To solve the equation for the nuclear fission process:
[tex]\[ {}_{92}^{235} \text{U} + {}_{0}^{1} \text{n} \rightarrow {}_{56}^{139} \text{Ba} + {}_{B}^{A} \text{C} + 3{}_{0}^{1} \text{n} \][/tex]
we will apply the principles of conservation of mass number (A) and atomic number (Z).
### Step-by-Step Solution:
1. Conservation of Mass Number (A):
The total mass number before the reaction must equal the total mass number after the reaction. Before the reaction, we have one Uranium nucleus and one neutron:
[tex]\[ A_U + A_n = 235 + 1 = 236 \][/tex]
After the reaction, we have Barium, the unidentified element, and three neutrons:
[tex]\[ 236 = 139 + A_C + 3 \][/tex]
Simplifying this, we get:
[tex]\[ 236 = 139 + A_C + 3 \][/tex]
[tex]\[ 236 = 142 + A_C \][/tex]
[tex]\[ A_C = 236 - 142 = 94 \][/tex]
So, the mass number [tex]\( A \)[/tex] of the unidentified element [tex]\( C \)[/tex] is 94.
2. Conservation of Atomic Number (Z):
Similarly, the total atomic number before the reaction equals the total atomic number after the reaction. Before the reaction, the atomic numbers are:
[tex]\[ Z_U + Z_n = 92 + 0 = 92 \][/tex]
After the reaction, the atomic numbers are:
[tex]\[ 92 = 56 + Z_C + 0 \][/tex]
Thus, we can solve for the atomic number [tex]\( Z_C \)[/tex]:
[tex]\[ 92 = 56 + Z_C \][/tex]
[tex]\[ Z_C = 92 - 56 = 36 \][/tex]
So, the atomic number [tex]\( Z \)[/tex] of the unidentified element [tex]\( C \)[/tex] is 36.
3. Identifying Element [tex]\( C \)[/tex]:
Using the periodic table, we look up the element with an atomic number of 36. The element with atomic number 36 is Krypton (Kr).
### Final Reaction:
The complete nuclear fission reaction with the identified element is:
[tex]\[ {}_{92}^{235} \text{U} + {}_{0}^{1} \text{n} \rightarrow {}_{56}^{139} \text{Ba} + {}_{36}^{94} \text{Kr} + 3 {}_{0}^{1} \text{n} \][/tex]
Now let's fill in the unknowns based on our findings:
- [tex]\( A = 94 \)[/tex]
- [tex]\( B = 36 \)[/tex]
- [tex]\( C = \text{Kr} \)[/tex]
Therefore, our nuclear fission equation is:
[tex]\[ {}_{92}^{235} \text{U} + {}_{0}^{1} \text{n} \rightarrow {}_{56}^{139} \text{Ba} + {}_{36}^{94} \text{Kr} + 3 {}_{0}^{1} \text{n} \][/tex]
The final answers are:
- [tex]\( A = 94 \)[/tex]
- [tex]\( B = 36 \)[/tex]
- [tex]\( C = \text{Kr} \)[/tex]
