Chapter 21, Problem 21.47QA

Interpretation Introduction

To find:

the missing nuclide formed in the given fission reactions.

Answer to Problem 21.47QA

Solution:

a) $U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Zr\mathrm{ }+\mathrm{ }Te\mathrm{ }+\mathrm{ }\mathrm{ }2\mathrm{ }n$

b) $U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Nb\mathrm{ }+\mathrm{ }Sb\mathrm{ }+\mathrm{ }\mathrm{ }4\mathrm{ }n$

c) $U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Rb\mathrm{ }+\mathrm{ }Cs\mathrm{ }+\mathrm{ }\mathrm{ }3\mathrm{ }n$

Explanation of Solution

1) Concept:

In the given reactions, the nuclide captures neutrons, and new nuclides are produced. The sum of the mass numbers (superscripts) of the two nuclei on the left side of the reaction must be equal to the sum of mass numbers of the species on the right side. Similarly, the sum of the atomic numbers (subscripts) of the two nuclei on the left side of the reaction must be equal to the sum of the atomic numbers of the species on the product side. On the basis of the atomic number of the product formed, we can determine the identity of the unknown nuclide.

2) Calculations:

a) $\mathrm{ }\mathrm{ }\mathrm{ }U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Zr\mathrm{ }+\mathrm{ }X\mathrm{ }+\mathrm{ }\mathrm{ }2\mathrm{ }n$

Sum of mass numbers: 

$235+1=96+A+2 \times \left(1\right)$

$236=96+A+2$

$236=98+A$

$A=236-98=138$

Sum of atomic numbers:

$92+0=40+Z+2 \times \left(0\right)$

$92=40+Z+0$

$Z=92-40 =52$

The atomic number 52 corresponds to the element Tellurium, $\mathrm{T}\mathrm{e}.$

Thus, the nuclide formed is $Te.$

$U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Zr\mathrm{ }+\mathrm{ }Te\mathrm{ }+\mathrm{ }\mathrm{ }2\mathrm{ }n$

b) $U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Nb\mathrm{ }+\mathrm{ }X\mathrm{ }+\mathrm{ }\mathrm{ }4\mathrm{ }n$

Sum of mass numbers: 

$235+1=99+A+4 \times \left(1\right)$

$236=99+A+4$

$236=103+A$

$A=236-103=133$

Sum of atomic numbers:

$92+0=41+Z+4 \times \left(0\right)$

$92=41+Z+0$

$Z=92-41 =51$

The atomic number 51 corresponds to the element Antimony, $\mathrm{S}\mathrm{b}.$

Thus, the nuclide formed is $Sb$.

$U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Nb\mathrm{ }+\mathrm{ }Sb\mathrm{ }+\mathrm{ }\mathrm{ }4\mathrm{ }n$

c) $U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Rb\mathrm{ }+\mathrm{ }X\mathrm{ }+\mathrm{ }\mathrm{ }3\mathrm{ }n$

Sum of mass numbers: 

$235+1=90+A+3 \times \left(1\right)$

$236=90+A+3$

$236=93+A$

$A=236-93=143$

Sum of atomic numbers:

$92+0=37+Z+3 \times \left(0\right)$

$92=37+Z+0$

$Z=92-37 =55$

The atomic number 55 corresponds to the element Cesium,$\mathrm{C}\mathrm{s}$.

Thus, the nuclide formed is $Cs$.

$U\mathrm{ }+\mathrm{ }n\mathrm{ }\mathrm{ }\mathrm{ }\to \mathrm{ }\mathrm{ }Rb\mathrm{ }+\mathrm{ }Cs\mathrm{ }+\mathrm{ }\mathrm{ }3\mathrm{ }n$

Conclusion:

The sum of the mass numbers of all the species on both sides of the reaction must be equal. The same applies for the atomic numbers.