Let's solve each nuclear reaction step-by-step to find the missing values.
### First Reaction:
[tex]\[ {}_1^2 H + {}_1^2 H \rightarrow {}_E^0 He + {}_0^1 n \][/tex]
1. Atomic Numbers:
- The sum of the atomic numbers on the left (Reactants) must equal the sum on the right (Products).
- Left side: 1 (for Hydrogen) + 1 (for Hydrogen) = 2.
- Right side: E (for Helium) + 0 (for neutron) = E.
- Therefore, E = 2.
2. Mass Numbers:
- The sum of the mass numbers on the left must equal the sum on the right.
- Left side: 2 (for Hydrogen) + 2 (for Hydrogen) = 4.
- Right side: 0 (for Helium) + 1 (for neutron) = 1.
- Therefore, Helium must have a mass number that, when added to 1 (mass of the neutron), equals 4.
- Hence, the mass number of Helium is 3 (since 4 = 3 + 1).
### Second Reaction:
[tex]\[ {}_{92}^{238} U \rightarrow {}_G^{234} Th + {}_2^4 He \][/tex]
1. Atomic Numbers:
- The difference in atomic numbers from Uranium to Thorium plus Helium must balance.
- Uranium has an atomic number of 92.
- Helium (α-particle) has an atomic number of 2.
- Therefore, Thorium, G = 92 - 2 = 90.
2. Mass Numbers:
- The difference in mass numbers from Uranium to Thorium plus Helium must balance.
- Uranium has a mass number of 238.
- Helium (α-particle) has a mass number of 4.
- Therefore, Thorium has a mass number, G = 238 - 4 = 234.
### Final Results:
- First Reaction:
[tex]\[ {}_1^2 H + {}_1^2 H \rightarrow {}_2^3 He + {}_0^1 n \][/tex]
[tex]\[ E = 2 \][/tex]
- Second Reaction:
[tex]\[ {}_{92}^{238} U \rightarrow {}_{90}^{234} Th + {}_2^4 He \][/tex]
[tex]\[ G = 90 \][/tex]
Thus:
- D: [tex]\[\][/tex]
- F: [tex]\[\][/tex]
- E: [tex]\(2\)[/tex]
- G: [tex]\(90\)[/tex]
