(B) The given nuclear reaction is $_4^9Be + _1^1H \to X + _2^4He$. Applying the law of conservation of mass number: $9 + 1 = A + 4$,which gives $A = 6

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(B) The given nuclear reaction is $_4^9Be + _1^1H \to X + _2^4He$. Applying the law of conservation of mass number: $9 + 1 = A + 4$,which gives $A = 6$. Applying the law of conservation of atomic number: $4 + 1 = Z + 2$,which gives $Z = 3$. Therefore,the product $X$ is $_3^6Li$.

Class 12 Chemistry Nuclear Chemistry Artificial transmutation

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In the nuclear reaction $_4^9Be(p, \alpha)X$,the $X$ is

A
$_2^4He$
B
$_3^6Li$
C
$_3^7Li$
D
$_4^8Be$

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In the nuclear reaction $_4^9Be(p, \alpha)X$,the $X$ is

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What is $X$ in the following reaction?
${}_{12}Mg^{26} + {}_{1}H^{2} \longrightarrow {}_{12}Mg^{27} + X$