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How to convert C-12 / C-13 to C-1

  1. Feb 5, 2007 #1
    How to convert C-12 / C-13 to C-1, does any one know how this can be done, how much energy will I need to bang out a neutron?
     
  2. jcsd
  3. Feb 5, 2007 #2

    Astronuc

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    In
    what is C-1?

    If the nuclide C-12 absorbs a neutron, it becomes C-13. C-13 is stable.

    Knocking out a neutron of C-12 would produce C-11, which transforms by electron capture (EC) to B-11. One could use a (p,n) reaction.

    Determine the binding energy of the last neutron.
     
  4. Feb 6, 2007 #3

    Gokul43201

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    There's no such thing as C-1. C has 6 protons.
     
  5. Feb 6, 2007 #4
    Carbon-12 contains 6 protons, 6 neutrons and 6 electrons.

    If I must take a shot, C-6 will be the highest ?
     
    Last edited: Feb 6, 2007
  6. Feb 6, 2007 #5
    Firstly, you won't get the equipment needed to do such nuclear reactions. It's expensive, complicated, dangerous and can be used to make nuclear weapons materials.

    Secondly, you can't make Carbon-6, the lack of neutrons means it would be too unstable to exist. I doubt you can get Carbon-10, any decay process would favour Boron instead.

    You still don't seem to have realised that if such processes were possible and simple, someone else would have realised it by now. You're showing that you haven't studied nuclear physics at all and are just stumbling about in the dark unfortunately.
     
  7. Feb 6, 2007 #6
    What happens to H-6 then it’s on decay mode?

    He-6 becomes H-6, but H-6 can becomes nothing?
     
  8. Feb 6, 2007 #7

    Astronuc

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    C12 (6p, 6n) and C13 (6p, 7n) are the stable isotopes of carbon. The other isotopes are unstable or radioactive.

    What is it that one is trying to do by changing the number of neutrons in the carbon nucleus?
     
  9. Feb 6, 2007 #8
    Ok the big idea is to make Carbon decay to Hydrogen, this is part of one of my projects, and I doubt that it will take as much as 8MeV gammas to do the job? Because it’s all light materials
     
  10. Feb 6, 2007 #9

    Gokul43201

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    You're doing a project that involves nuclear fission? And you're doing it without learning that a nucleus is?

    I fear this is falling on deaf ears, but your approach to science is completely backwards. You start at the basics. You don't stumble upon them in the process of building a nuclear bomb!
     
  11. Feb 6, 2007 #10

    Astronuc

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    Carbon won't decay to hydrogen. Rather, one could knock out a proton, by a sufficiently energetic (n,p) reaction, but that leaves a boron nucleus. One could knock out another proton and obtain Be, and so on throught Li, He and finally H. However, when all is said and done, that's a lot of energy expended for little gain.

    Back to the (n,p) reaction, that is how one could produced hydrogen, but then the nuclear energy input greatly exceeds any chemical energy.

    Well He-6 apparently decays by beta emission to Li-6. H-6, H-5 and H-4 decay by neutron emission, at least according the Chart of Nuclides on BNL's website. I would be interested in how someone made H-4, H-5 or H-6, as these would tend to be highly unstable with very short half-lives.
     
  12. Feb 6, 2007 #11
    Yip, dam, I understand this now :frown:

    So He-6 is in energy absorbing state again, and then it goes to Li-6, wierd?
     
    Last edited: Feb 6, 2007
  13. Feb 6, 2007 #12
    Why is it weird? What's weird is expecting a light nucleus to exist with twice as many neutrons as protons. Though the heavy nuclei do it, it's highly unstable for light ones.
     
  14. Feb 6, 2007 #13
    I’ve just found out about decay, that O-15 decays to N-15, and now it’s in the opposite way like He-6 go to Li-6
    And I’m in my early 19, a young man, with a lot of mental vomit like russ said
     
  15. Feb 6, 2007 #14
    Jacquesl how about reviewing the basics.

    Do you understand how to calculate the Q value for a decay or reaction?

    Do you know that you need a positive Q value for a decay to be possible?

    For instance calculate the Q value for the hypothetical reaction you are interested in.

    12C ---> 6H+6n

    It will turn out to be a quite large negative value. Thats the minimum energy you would have to put in for it to happen(ignoring whatever process you want to use).
     
  16. Feb 6, 2007 #15
    Nope

    Nope
     
    Last edited: Feb 6, 2007
  17. Feb 6, 2007 #16
    The Q value is the energy difference betwen what you begin with and what you end with. Its very simple.

    In the example above you would calculate it as

    Q=[M(12C)-6M(H)-6M(n)]*C^2

    Where M denotes atomic mass.

    For a decay to be possible the Q value has to be positive, if its not positive that particular decay is not energeticaly favorable and you need to supply energy to it. Decays happen because the nuclei wants to progress to the configuration with the lowest energy.
     
  18. Feb 6, 2007 #17
    Lol, I not so sure man

    What does Q symbol stand for, I know it’s energy now, but, Joule, eV, Coulomb or Watts ?
    In my electronic book, my have a formula of Q = C x V
    that C is for Capacitance and the V like you know, Volt
    In the book it describes as says something like, Q = Charge


    Q=[M(12C)-6M(H)-6M(n)]*C^2

    Q = Charge
    M = Mass
    (n) = neutrons
    12C = C-12, carbon / (12XC), what must C value be
    And that H what value must you put in there?
     
    Last edited: Feb 6, 2007
  19. Feb 6, 2007 #18
    My idea is to take the CO2 and make the C turn into H via neutron capture and then you just use that O2 + H = Energy
    If you turn C to H there will be some energy release I’m not yet 100% sure where, but somewhere there will be a gain, because it took energy to make H-1 turn to C

    So I want to take C-12 / C-13 and convert the isotopes to C-1, but now that wont be the case, I cant make it C-1, I can only make it C-8, so that sucks
     
  20. Feb 6, 2007 #19

    After reading this whole thing... I still don't get whata C-1 is? I think I know what he means when he puts C-12 or C-13, but when I see C-1 I am very very confused. I think you need to pick up an introductory physics book and read the intro of nuclear physics. Then buy a nuclear physics book and work through it. I suggest the book by Krane as a nice starting point.

    Like everyone else has said. Lets start from the basics. Cause you are not speaking science when you write something like C-1. And that makes it really hard on the rest of us.
     
  21. Feb 6, 2007 #20
    I dont mean this in a offensive way. But you need to read up one some basic physics.

    The Q value has nothing at all to do with electromagnetism. It is the net energy release from the decay. It is just E=mc^2 applied to the decay situation.

    C at the end is the speed of light.
    By M(12C) I mean the atomic mass of carbon 12. By 6M(H) I mean 6 times the atomic mass of hydrogen.

    The energy can be of any unit depending on what units you put into the formula(this hold for every formula anywhere in physics).
    Btw Coloumb and Watts are not even units of energy.

    This above is mostly just jibberish. There is no logic behind it. You are mixing chemistry with nuclear physics in a way that makes no sense at all.

    There is no such thing as C-1. You need to understand the notation.
     
    Last edited: Feb 6, 2007
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