Question

Complete the following fission reaction.

${}_0{}^{1}n+{}_{92}{}^{235}U\to {}_{50}{}^{131}Sn+?+2{}_0{}^{1}n+Energy$

Short answer

The fission reaction is${}_{92}{}^{235}U+{}_0{}^{1}n\to {}_{50}{}^{131}Sn+{}_{42}{}^{105}Mo+2{}_0{}^{1}n+Energy$.

Step-by-step solution

Introduction

Create a fission reaction that is incomplete.

${}_{92}{}^{235}U+{}_0{}^{1}n\to {}_{50}{}^{131}Sn+?+2{}_0{}^{1}n+Energy$

Step 2 Determine the mass number that is missing.

$235+1=131+?+2$

$236=133+?$

$236=131+?$

$105=?$(number of new nuclei in mass)

Explanation

Determine the atomic number that is missing.

$92+0=50+?+0$

$92-50=?$

$42=?$(new nucleus' atomic number)

Step 4 Determine the new nucleus's symbol - the element with atomic number $42$on the periodic table is molybdenum, $Mo$; the symbol for this isotope of $Mo$is written aslocalid="1652247286343" ${}_{42}{}^{105}Mo$

Step 5 Bring the fission reaction to a conclusion.

${}_{92}{}^{235}U+{}_0{}^{1}n\to {}_{50}{}^{131}Sn+{}_{42}{}^{105}Mo+2{}_0{}^{1}n+Energy$