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Alpha Decay extra electrons

  1. Apr 13, 2008 #1
    Alpha Decay "extra" electrons

    I am sure this question has been asked before, and I have searched many sources for a good answer, but have yet to be satisfied with what i have read.

    Question 1 deals with this version of the decay formula
    238/92 U --> 234/90 Th + (alpha)

    Its seems incomplete. There are 92 electrons on the left but only 90 on the right since alpha is just a nucleus.

    Why is this reaction never written as:
    238/92 U --> 234/90 Th + (alpha) +2 (1/0 e)
    As I have read that there are 2 atomic electrons stripped that then get attached to another atom soon after, shouldnt they be accounted for in the original reaction and not just missing ???

    When determining energy released, rather than using the alpha particle (He nucleus) mass, we use the full Helium atom mass to account for the two "extra" electrons, but why doesnt the reaction simply have 2 extra electrons as a product to tell the story more accurately

    My second question is basically the same one from a different view point
    Often the reaction is written as:
    238/92 U --> 234/90 Th + 4/2 He
    why isnt it written
    238/92 U --> 234/90 Th + 4/2 He (2+) + 2 (1/0 e)

    Is the revised reaction I wrote not what actually happens? Traditionally the extra 2 electrons are lumped into the helius nucleus to simplify and everything balances, but it has always seemed to be skipping a step in the process?


    Maybe I am missing something entirely, thanks in advance for the clarification if you have it.

    p.s. A similiar question can be asked about beta decay
    normally shown 14/6 C --> 14/7 N + Beta
    why not ?? 14/6 C --> 14/7 N (1+) + Beta
  2. jcsd
  3. Apr 13, 2008 #2


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    One should account for the electrons, which one would do if one uses atomic masses for all the nuclides. So if one uses atomic masses for 238/92 U and 234/90 Th, then one needs to use the atomic mass for He, which includes the mass of the electrons.

    Each electron has a rest mass of 0.511 MeV, so in alpha decay equations, one has to be consistent with mass, either nuclear or atomic, but not mixed.

    In the beta decay, the nucleus looses the beta particle, but since Z increases by one, it gains an electron from another atom, and there is a cascade of ions tranfering electrons until the emitted beta neutralizes when it finally comes to rest, so to speak.
  4. Apr 13, 2008 #3
    Thanks for the insight ... I see that mathematically it can be accounted for in a nuclide way or total atomic way, but which one is more accurate to represent the true reaction we are seeing or looking at.

    Is the true alpha decay reaction we are looking at simply an analysis of the nuclides. Meaning, the reaction describes Uranium nucleus decays to thorium nucleus + helium nucleus and in reality the electrons are not involved in the reaction (but in practice we use the full atom masses because they are convenient and the electron masses essentially cancel out) ???

    I guess the question is what happens with the electrons that are present while this process is happening. Clearly for a Uranium nucleus to decay it must be part of a uranium atom which has electrons in its shells ... as it decays a single uranium atom (or nuclei ?) turns into thorium and in 4.5 billion years half of all the uranium atoms will now be thorium, so in that time while uranium nuclei are becoming thorium nucli (2 less electrons for each throium) atom, where are all of the extra electrons that were once part of the uranium atoms?
  5. Apr 14, 2008 #4


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    Th is left as a 2- ion in the reaction first posted.
  6. Apr 14, 2008 #5


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    Alpha emission is a nuclear process. The alpha (Z=2) leaves the nucleus, and since the atomic number of the original nuclear decreases by 2, so two electrons leave that atom. Somewhere the alpha particle slows down and stops, and it becomes a helium atom by picking up 2 electrons. Charge neutrality is maintained. So in a uranium deposit, one will not only find various decay products, like Th, Pa, Ac, Ra, . . . (in proportions related to their respective half-lives), but one will find He.

    We generally use atomic masses because that's the physical nature of elements. It's impossible to isolate nuclei of most elements, practically only the lightest elements and then only a few particles.

    There are heavy ion accelerators, but they do not strip all electrons from an nucleus.

