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Decay of U238

  1. Jan 8, 2008 #1
    Hi i got a short question if i know the following

    U238 decays spontaniously to 2 Neutrons and Tellurium136 and some other isotope.

    How can i tell what other isotop is produced.
    Sure i know about baryon number conservation lepton number conservation can i just
    say ok i got 238 baryons on the right so my new isotop has to have 100 baryons left and because of lepton number conservation i still need to have the same number of electrons and hence protons ( cause of charge conservation) so that i get Zirconium( 40 Protons and the isotop with mass 100 ) ?
    is that the correct logic ?
     
  2. jcsd
  3. Jan 8, 2008 #2

    Astronuc

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    Staff: Mentor

    Logic is correct.

    Both Z (atomic number) and A (atomic mass) must be conserved.

    Start with U-238 (Z=92, A=238) which spontaneously fissions (rather than decay) to

    2 n (Z=0, A=1), i.e. no charge

    and Te (Z=52, A=136) and X

    Z(X) = 92 - 2 (0) - 52 = 40
    A(X) = 238 - 2 (1) - 136 = 100

    So X must be (Z=40, A=100), and Z = 40 => Zr and Zr100, since A=100.

    www.webelements.com is a good site for quick references on elements.

    http://www.nndc.bnl.gov/chart/ is a good reference on the chart of nuclides. Use zoom feature top right to see details of radionuclides.
     
  4. Jan 8, 2008 #3
    cool thanks :)
     
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