and strong nuclear forces for me? I can't quite get them straight.
     
  7. May 28, 2009 #6
    It is not correct to conclude: "so the nucleus is actually the most probable point in space for a K-shell electron to be located at." The space is not 1D but 3D, so the radial probability density is proportional to r2exp(-r)dr. The average r is not zero but the Bohr radius r0.

    For atoms with more than one electron, the lowest orbit is smaller than the Bohr radius r0 (the nucleus charge Z>1 is involved in ψ0(r)).

    Bob_for_short.
     
    Last edited: May 28, 2009
  8. May 28, 2009 #7

    clem

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    The strong nuclear force holds the neutrons and proton together in the nucleus. It has a range of about 2 fm, and does not interact at all with the electron. The weak force is responsible for beta decay and K-capture. It is an effective contact interaction. At atomic and nuclear energies, it converts particles into other particles, and does not act like a force that holds things together.
     
  9. May 28, 2009 #8
    The force that binds protons and neutrons together is actually a remnant of the strong force that binds de quarks inside the neutron and the proton.

    Particles can be classified between leptons and hadrons.

    Leptons don't feel the strong force, and hadrons do.

    Electrons are a type of leptons, so they don't feel the strong force.

    The strong force is mediated by gluons (eight masless gluons), while the weak force is mediated by the [tex]W^+, W^-[/tex] and [tex] Z^0 [/tex] bosons (massive).
     
  10. Jun 2, 2009 #9

    alxm

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    It's entirely correct. The most probable 3d point in space is [tex]\overrightarrow{r}=0[/tex] (vector!). The radial probability density, on the other hand, is one-dimensional. It does not express the probability of a point in space but the sum of the probabilities on the surface of a sphere with radius [tex]r[/tex]. (scalar!)

    The radial probability distribution doesn't represent any single point in space, but whole areas of space. You seem to be the dimensionally-confused one here. A surface is not a point.
     
    Last edited: Jun 2, 2009
  11. Jun 2, 2009 #10
    If an average is equal to zero, it does not mean that the zero is most probable point.

    Any point in 3D is characterized with two angles and the distance from the origin. So any probability dw to find a particle in an elementary volume dV is the product of the wave function squared |ψ(r)|2 and dV=r2dr⋅d(cos(θ))⋅dφ. Due to the factor r2 this probability tends to zero at r=0. So the most probable distance is not zero but ~a0.

    Bob.
     
  12. Jun 2, 2009 #11

    alxm

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    I never said the most probable distance was zero. The most probable distance is indeed the Bohr radius, and the most probable point is r=0. These quantities have different dimensions; the radial wave function and its square has the dimension of a 3d point - an infinitesimal volume element. The radial probability distribution, which you're referring to, is one-dimensional quantity - an infinitesimal line segment.

    What's so difficult to grasp about this? If I paint a set of spheres with an amount of paint that's exp(-r) per area unit (and somehow infinitely thin regardless of amount), then the point that has the greatest amount of paint isn't the same thing as the sphere that has the maximum amount of paint on it.
     
  13. Jun 2, 2009 #12
    You wrote that the nucleus was the most probable point for the electron. The probability to find the atomic electron within the nucleus of finite size Rn is proportional to the volume ratio of the nucleus ~Rn3 to the atomic volume ~a03. It is very very small: (Rn/a0)3<<1. Your statement based on the wave function r-dependence was misleading.

    Bob.
     
  14. Jun 2, 2009 #13

    alxm

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    Yes, and r=0 is the most probable point. It's a correct statement, and I worded it the way I did well aware of the fact that it's not the most probable radius.

    I don't view this as misleading. Rather the contrary - people tend to make the mistake of assuming that the radially-summed probability distribution, by which I mean the radial probability summed over the surface of a sphere of radius r, is the wave function (or square of it). And hence, that the electron has a zero probability of being at r=0. That's incorrect, since it comes about simply from the fact that a sphere with radius zero has zero surface area, not because the probability is zero there (unlike all other orbitals which do have a node there). If you were to take an arbitrarily small 'box' of space and place it so as to maximize the electronic density within, it would be placed at the nucleus, not out at a0.

    The fact that the nucleus is very small isn't really here nor there IMO, it helps explain why K-shell capture doesn't happen that much. (Relatively speaking.) My point was simply that an s-orbital electron has no problems at all getting near the nucleus, something which isn't true of electrons in any other orbital.
     
  15. Jun 2, 2009 #14

    jtbell

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    To see this point more clearly, imagine a uniform probability distribution, [itex]P(\vec r) = \rho[/itex] (a constant), throughout a sphere of radius [itex]a[/itex]. In this case, the particle is equally likely to be found at any point inside the sphere. Nevertheless the probability to find the particle at radius [itex]r[/itex] from the center is zero for [itex]r = 0[/itex] and maximum at [itex]r = a[/itex] (the surface of the sphere). Crudely speaking, as [itex]r[/itex] increases, there are more points with that radius, hence a larger probability for that radius, even though the probability at each point is the same. (This argument becomes more rigorous if you think in terms of thin spherical shells with different radii [itex]r[/itex] but the same thickness [itex]dr[/itex].)
     
    Last edited: Jun 3, 2009
  16. Jun 3, 2009 #15
    Recall that in Maxwell's distribution the most probable velocity (vector) is v=0, but the most probable speed (scalar) v~(T/m)1/2
     
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