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U-238 Decay chain program

  1. Jul 17, 2010 #1
    I wrote a program to find the percent of each element in the decay chain for U238 after a certain amount of time. I used the Bateman equations for serial decay chain below:
    [tex]
    N_n(t)= \frac{N_1(t)}{\lambda_n } \sum_{i=0}^n \lambda_i \alpha_i \exp({-\lambda_i t})
    [/tex]
    [tex]
    \alpha_i=\prod_{\substack{j=1 \\ j\neq i}}^n \frac{\lambda_j}{\lambda_j-\lambda_i}
    [/tex]
    I have a working program but I don't know if the numbers are right. This is the output after 4.468e9 year, the half-life of U238:
    U-238 50.0%
    Th-234 7.38402118408e-10%
    Pa-234m 2.4681490824e-14%
    Pa-234 8.55341319395e-12%
    U-234 0.0027474651976%
    Th-230 0.000843614748358%
    Ra-226 1.79287783423e-05%
    Rn-222 1.17157024517e-10%
    Po-218 6.59639173904e-14%
    At-218 5.31967076451e-16%
    Rn-218 1.24125650799e-17%
    Pb-214 5.70268705015e-13%
    Bi-214 4.23445792721e-13%
    Po-214 5.82681270543e-20%
    Tl-210 2.76622879307e-14%
    Pb-210 2.4957038641e-07%
    Bi-210 1.536048551e-10%
    Po-210 4.2400210405e-09%
    Tl-206 8.93491905013e-14%
    Pb-206 49.9963907364%

    I know that the 50% will be U238 but will 49.996% of the atoms be really be Pb206? The half-life of U238 is very long and the next longest in the chain is more than 4 magnitudes smaller, U234 with a half-life of 245500 years. Does any one know if these results look about right.
     
  2. jcsd
  3. Jul 17, 2010 #2

    mathman

    User Avatar
    Science Advisor

    Qualitatively it looks right, since U238 has (as you noted) a half life more than 4 magnitudes smaller. As a result you should expect that almost all the decayed U238 atoms would have already ended up as Pb206 - they don't spend much time in between.
     
  4. Jul 29, 2010 #3
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