Question

In a nuclear fission reaction a heavy nucleus divides to form smaller nuclei and one or more neutrons. Many nuclei can undergo fission, but the fission reactions of uranium-235 and plutonium-239 are the principal ones that generate energy in nuclear power plants. This problem deals with balancing the fission reaction of the uranium-235 isotope as it undergoes bombardment from a neutron.
A. 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.
B. In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs:
235 92U+10n→AZSr+143 54Xe+310n
Enter the isotope symbol for the strontium (Sr) nucleus in this reaction.

Answer 1

Answer:

$\frac{235}{92} U +$$\frac{1}{0} n$ -------> $\frac{139}{36} Ba$ + $\frac{94}{56} Kr + 3\frac{1}{0} n$

$\frac{235}{92} U + \frac{1}{0} n -------> \frac{90}{38} Sr + \frac{143}{54} Xe + 3\frac{1}{0} n$

Explanation:

In equation 1, equating the mass number (A) on both sides.

A = 235 + 1 = A + 94 + 3*1

236 = A + 94 + 3

A = 236 - 94 = 3

A = 139  

Equate the atomic numbers on both sides

92 + 0 = Z + 36 + 3*0

92 = Z + 36

Z = 92 - 36

Z = 56

In reaction 2, equating the mass number on both sides

235 + 1 = A + 143 + 3 *1

236 = A + 143 + 3

236 = Z + 146

Z = 90

Equatoing the atomic number of both sides

92 + 0 = Z + 54 + 3*0

92 = Z + 54

Z = 92 - 54

Z = 38.