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Understanding Beta Decay

  1. Nov 9, 2009 #1
    I was reading a text, describing the following disintegration:

    [itex]^{234}_{90}Th\xrightarrow{}~^{0}_{-1}e+~^{234}_{91}Pa[/itex]

    However, I prefer to understand it as,

    [tex]^{234}_{90}Th\xrightarrow{}~e^{-}+~^{234}_{91}Pa[/tex]

    Of course, the above two equations are missing out the antineutrino. Could someone please enlighten me why,

    [tex]e^{-}\equiv~^{0}_{-1}e[/tex]

    Am I supposed to read it as, p+n =0 but it contains a 'negative' proton number, i.e. a negative proton = an electron (although an electron has a mass of 1/1840 of that of a proton)

    Cheers,
    Mike.
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2

    mgb_phys

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    In a beta decay a neutron converts into a proton and an electron.
    The electron (the beta) is emitted but the proton remains in the nucleus changing the element to the next along in the table
     
  4. Nov 9, 2009 #3
    Oh no mgb_phy, that bit is quite clear. My question was more regarding this notation:

    [tex]e^{-}\equiv~^{0}_{-1}e[/tex]
     
  5. Nov 9, 2009 #4

    clem

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    The text is using a notation where the left superscript is the mass number, which is zero for an electron since the mass number is a roundoff of the mass in units of the nucleon mass to the nearest integer. The left subscript is the charge. It is not my favorite notation, but it agrees with rough conservation laws.
     
  6. Nov 9, 2009 #5
    Is this a rounding of off,

    [tex]m_e=\dfrac{m_p}{1836} = 9.109382\times 10^{-31}\approx~0[/tex]

    I agree, I too prefer simply using [tex]e^-[/tex]. I find [tex]^{0}_{-1}e [/tex] rather tedious.
     
  7. Nov 9, 2009 #6

    bcrowell

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    Putting the subscript and superscript in front of the e is totally nonstandard notation. I think the author of the textbook just wanted to make it easier to see how the masses and charges add up.
     
  8. Nov 9, 2009 #7
    Now that I think about it, you're quite right.

    [tex]^{234}_{90}Th\xrightarrow{}~^{0}_{-1}e+~^{234}_{91}Pa [/tex]

    From a purely 'summation' perspective one would read the right hand side as 91 - 1 = 90, what we started with. It's just rather 'fiddly' imho, as the electron is simply a result of the neutron converting into a proton in the nucleus. It's just with the above notation, the 'lost' proton is simply accounted for by the subscript '-1'.

    Thanks to everyone for their comments and advice :)
     
  9. Nov 9, 2009 #8

    mgb_phys

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    Sorry, Yes - the author is writing an electron as if it were an isotope. Never seen it written like this before but it makes a certain amount of sense I suppose
     
  10. Nov 9, 2009 #9

    arivero

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    Eh no!!!!!!

    Zero is not the same that 0 eV or 0 Kgr.

    The rounding off is

    [tex]\dfrac{me}{m_p}\approx~\dfrac{1}{1836} \approx~0 [/tex]
     
  11. Nov 9, 2009 #10
    Ah! I completely forgot the units; thanks for pointing that out.
     
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