For the nuclear reaction,_{12}^{24}Mg + _{1}^ | vedclass

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(A) In a nuclear reaction,the sum of atomic numbers $(Z)$ and the sum of mass numbers $(A)$ must be conserved on both sides of the equation.For the gi

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$_{11}^{22}Na$

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(A) In a nuclear reaction,the sum of atomic numbers $(Z)$ and the sum of mass numbers $(A)$ must be conserved on both sides of the equation.For the given reaction: $_{12}^{24}Mg + _{1}^{2}D 	o _{2}^{4}He + _{Z}^{A}X$Sum of atomic numbers $(Z)$: $12 + 1 = 2 + Z \implies Z = 11$.Sum of mass numbers $(A)$: $24 + 2 = 4 + A \implies A = 22$.The element with atomic number $11$ is Sodium $(Na)$.Therefore,the missing nuclide is $_{11}^{22}Na$.

For the nuclear reaction,$_{12}^{24}Mg + _{1}^{2}D \to _{2}^{4}He + ?$,the missing nuclide is

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