Determine the missing particles in the following nuclear reactions: a) $$^{239}_{94}Pu + ^{4}_2He \rightarrow ? + ?^1_0n$$ б) $$^2_1H + ^3_1D \rightarrow ^1_1H + ?$$ S2 a) $$? + ^1_1H \rightarrow ^{24}_{11}Na + ^4_2He$$ б) $$^{27}_{13}Al + ^1_0n \rightarrow ^{26}_{12}Mg + ?$$

Here are the completed nuclear reactions:

• a) $$^{239}_{94}Pu + ^4_2He \rightarrow ^{242}_{96}Cm + ^{1}_0n$$
• б) $$^2_1H + ^3_1D \rightarrow ^1_1H + ^4_2He$$
• a) $$^{23}_{11}Na + ^1_1H \rightarrow ^{24}_{11}Na + ^4_2He$$
• б) $$^{27}_{13}Al + ^1_0n \rightarrow ^{28}_{13}Al + ^1_0n$$ or $$^{27}_{13}Al + ^1_0n \rightarrow ^{26}_{12}Mg + ^2_1H$$ (The problem statement for the last reaction seems to have a typo with Mg having an atomic number of 12 and mass number of 26, and also the reaction is incomplete. Two possible completions are provided.)

Explanation of Principles: Nuclear reactions must conserve both the total mass number (superscript) and the total atomic number (subscript) on both sides of the equation. The missing particles are determined by balancing these numbers. * Conservation of Mass Number: The sum of the superscripts on the left side must equal the sum of the superscripts on the right side. * Conservation of Atomic Number: The sum of the subscripts on the left side must equal the sum of the subscripts on the right side. The subscript also represents the number of protons, which defines the element. Step-by-step solution for the first reaction (a): * Reaction: $$^{239}_{94}Pu + ^4_2He \rightarrow ? + ?^1_0n$$ * Balance Mass Number: $$239 + 4 = X + 1 ightarrow X = 242$$ * Balance Atomic Number: $$94 + 2 = Y + 0 ightarrow Y = 96$$ * Identify the missing particle: An element with atomic number 96 is Curium (Cm). So the missing particle is $$^{242}_{96}Cm$$. Step-by-step solution for the second reaction (б): * Reaction: $$^2_1H + ^3_1D \rightarrow ^1_1H + ?$$ * Balance Mass Number: $$2 + 3 = 1 + X ightarrow X = 4$$ * Balance Atomic Number: $$1 + 1 = 1 + Y ightarrow Y = 1$$ * Identify the missing particle: An element with atomic number 1 is Hydrogen (H). So the missing particle is $$^4_2He$$ (Helium, due to atomic number 2). Step-by-step solution for the third reaction (a): * Reaction: $$? + ^1_1H \rightarrow ^{24}_{11}Na + ^4_2He$$ * Balance Mass Number: $$X + 1 = 24 + 4 ightarrow X = 27$$ * Balance Atomic Number: $$Y + 1 = 11 + 2 ightarrow Y = 12$$ * Identify the missing particle: An element with atomic number 12 is Magnesium (Mg). So the missing particle is $$^{27}_{12}Mg$$. Step-by-step solution for the fourth reaction (б): * Reaction: $$^{27}_{13}Al + ^1_0n \rightarrow ^{26}_{12}Mg + ?$$ * Balance Mass Number: $$27 + 1 = 26 + X ightarrow X = 2$$ * Balance Atomic Number: $$13 + 0 = 12 + Y ightarrow Y = 1$$ * Identify the missing particle: An element with atomic number 1 and mass number 2 is Deuterium ($$^2_1H$$). Therefore, the missing particle is $$^2_1H$$. Note: The image had a possible typo for the last reaction, where the product Mg had a mass number of 26 and atomic number 12. This is consistent with the reaction $$^{27}_{13}Al + ^1_0n \rightarrow ^{28}_{13}Al$$ and subsequent decay, or it could be intended as $$^{27}_{13}Al + ^1_0n \rightarrow ^{26}_{12}Mg + ^2_1H$$. I have provided the solution for the latter which results in a valid nuclear reaction. If the intended product was $$^{28}_{13}Al$$, then a neutron ($$^1_0n$$) would be emitted as a byproduct: $$^{27}_{13}Al + ^1_0n \rightarrow ^{28}_{13}Al + ^0_0́̄̅