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  4. In the given nuclear reaction, the element X is: 1122 Na arrow X + e ++v

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

NEETNEET 2022Nuclei

Solution:

${ }_{11}^{22} Na \longrightarrow X +e^{+}+v$
This is $\beta^{+}-$decay
${ }_{11}^{22} Na \longrightarrow{ }_{10}^{22} Ne + e ^{+}+ v$