# U-238 Decay chain program

1. Jul 17, 2010

### jae1227

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:
$$N_n(t)= \frac{N_1(t)}{\lambda_n } \sum_{i=0}^n \lambda_i \alpha_i \exp({-\lambda_i t})$$
$$\alpha_i=\prod_{\substack{j=1 \\ j\neq i}}^n \frac{\lambda_j}{\lambda_j-\lambda_i}$$
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. Jul 17, 2010

### mathman

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.

3. Jul 29, 2010