In the following nuclear reaction, x is: { }_ | vedclass

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Question

(C) In a nuclear reaction, both the mass number and the atomic number are conserved.Given reaction: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n

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${ }_{15} P^{30}$

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(C) In a nuclear reaction, both the mass number and the atomic number are conserved.Given reaction: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n^1 + X$Conservation of mass number $(A)$:$27 + 4 = 1 + A_X$$31 = 1 + A_X \Rightarrow A_X = 30$Conservation of atomic number $(Z)$:$13 + 2 = 0 + Z_X$$15 = Z_X$Since the atomic number is $15$, the element is Phosphorus $(P)$.Therefore, $X = { }_{15} P^{30}$.Hence, option $C$ is correct.

In the following nuclear reaction, $x$ is: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n^1 + X$

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