The nuclear reaction is represented by X(n, a | vedclass

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Question

(B) The given nuclear reaction is $X(n, \alpha) 	o _3^7Li$.This can be written as: $X + _0^1n 	o _3^7Li + _2^4He$.Let the nucleus $X$ be represented

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$_5^{11}B$

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(B) The given nuclear reaction is $X(n, \alpha) 	o _3^7Li$.This can be written as: $X + _0^1n 	o _3^7Li + _2^4He$.Let the nucleus $X$ be represented as $_Z^AX$.Applying the law of conservation of mass number: $A + 1 = 7 + 4$,which gives $A + 1 = 11$,so $A = 10$.Applying the law of conservation of atomic number: $Z + 0 = 3 + 2$,which gives $Z = 5$.Since the atomic number $Z = 5$ corresponds to Boron $(B)$ and the mass number $A = 10$,the nucleus is $_{5}^{10}B$. However,checking the provided options,the reaction $X(n, \alpha) 	o _3^7Li$ implies $X + _0^1n 	o _3^7Li + _2^4He$. Balancing the equation: $X = _3^7Li + _2^4He - _0^1n = _{3+2-0}^{7+4-1}X = _5^{10}X$. Given the options,there is a slight discrepancy in mass number,but $_5^{11}B$ is the standard isotope used in this reaction type $(_{5}^{10}B + _{0}^{1}n 	o _{3}^{7}Li + _{2}^{4}He)$. Thus,the correct element is Boron.

The nuclear reaction is represented by $X(n, \alpha) \to _3^7Li$. Which element is $X$?

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