In the nuclear reaction: X(n, alpha) {_3Li^7} | vedclass

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(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.This can be written as: $_ZX^A + _0n^1 	o _3Li^7 + _2He^4$.According to the law of conserva

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$_5B^{10}$

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(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.This can be written as: $_ZX^A + _0n^1 	o _3Li^7 + _2He^4$.According to the law of conservation of mass number and atomic number:For atomic number $(Z)$: $Z + 0 = 3 + 2 \implies Z = 5$.For mass number $(A)$: $A + 1 = 7 + 4 \implies A + 1 = 11 \implies A = 10$.Thus,the nucleus $X$ is $_5X^{10}$,which corresponds to Boron-$10$ $(_{5}B^{10})$.

In the nuclear reaction: $X(n, \alpha) {_3Li^7}$,the term $X$ will be

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