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Question about Nuclear fission in general, here. If I have the fission of U-235 and I know that one of the products is Cs-137, is there a way of figuring out the other product? Should there be multiple possibilities? I understand that total number of Nucleons is preserved, but I'm supposing that I need to take into account that neutrons alone are also released? I'm almost certain that if I can work out the other product then I won't be able to know the isotope that it is, but can I even work out the atomic number of the other element? If so, how?

Many thanks

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    $\begingroup$ Note that the direct fission yield of Cs-137 from thermal fission of U-235 is relatively low. The cumulative yield of Cs-137 is mainly due to the fission products Xe-137, I-137, and Te-137, which decay to Cs-137. $\endgroup$
    – user59991
    Commented Aug 24, 2015 at 14:04
  • $\begingroup$ Do you have a source for that information which I could also use? Or is that just your own knowledge? I've been looking for a place which would just display common fission products for various isotopes but have come up dry so far. $\endgroup$
    – Matt
    Commented Aug 24, 2015 at 14:21
  • $\begingroup$ Fission product yields are available from various databases, for example here. You may want to compare the cumulative fission yield of Cs-137 (6.2 %) with the independent fission yield of Cs-137 (0.06 %). $\endgroup$
    – user59991
    Commented Aug 24, 2015 at 14:29

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Fission reaction produce a distribution of several different nuclei. Wikipedia (User:JWB) has a nice graph that shows the relative probabilities.

However, in the case of a single reaction, where it is given that a Cs-137 nucleus is produced, you can probably be more specific because there are correlations between the fission products, but it will require delving into the nuclear data - https://www-nds.iaea.org/

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  • $\begingroup$ I would say that this answer is better than mine, so it's better to rely upon it $\endgroup$
    – Chaosit
    Commented Aug 24, 2015 at 8:58
  • $\begingroup$ Good answer. And with a useful link too! Thanks a lot. $\endgroup$
    – Matt
    Commented Aug 24, 2015 at 9:17
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As indicated in the answer to this physics.stackexchange question the total amount of nucleons is preserved during fission.

As a result the atomic number of a daughter product can be predicted if another was already known. In the case of $U_{92}^{235}$ fissioning to $Cs_{55}^{137}$, we know the atomic number of the second fission product is given by:

$92=55+Z$, i.e. $Z=37$, which is the atomic nummber of Rb (rubidium).

Now as to the atomic weight of the Rb isotope, also use preservation of neutrons.

A $U_{92}^{235}$ nucleus contains $235-92=143$ neutrons and a $Cs_{55}^{137}$ nucleus contains $137-55=82$ neutrons. The difference is $143-82=61$ neutrons to be allocated.

Assume that $2$ to $4$ free neutrons also resulted from the fission then that leaves $57$ to $59$ neutrons for the Rb isotope so its atomic weight is probably $94$ to $98$.

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Actually fission processes are stochastic, so we can't predict exact products of it. The most probable fission products are Cs-137 and Sr-90. The total amount of neutrons emitted is also quite unpredictable, but as I recall it usually lies in the range between 2 and 4.

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