(a) – (d) Complete the following table, which refers to the generalized fission reaction

${}^{\mathbf{235}}\mathbf{U}\mathbf{+}\mathbf{n}\mathbf{\to }\mathbf{X}\mathbf{+}\mathbf{Y}\mathbf{+}\mathbf{bn}$

Consider the given chemical equation as:

${}^{235}U+n\to X+Y+bn$.

From the atomic and mass numbers given, determine the unknown isotope as well as the value of b.

Note that the sum of superscripts (mass numbers A) must balance, as well as the sum of Z values (where reference to Appendix F or G is helpful). A neutron has Z=0 and A=1. Uranium has Z=92.

Since xenon has Z=54, then “Y” must have Z=92-54=38, which indicates the element strontium. The mass number of “Y” is

235+1-140-1=95

So “Y” is ${}^{95}Sr$.

Iodine has Z=53, so “Y” has Z=92-53=39, corresponding to the element yttrium (the symbol for which, coincidentally, is Y).

Since 235+1-139-2=95, then the unknown isotope is ${}^{95}Y$.

The atomic number of zirconium is Z=40. Thus,92-40-2 , which means that “X” has Z=52 (tellurium). The mass number of “X” is 235+1-100-2=134, so we obtain ${}^{134}Te$.

Examining the mass numbers solve as:

$$\begin{gathered} b=235+1-141-92 \\ =3 \end{gathered}$$

The completed table is shown below: