IDNLearn.com: Your go-to resource for finding expert answers. Find the information you need quickly and easily with our comprehensive and accurate Q&A platform.
Sagot :
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]
#### 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]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
