re: Are you sure?
If all the "daughter" isotopes decay to nothing in the "burning" process, then why is it possible to construct a "breeder reactor," and why are the "spent" fuel rods more radioactive than the same fuel rods when they were inserted in the reactor?
If the reactor design is much more economical of neutrons, enough U-238 can be converted to plutonium so that after a fuel cycle there is more fissionable material than there was in the original fuel rods in the reactor. Such a design is called a breeder reactor. Breeder reactors essentially use U-238 as fuel, and there is 140 times as much of it as there is U-235.
A fission gives off roughly 200 MeV. The fission fragments carry off 82% of the energy in the form of kinetic energy. Prompt neutrons carry off another 2.5%, prompt gammas carry off 3.5%, beta decay accounts for 4%, delayed gammas (from radioactive decay) account for 3%, and neutrinos carry off the remaining 5%. The neutrinos and their energy are lost, since the probability of interaction with neutrinos is so small.
From; http://www-formal.stanford.edu/jmc/progress/nuclear-supplement.html
Nuclear energy is released by one of three nuclear reactions:
Fusion, the fusing together of atomic particles.
Fission, the breaking of the binding forces of an atom's nucleus.
Decay, is a term used for the slower natural fission process of a nucleus breaking down into a more stable form.
From; http://en.wikipedia.org/wiki/Nuclear_energy