In the nuclear reaction _7N^{14} + _2He^4 to | vedclass

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(A) In a nuclear reaction,both the mass number $(A)$ and the atomic number $(Z)$ must be conserved on both sides of the equation.For the reaction $_7N

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$_8O^{17}$

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(A) In a nuclear reaction,both the mass number $(A)$ and the atomic number $(Z)$ must be conserved on both sides of the equation.For the reaction $_7N^{14} + _2He^4 	o _ZX^A + _1H^1$:Sum of mass numbers on the left side: $14 + 4 = 18$.Sum of mass numbers on the right side: $A + 1 = 18$,which gives $A = 17$.Sum of atomic numbers on the left side: $7 + 2 = 9$.Sum of atomic numbers on the right side: $Z + 1 = 9$,which gives $Z = 8$.Therefore,the product nucleus is $_8O^{17}$.

In the nuclear reaction $_7N^{14} + _2He^4 \to _ZX^A + _1H^1$,the product nucleus $_ZX^A$ formed is:

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