In the given nuclear reaction, the element $X$ is $^{22}_{11} ext{Na} ightarrow X + e^+$

Question

In the given nuclear reaction, the element $X$ is $^{22}_{11}	ext{Na} ightarrow X + e^+$

Accepted Answer

In a positron emission ($e^+$), the atomic number decreases by 1 and the mass number remains the same. $^{22}_{11}	ext{Na} ightarrow ^{22}_{10}	ext{Ne} + ^{0}_{+1}e$. Thus, $X$ is $^{22}_{10}	ext{Ne}$.

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