In the given nuclear reaction,the element X i | vedclass

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(B) The given reaction is ${ }_{11}^{22} Na ightarrow X + e ^{+} + 
u$.This represents a $\beta^{+}$ decay (positron emission).In $\beta^{+}$ decay

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${ }_{10}^{22} Ne$

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(B) The given reaction is ${ }_{11}^{22} Na ightarrow X + e ^{+} + 
u$.This represents a $\beta^{+}$ decay (positron emission).In $\beta^{+}$ decay,the atomic number $Z$ decreases by $1$ while the mass number $A$ remains constant.For the parent nucleus ${ }_{11}^{22} Na$,the atomic number $Z = 11$ and the mass number $A = 22$.After emitting a positron $(e^{+})$,the new atomic number $Z' = 11 - 1 = 10$.The mass number remains $A' = 22$.The element with atomic number $10$ is Neon $(Ne)$.Therefore,the reaction is ${ }_{11}^{22} Na ightarrow { }_{10}^{22} Ne + e ^{+} + 
u$.Thus,$X$ is ${ }_{10}^{22} Ne$.

In the given nuclear reaction,the element $X$ is:
${ }_{11}^{22} Na \rightarrow X + e ^{+} + \nu$

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In the given nuclear reaction,the element $X$ is:${ }_{11}^{22} Na \rightarrow X + e ^{+} + \nu$