Question

When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation of two new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U+10n→AZBa+9436Kr+310n Enter the isotope symbol for the barium (Ba) nucleus in this reaction.

Answer 1

Answer:

$^{139}_{56}Ba$

Explanation:

The fission reaction taking place given in the question is:

$^{235}_{92}U+^{1}_{0}n\rightarrow ^{A}_{Z}Ba+^{94}_{36}Kr+3^{1}_{0}n$

The mass is conserved in the reaction.

So,

235 + 1 = A + 94 + 3×1

A = 139 g/mol

Also, number o protons remain same. So,

92  = Z + 36

Z = 56

The isotopic symbol is $^{139}_{56}Ba$