Sure, let's solve the nuclear fission reactions step by step.
#### Reaction A
The first reaction is given as:
[tex]\[ {}_{92}^{233}U + {}_0^1n \rightarrow {}_{30}^0Kr + {}_{50}^1Ba + 3 \, {}_0^1n \][/tex]
To complete this reaction, we need to verify both the mass numbers (top numbers) and atomic numbers (bottom numbers) are conserved.
1. Conserving Atomic Numbers:
- Left Side (Reactants): [tex]\( 92 + 0 = 92 \)[/tex]
- Right Side (Products so far): [tex]\( 30 + 50 + (3 \times 0) = 80 \)[/tex]
To balance the atomic numbers on the right side:
[tex]\[ 92 = 80 + x \Rightarrow x = 12 \][/tex]
So, the unknown particle should have an atomic number of 12, which corresponds to Magnesium (Mg).
2. Conserving Mass Numbers:
- Left Side (Reactants): [tex]\( 233 + 1 = 234 \)[/tex]
- Right Side (Products so far): [tex]\( 0 + 1 + (3 \times 1) = 4 \)[/tex]
To balance the mass numbers on the right side:
[tex]\[ 234 = 0 + 1 + (3 \times 1) + y \Rightarrow y = 233 \][/tex]
So, the unknown particle should have a mass number of 233.
Thus, we can complete the reaction as follows:
[tex]\[ {}_{92}^{233}U + {}_0^1n \rightarrow {}_{30}^{233}Kr + {}_{50}^{1}Ba + {}_{12}^{0}Mg + 3 \, {}_0^1n \][/tex]
But this is not seeming likely, let's correct it:
[tex]\[ {}_{92}^{233}U + {}_0^1n \rightarrow {}_{36}^{95}Kr + {}_{56}^{137}Ba + 3 \, {}_0^1n \][/tex]
This balances both atomic and mass numbers:
- Atomic numbers: [tex]\( 92 + 0 = 36 + 56 + 0 = 92 \)[/tex]
- Mass numbers: [tex]\( 233 + 1 = 95 + 137 + 3 \times 1 = 234 \)[/tex]
#### Reaction B
The second reaction is given as:
[tex]\[ {}_{94}^{239}Pu + {}_0^1n \rightarrow {}_c^8Ba + {}_{38}^{91}Sr + 3 \, {}_0^1n \][/tex]
Similar to reaction A, we will ensure the conservation of atomic and mass numbers.
1. Conserving Atomic Numbers:
- Left Side (Reactants): [tex]\( 94 + 0 = 94 \)[/tex]
- Right Side (Products so far): [tex]\( c + 38 + (3 \times 0) = c + 38 \)[/tex]
To balance the atomic numbers on the right side:
[tex]\[ 94 = c + 38 \Rightarrow c = 56 \][/tex]
So, the unknown particle should have an atomic number of 56, which corresponds to Barium (Ba).
2. Conserving Mass Numbers:
- Left Side (Reactants): [tex]\( 239 + 1 = 240 \)[/tex]
- Right Side (Products so far): [tex]\( 8 + 91 + (3 \times 1) = 100 + 3 = 103 \)[/tex]
To balance the mass numbers on the right side:
[tex]\[ 240 = x + 91 + 8 = 240 \][/tex]
So, the unknown particle should have a mass number of 141.
Thus, we can complete the reaction as follows:
[tex]\[ {}_{94}^{239}Pu + {}_0^1n \rightarrow {}_{56}^{141}Ba + {}_{38}^{91}Sr + 3 \, {}_0^1n \][/tex]
Both atomic and mass numbers are conserved here.
- Atomic numbers: [tex]\( 94 + 0 = 56 + 38 + (3 \times 0) = 94 \)[/tex]
- Mass numbers: [tex]\( 239 + 1 = 141 + 91 + (3 \times 1) = 240 \)[/tex]
Hence, the correctly completed reactions are:
A:
[tex]\[ {}_{92}^{233}U + {}_0^1n \rightarrow {}_{36}^{95}Kr + {}_{56}^{137}Ba + 3 \, {}_0^1n \][/tex]
B:
[tex]\[ {}_{94}^{239}Pu + {}_0^1n \rightarrow {}_{56}^{141}Ba + {}_{38}^{91}Sr + 3 \, {}_0^1n\][/tex]
