Re: Why, when uranium is used in nuclear power plants is lead not a by-product?

First off, let me say I'm sorry, I have no idea why this never posted when
I answered it several months ago, but I'll answer again!  

I will assume that you mean when uranium is fissioned in a NPP, why is lead
not a byproduct.  The answer is rather painfully complex if you care about
lots of math but easy to understand conceptually.  When U-235 absorbs a
thermal neutron (thermal in this sense means that it is traveling slowly
for a subatomic particle, 0.025eV or about 2500m/s) it breaks into two
parts plus (on average) 2.43 additional free neutrons, and a few other
significantly less energetic things (we'll ignore the other things for the
sake of this explanation).  The extra neutrons and the two fragments of the
original U-235 atom must conserve momentum as they fly apart.  Because of
this, it is energetically preferable for the masses of the large fragments
to be similar in mass.  This could jovially be described as the "atoms are
lazy principle."  If the masses were significantly different, the lighter
one would have to go much faster to compensate, nature prefers balance so
that isn't really probable.  There is an excellent graph in Basic Nuclear
Engineering, 4th Ed. by Foster and Wright, p. 75.  The graph itself has a
reference: [ANL 5800, p.11], ANL refers to Argonne National Laboratories,
but I could not find the reference quickly at the ANL website
(http://www.anl.gov/).   

The graph is of the probability, or yield, versus the mass number of the
fragment.  We always called it the dromedary curve (two humped camel) in
college.  The two peaks correspond to roughly a mass of 92 and 138, which
you will notice is just a little less than 235.  The yield for mass numbers
73 and 160 are less than 0.0001%.  The mass number for the stable isotopes
of lead are 206, 207 and 208.  So lead is not produced in a fission
reaction.  To get a real feel for this, Knolls Atomic Power Laboratory puts
out a publication called the Chart of the Nuclides
(http://www.kapl.gov/whatwedo/chartofnuclides.html) which contains a
concise synopsis of the nuclear data for every isotope of every element
that is known.  (They've been doing this for the US Navy for a very long
time.)  It also has probabilities for the fission yield for U-235, U-233
and Pu-239.  You can also do a web search for "chart of the nuclides" or
just "nuclides" and get some free information.  

Now to question two, lead and uranium deposits.  There are two answers to
this.  Where there is uranium, there is lead, but there isn't much lead
compared to the quantity of uranium.  In this case, the lead is there from
decaying uranium.  In the case of major deposits, however, the chemistry
and geology that leads to uranium deposits differs significantly from that
which produces a lead ore.  Lead prefers to occur naturally with silver and
zinc while uranium tends to prefer arsenic.  Lead is also more abundant
than uranium by a large margin.  For more on this, I would have a look at
the US Geological Survey website (http://www.usgs.gov/) and search on lead
and uranium.  

I hope this answers your questions.  If you end up with more, please submit
another question to MadSci, and I promise it won't take as long to get back
to you the second time! 

Scott Kniffin
NASA GSFC Code 561.4
Sr. Engineer, Radiaiton Effects and Analysis Group