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Thread: Proton --> Neutron transformation in electron capture.

  1. #1 Proton --> Neutron transformation in electron capture. 
    Member epidecus's Avatar
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    I was pulled in physics on the topic of "nuclear decay by K-electron capture" in chemistry class. My textbook didn't tell me much, so I did some research. I want to know if I understand the process correctly, as well as answers to a few questions. I'm in an introductory chemistry course diving into particle physics, so simpler terms are appreciated. Alright, basically:

    1. We have an atom where an electron in the K (innermost) energy level "falls" into the nucleus.

    2. The falling electron interacts with one of the protons via the weak force / interaction.

    A proton consists of 3 quarks: up, up, down. A neutron is similar with 3 quarks except it is: up, down, down. In the weak interaction between the proton and the electron, one of two things happens. The interaction...

    3.i) ...causes the electron to turn one of the "up" quarks in the proton into a "down" quark. The resulting composite particle is now "up, down, down" i.e. the proton became a neutron. Also, while the quark conversion (up-->down) was happening, a W+ boson was released, being absorbed by the electron making it an electron neutrino.

    3.ii) ...causes the electron to release a W- boson, turning the electron into an electron neutrino. The W- boson is absorbed by one of the "up" quarks in the proton, making it into a "down" quark. The proton becomes a neutron.

    4) A proton became a neutron ---> The mass number does not change, but the atomic number decreases by one. The product is the original element's predecessor by atomic number.
    My textbook only mentioned the first and last steps. I had to research the intermediates, and now I understand why the authors decided to exclude that information from an introductory chemistry textbook. Just a couple of questions, though...

    A) Is all of this essentially correct? I'm mainly worried about the particle interactions (steps 2 and 3).

    B) Why does #1 happen? What makes an electron in the K-shell fall into the nucleus?
    Last edited by epidecus; 08-26-2013 at 03:47 AM. Reason: Typos.
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  2. #2  
    mvb
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    First of all, you did very well in figuring out from references what was going on. There is very little left to say.

    The two possible transformation orders you found [3i and 3ii], both occur. More precisely, an up-quark in the proton can first emit a W+ boson, becoming a down-quark and turning the proton into a neutron. The W+ then goes on to change the electron into an electron neutrino. But it is also possible for the electron to emit a W- boson, becoming an electron neutrino, and for the W- to go on to change the up-quark into a down quark. Quantum mechanics allows for both processes, and in any given nuclear there is no way to determine which occurred.

    Now the electron must be very near the proton for this to happen, as you have understood. However, the wave function of the electron, that is the probability of the electron being at a given location, extends into the nucleus at all times. So at any time there is a known probability of the electron being close to a proton. The electron doesn't have to fall into the nucleus, it is there all the time.

    Just in case any readers of this response are not aware of quantum probabilities, the electron is not a familiar, classical object that has a single, knowable location. All that quantum mechanics allows us to calculate is a function, the "wave function," whose square gives the probability of finding the electron at each point. If you get a number of identical atoms and find where the location is in each one, you will get different answers for each atom, and there is nothing about the atom that tells you which one will have its electron at which point. All you can determine in advance is the probability that the electron is at a given point.
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    Quote Originally Posted by mvb View Post
    First of all, you did very well in figuring out from references what was going on. There is very little left to say.
    Initially, all I was able to find was post #4 from this. It was the only one that seemed solid as an answer and explanatory in the process. I checked a few of the details elsewhere, though I found little (which is why I asked for confirmation).

    Now the electron must be very near the proton for this to happen, as you have understood. However, the wave function of the electron, that is the probability of the electron being at a given location, extends into the nucleus at all times. So at any time there is a known probability of the electron being close to a proton. The electron doesn't have to fall into the nucleus, it is there all the time.
    If the exact location of an electron at a given time within the cloud cannot be deduced, then we don't know when it will actually come close enough to the nucleus to allow for capture. Does this mean that this type of decay happens (to our understanding) spontaneously and arbitrarily?
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  4. #4  
    mvb
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    Quote Originally Posted by epidecus View Post
    If the exact location of an electron at a given time within the cloud cannot be deduced, then we don't know when it will actually come close enough to the nucleus to allow for capture. Does this mean that this type of decay happens (to our understanding) spontaneously and arbitrarily?
    Yes, exactly. What happens in electron capture is that at any given time there is some probability of the reaction taking place. This probability comes from

    (the probability that the process involving the W takes place given that the electron is in a given spot) X
    (the probability that the electron is in that spot)

    summed over all possible spots. You cannot predict the time the decay will occur; you can only calculate the probability. In addition, the probabilities are generally not time-dependent, so the probability of decay at any time is the same. As a result, the likelihood that the atom has not yet decayed is an exponential function of time. This is why you see mentions of the half-life of the decay of the nucleus. The half-life is the length of time required for half of a (large) set of nuclei to decay.
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    Junior Member tcrosa27's Avatar
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    This is sooooooooooooooooooo awesome!
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  6. #6  
    Member epidecus's Avatar
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    Quote Originally Posted by tcrosa27 View Post
    This is sooooooooooooooooooo awesome!
    Indeed

    Also, @mvb. I left you a couple likes, but it appears I forgot something... Thanks!
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    Junior Member butchtroll's Avatar
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    I had questions about the strong and weak nuclear forces and I did some work on a standard model that did not require them. I did come up with some ideas! Relax, I am not preparing to elaborate on them... The interesting thing is, that when I embarked on this journey it seemed that quantum physics was a farce, now I find myself returning to it from the opposite direction! That is working from the most elementary particle up! I still have doubts about the existence of the nuclear forces (and the neutron for that matter).
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  8. #8  
    mvb
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    Quote Originally Posted by epidecus View Post
    Indeed

    Also, @mvb. I left you a couple likes, but it appears I forgot something... Thanks!
    And thanks to you also. It feels soooo good to have said something useful to someone who really wants to learn.
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  9. #9  
    Member epidecus's Avatar
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    Quote Originally Posted by mvb View Post
    And thanks to you also. It feels soooo good to have said something useful to someone who really wants to learn.
    And I'm glad I can contribute to your happiness Not everyone here is a crank trying to promote pseudoscience, fortunately and unfortunately.
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