Question

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

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

Short answer

Consider the given chemical equation as:

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

Step-by-step solution

Given

Consider the given chemical equation as:

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

Determine the concept

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

(a) Figure out the unknown isotope

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 ${}^{95}Sr$.

(b) Determine out the unknown isotope

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 ${}^{95}Y$.

(c) Determine out the unknown isotope

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 ${}^{134}Te$.

(d) Determine the value of b

Examining the mass numbers solve as:

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

The completed table is shown below: