Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Forces are actually involved in radioactivity

  1. Dec 22, 2005 #1
    What forces are actually involved in radioactivity other than the weak nuclear force?
    And would anyone know any nice animations that one could view on the net of radioactive decay processes?
    Thanks
     
  2. jcsd
  3. Dec 22, 2005 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Alpha decay involves strong nuclear force as well as electromagnetic, not weak force.
     
  4. Dec 22, 2005 #3
    As far as I can see, when the strong nuclear force applied by the neutrons in a nucleus can no longer overcome the force of the protons repelling each other, alpha decay occurs.
    So I guess it involves EM and strong nuclear force in this case.

    Beta particles are emitted when there are too many neutrons and a beta particle is emitted when a neutron decays into a proton and an electron (the particle).

    Correct me if I'm wrong >_<
     
  5. Dec 22, 2005 #4
    Is the weak nuclear force only involve the decay of quarks and leptons?
     
    Last edited: Dec 22, 2005
  6. Dec 22, 2005 #5

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    No.. the strong force is many, many times stronger than the coulomb repulsion. The stability or instability of an atom or nucleus has never been influence by the EM forces. If that were the case, there's no reason why adding an electrically neutral neutron to a nucleus would make it unstable.

    Yikes. You need to look up "weak interactions".

    Zz.
     
  7. Dec 22, 2005 #6
    Haha... well, I haven't studied this stuff yet. :P
    Ah, I see. So can you clarify it a little for me?
     
    Last edited: Dec 22, 2005
  8. Dec 22, 2005 #7
    And a positron is emitted when a proton decays to a neutron?
     
  9. Dec 22, 2005 #8
    With a neutrino...
     
  10. Dec 22, 2005 #9

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Er... where do you get this stuff?

    A proton decay, if it actually happens, will not be via a channel that has a neutron has a daughter product. This will severely violates conservation of energy/mass. If the Standard Model prediction is correct (where's Benlillie when I need him?), I believe the dominant channel for a proton decay is to a positron and a pi0 meson.

    Zz.
     
  11. Dec 22, 2005 #10

    Astronuc

    User Avatar

    Staff: Mentor

  12. Dec 22, 2005 #11

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Beta decay is the reaction: neutron -> proton + e- + electron anti-neutrino.

    - Warren
     
  13. Dec 22, 2005 #12

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Where exactly on that page is this described? That page only describe beta decay. This is not proton decay.

    Zz.
     
  14. Dec 22, 2005 #13

    Astronuc

    User Avatar

    Staff: Mentor

    If one uses this link - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c3 - one had to look at the frame (plate) below entitled "Positron and neutrino" which should be http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c4, but I had problems with the link, but now it seems to work.

    But I like Warren's equation [itex] p\,\rightarrow\,n\,+\,e^+\,+\,\nu_e[/itex]

    Here's how a chemistry site writes it - http://dbhs.wvusd.k12.ca.us/webdocs/Radioactivity/Writing-Positron-EC.html
     
    Last edited: Dec 22, 2005
  15. Dec 22, 2005 #14
    Yes, there are two types of "beta decay", b- decay (also called nagatron decay) and b+ decay (also called positron decay). Positron decay involves [P] ---> [N] + [b+] + [neutrino]. This decay of course does not happen to a free proton, only to protons within isotopes. Thus, one isotope decays into another via positron decay--it is not a free proton involved in this type of decay. It is energetically only possible when the mass of the parent isotope is greater than the mass of the daughter isotope by two electron masses. As far as is known, the "free" proton is very, very stable (never decays !--or at least it has never been experimentally observed). Also, positron decay is very well known and used in medical research such as PET scan--see this link:http://www.radiologyinfo.org/content/petomography.htm
    It is a hypothesis of some nucleon cluster models that positron decay would be predicted in those isotopes that have a [PP] halo structure in outer nuclear shell, which is very, very unstable. One can see how positron decay of one of the halo [P] nucleons would result in a very stable [NP] cluster structure (called the deuteron).
     
  16. Dec 22, 2005 #15

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Maybe I'm missing something, but I still don't see a proton decay channel being shown in that page. And chroot's "equation" is for a typical beta decay. Why are you able to use that as your proton decay? If you look closely, there's a major energy conservation violation on the left side when compared to the right side of the "equation". The starting point (proton) has less mass than the daughter particles. I can understand if the parent particle has a greater mass than the sum of the daughter particles, but for the other way around, something else has to come in and participate. This is not the proton decay.

    Zz.
     
  17. Dec 22, 2005 #16
    The proton decay channel only happens in nuclei- not in free space. Energy is conserved due to the proton to neutron ratio being too high in the nuclei and it becomes energetically favorable for the nuclei as a whole. This site says it never happens naturally- that is outside of a man made reactor.

    http://www.antonine-education.co.uk/Physics_A2/Module_5/Topic_4/TOPIC_4.HTM
     
    Last edited: Dec 22, 2005
  18. Dec 22, 2005 #17

    Astronuc

    User Avatar

    Staff: Mentor

    That is true about proton decay in free space. Certainly to make this rigorously correct, I would have to attach some number of additional nucleons to both sides of the equation. I did not mean to imply that a free proton spontaneously decay by positron emission.

    [itex] p\,(in\,_{12}Mg^{23})\,\rightarrow\,n\,(in\,_{11}Na^{23})\,+\,e^+\,+\,\nu_e[/itex]

    Mg 23 - 22.994123669 amu
    Na 23 - 22.98976928087 amu and the Na is lighter despite the fact that Na has more neutrons - the difference obviously being the binding energy.
     
  19. Dec 23, 2005 #18
    This web site is incorrect. It is known that K-40 (a natural isotope) has a small % decay mode for positron decay to Ar-40--see below from wikipedia:http://en.wikipedia.org/wiki/Potassium
    "There are seventeen known isotopes of potassium. Three isotopes occur naturally: K-39 (93.3%), K-40 (0.01%) and K-41 (6.7%). Naturally occurring K-40 decays to stable Ar-40 (11.2%) by electron capture and by positron emission, and decays to stable Ca-40 (88.8%) by beta decay; K-40 has a half-life of 1.250 × 109 years."
     
  20. Dec 26, 2005 #19
    Ok so getting to the point, the only forces involved with radioactivity of an atom are the weak nuclear and the strong nuclear forces?
    How exactly does the strong nuclear force come into it?
     
  21. Dec 26, 2005 #20

    Astronuc

    User Avatar

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Forces are actually involved in radioactivity
  1. Radioactive decay (Replies: 4)

  2. Radioactive elements (Replies: 3)

  3. About radioactivity (Replies: 6)

Loading...