To solve these nuclear fission reactions, we need to ensure the conservation of both mass number (the sum of protons and neutrons) and atomic number (the number of protons) on both sides of the reaction.
Let's analyze each reaction step-by-step.
### Reaction 1:
[tex]\[ {}_{92}^{235}U + {}_0^1n \rightarrow {}_{36}^{90}Kr + {}_{56}^\wedge Ba + 3{}_0^1n \][/tex]
First, we sum the mass numbers and atomic numbers on the reactant side:
- Reactant mass number: [tex]\( 235 (U) + 1 (n) = 236 \)[/tex]
- Reactant atomic number: [tex]\( 92 (U) + 0 (n) = 92 \)[/tex]
Next, we balance these with the products:
- Product known mass: [tex]\( 90 (Kr) \)[/tex]
- Product known atomic number: [tex]\( 36 (Kr) \)[/tex]
Including the barium (Ba) and neutrons in the product side:
- Number of neutrons = 3, so their combined mass number: [tex]\( 3 \times 1 = 3 \)[/tex]
- Their combined atomic number: [tex]\( 3 \times 0 = 0 \)[/tex]
Solving for Barium:
We need total mass number of products equal to the reactants, 236:
[tex]\[ 90 (Kr) + \wedge (Ba) + 3 (neutrons) = 236 \][/tex]
So:
[tex]\[ \wedge = 236 - 93 = 143 \][/tex]
We need total atomic number of products to equal reactants, 92:
[tex]\[ 36 (Kr) + \square = 92 \][/tex]
So:
[tex]\[ \square = 92 - 36 = 56 \][/tex]
The completed reaction is:
[tex]\[ {}_{92}^{235}U + {}_0^1n \rightarrow {}_{36}^{90}Kr + {}_{56}^{143}Ba + 3{}_0^1n \][/tex]
### Reaction 2:
[tex]\[ {}_{94}^{239}Pu + {}_0^1n \rightarrow {}_c^\wedge Ba + {}_{38}^{92}Sr + 3{}_0^1n \][/tex]
First, we sum the mass numbers and atomic numbers on the reactant side:
- Reactant mass number: [tex]\( 239 (Pu) + 1 (n) = 240 \)[/tex]
- Reactant atomic number: [tex]\( 94 (Pu) + 0 (n) = 94 \)[/tex]
Next, we balance these with the products:
- Product known mass: [tex]\( 92 (Sr) \)[/tex]
- Product known atomic number: [tex]\( 38 (Sr) \)[/tex]
Including the barium (Ba) and neutrons in the product side:
- Number of neutrons = 3, so their combined mass number: [tex]\( 3 \times 1 = 3 \)[/tex]
- Their combined atomic number: [tex]\( 3 \times 0 = 0 \)[/tex]
Solving for Barium:
We need total mass number of products equal to the reactants, 240:
[tex]\[ 92 (Sr) + \wedge (Ba) + 3 (neutrons) = 240 \][/tex]
So:
[tex]\[ \wedge = 240 - 95 = 145 \][/tex]
We need total atomic number of products to equal reactants, 94:
[tex]\[ 38 (Sr) + \square = 94 \][/tex]
So:
[tex]\[ \square = 94 - 38 = 56 \][/tex]
The completed reaction is:
[tex]\[ {}_{94}^{239}Pu + {}_0^1n \rightarrow {}_{56}^{145}Ba + {}_{38}^{92}Sr + 3{}_0^1n \][/tex]
So, the finalized values for the second reaction are:
A: [tex]\({}_{50}^{\wedge}Ba\)[/tex]: [tex]$\rightarrow {}_{56}^{145}Ba$[/tex]
B: [tex]${}_{c}^{92}Sr$[/tex] [tex]$\rightarrow {}_{38}^{92}Sr$[/tex]
C: [tex]\({}_{0}^{3}n\)[/tex]
Thus, this ensures mass and atomic numbers are conserved in both reactions.
