Decay of U238

1. Jan 8, 2008

Mr.Brown

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. Jan 8, 2008

Astronuc

Staff Emeritus
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.

3. Jan 8, 2008

Mr.Brown

cool thanks :)