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Homework Help: Beta decay

  1. Oct 24, 2014 #1
    1. The problem statement, all variables and given/known data
    from the physics fact in the photo, i know that the daughter nucleus is a positive ion....
    but , referring to the part 2 , finding the mass loss in beta decay in terms of atomic mass , it says that A_(z+1) Y has extra one orbitting electron comapred to the parent nucleus.
    this is contrary to the physics fact above.....
    A_(z+1) Y is atom with proton number and number of electron = Z+1 or A_(z+1) Y is a positive nucleus ?

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Oct 24, 2014 #2
    and i dont understand part 2 ) ,
    why the mass defect is mx -(my-me) -me ?

    the extra orbittin elc\ectron mass is already deducted in the last part -me am i right?
    why the mass of elctron is deducted again in ( my-me )?
  4. Oct 24, 2014 #3


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    The atomic masses of [itex] ^A_Z X[/itex] and [itex] ^A_{Z+1} Y[/itex] are things you can look up in a table of atomic masses. For example, let's take carbon twelve, [itex] ^{12}_6 C [/itex]. You can look that up and you'll find that the atomic mass is exactly 6 grams per mole (and divide by Avogadro's number to find the mass of a single atom). But that number assumes that the [itex] ^{12}_6 C [/itex] atoms under consideration all have full shells of electrons (in this case 6). If you want to measure the atomic mass of a [itex] ^{12}_6 C ^+[/itex] ion, with only 5 electrons, you'll need to subtract off the mass of an electron from the value found in the table.

    The change in mass, [itex] \Delta m [/itex] is the atomic mass of the parent atom minus the atomic mass of everything involved after the decay.

    So what is left after the decay? The atomic mass of [itex] ^A_{Z+1} Y[/itex] with the electron subtracted off, and the atomic mass of the beta particle.

    I'm not sure I understand. [itex] m_e [/itex] is the electron mass.

    The byproducts of the beta decay consist of a positive ion and an electron. But the mass of [itex] ^A_{Z+1} Y[/itex], which you can look up in a table, is not the mass of a positive ion. It's the mass of a neutral atom. So to find the mass of the positive ion, you must take the mass of [itex] ^A_{Z+1} Y[/itex] and subtract off the mass of an electron.
  5. Oct 24, 2014 #4
    [itex]A_{Z X}[/itex] means [itex] ^{12}_6 C [/itex] atom here? [itex] ^A_{Z+1} Y[/itex] represent [itex] ^{12}_7 C ^+[/itex] ion am i right? to find the mass of [itex] ^{12}_6 C [/itex] atom, the mass os 1 electron is subtracted ... why you said that the mass is subtracted to find the mass of [itex] ^{12}_5 C ^+[/itex] ion?
  6. Oct 24, 2014 #5


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    I was just using carbon twelve as hypothetical example (and in retrospect, not such a good one).

    [itex] ^A_Z X [/itex] as expressed the problem can be any atom.

    Well, if the parent atom was carbon, then the daughter atom would be a nitrogen ion, [itex] ^{12}_7 N^+ [/itex]. Carbon isn't carbon if it has 7 protons. It's nitrogen. But let's pretend for the moment that the parent atom is some weird boron isotope with 7 neutrons, [itex] ^{12}_5 B [/itex], then the daughter ion after a beta decay would be [itex] ^{12}_6 C^+ [/itex].

    These are admittedly bad examples though since they are not typically atoms/isotopes one considers for beta decay.

    A better example would be carbon fourteen.

    [tex] ^{14}_6 C \ \ \rightarrow \ \ ^{14}_7 N^+ + e^- + \bar \nu [/tex]

    Where the [itex] \bar \nu [/itex] is an anti-neutrino. But we're neglecting the mass of the anti-neutrio for this exercise because it its mass is negligible.

    For example, here is a table of the masses of various atoms:


    It is assumed that the masses given are from atoms that are neutral (not ions). You won't easily find a table of masses of ionized atoms anywhere. So if you want to find the mass of a positive ion -- missing one and only one electron -- you can do that by looking up the mass of a neutral atom and subtracting off the mass of an electron.
  7. Oct 24, 2014 #6


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    Just to follow up for the sake of clarity:

    Continuing the carbon fourteed decay example, if you wanted to find the mass of the positive [itex] ^{14}_7 N^+ [/itex] ion, you could look up the atomic mass of the neutral [itex] ^{14}_7 N [/itex] atom in the table, http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl and then subtract off the mass of an electron.

    But as the book points out, it's not necessary to do this when finding the [itex] \Delta m [/itex] because when considering the mass of the beta particle, the same as the mass of the missing electron, they cancel out.
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