    This might be useful - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radser.html

    and - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html
    Last edited: Apr 14, 2008
  7. Apr 14, 2008 #6


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    EDIT: my misstake, this can't occur since Th can only bind 90 electrons =)
    But in beta+ decay, where the charge of the nuclei is increased, you'll get a positive ion
  8. Apr 14, 2008 #7
    Thank you Astronuc, your explanation of the alpha particle aquiring the stripped electrons makes it all work for me. I dont know why this is not explained in every text and online source as I have read a lot and this is the first I have heard of that idea

    My final question I suppose is why do you never see the full atomic particle reaction:
    238/92 U --> 234/90 Th + 4/2 He (2+) + 2 (1/0 e)

    Since doesnt this accurately show what truley happens:
    A Uranium Atom Decays resulting in a full thorium atom, plus an alpha (He nucleus) and 2 electrons. Then followup reactions occur where the Th decays further and the alpha particle combines with the two extra e's to become helium. Understandably, its easier to just combine the last two terms in to a full Helium atom, but doesnt that happen after this first reaction happens, and we want to specifically show its an alpha particle at first and not a helium atom. Thanks for all the insight, it is very useful.
  9. Apr 14, 2008 #8
    sorry here may be i did not understand the question here but how can the nucleus give electrons and it is not allowed the existance of an electron in the nucleus
    is that reaction happens between the nucleus and the electron cloud or there mechanism that give that 2 electron away from U.
  10. Apr 14, 2008 #9

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    There's nothing mysterious. The U atom has 92 electrons. The Th atom has 90, so at the end of the day you have a Th 2- ion and a He 2+ ion. Eventually, the charge migrates from the negative ion to the positive ion.
  11. Apr 14, 2008 #10
    This seems to contradict the earlier post where astronauc noted that when the uranium decays, "2 electrons leave" ... and then connect with the alpha particle to form a helium atom. So do two electrons "leave" (where do they go), or does the thorium become a -2 ion which later is neutralized when the He nuclius accepts electrons. In the end the result is the same, the He nucleus snags the two electrons to become a helium atom, but now the question is where does it get them from. Are they free floating, or are they ripped from the newly formed thorium ion, and is this process rapid .. does the alpha particle take from the throrium on its way out, or does it happen later, but nonetheless fairly rapidly ?
  12. Apr 14, 2008 #11

    Vanadium 50

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    The alpha particle is charged when it is emitted. The eventual neutralization happens more slowly. If you live in a cold climate, you know that you can carry quite a charge on yourself in winter before it is neutralized.
  13. Apr 15, 2008 #12
    Anwser to the first question:
    You must know that even though alfa particle has no electrons, it has 2 protons, because alfa particle is constituted from 2 neutrons and 2 elektrons, having mass of 4 and charge of 2+. Therefore in the equation you have on the LHS 92 , on the RHS you have 90 + 2 (from alfa particle). Thus you have equilibrium and no problems at all.

    Anwser to the second question:
    You must understand that in alfa decay only alfa particles are emitted while the electrons are beta particles. Furthermore, electrons have a charge of -1, never +1 as you state it. +1 charge is on positrons which are antiparticles of the electrons.
  14. Apr 15, 2008 #13


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    When the U (Z=92) transforms to Th (Z=90) as a result of alpha decay, the two electrons are lost to the nearby medium. The decay nuclide is like surrounded by similar material, or if at the surface of a solid, it is in contact with air or some gas. Likely though the alpha particle rips away some electrons on its way out of the nucleus, but those electrons are quickly recaptured.

    As charge particles (alpha and beta), and even gamma rays, pass through material, they interact with atomic electrons and excite or ionize atoms (hence the term - "ionizing radiation"). The range of an alpha particle in solid material is very small - on the order of microns, so it does not travel far from the nucleus from which it originated. Because it has a +2 charge, the alpha particle will ultimately attract two electrons. There will be an effective net migration of 2 electrons along the path the alpha particle took while slowing down.
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