To complete the given nuclear fission equation properly, let's break down the problem step by step, adhering to conservation laws:
1. Conservation of Mass Number (A):
Initially, we have uranium-235 [tex]\(({}_{92}^{235}U)\)[/tex] and a neutron [tex]\(({}_{0}^{1}n)\)[/tex]:
[tex]\[
235 + 1 = 236
\][/tex]
In the products, we have one barium-139 [tex]\(({}_{56}^{139}Ba)\)[/tex] and another unknown nucleus [tex]\(({}_{B}^{A}C)\)[/tex] along with three neutrons [tex]\((3 \times {}_{0}^{1}n)\)[/tex]:
[tex]\[
139 + A + (3 \times 1) = 139 + A + 3
\][/tex]
Since the total mass number before and after the reaction must be the same:
[tex]\[
236 = 139 + A + 3
\][/tex]
Solving for [tex]\(A\)[/tex]:
[tex]\[
A = 236 - 139 - 3 = 94
\][/tex]
2. Conservation of Atomic Number (B):
Initially, we have uranium-235 with atomic number 92 [tex]\(({}_{92}^{235}U)\)[/tex] and a neutron which has an atomic number of 0 [tex]\(({}_{0}^{1}n)\)[/tex]:
[tex]\[
92 + 0 = 92
\][/tex]
In the products, we have one barium nucleus with an atomic number of 56 [tex]\(({}_{56}^{139}Ba)\)[/tex] and another unknown nucleus [tex]\(({}_{B}^{A}C)\)[/tex], along with three neutrons [tex]\((3 \times {}_{0}^{1}n)\)[/tex]:
[tex]\[
56 + B + (3 \times 0) = 56 + B
\][/tex]
Since the total atomic number before and after the reaction must be the same:
[tex]\[
92 = 56 + B
\][/tex]
Solving for [tex]\(B\)[/tex]:
[tex]\[
B = 92 - 56 = 36
\][/tex]
3. Determining the Element (C):
From the periodic table, the element with atomic number 36 is krypton (Kr).
Therefore, the values for [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are:
- A: 94
- B: 36
- C: Kr
Combining these results, the completed nuclear fission equation looks like:
[tex]\[
{}_{92}^{235} U + {}_{0}^{1} n \rightarrow {}_{56}^{139} Ba + {}_{36}^{94} Kr + 3 {}_{0}^{1} n
\][/tex]
