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

Question

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

Aicharya · Accepted Answer

As positron is emitting, it is positive beta decay with equation, $X_{z}^{A} ightarrow Y_{z-1}^{A} + e^{+} + 
u$ Hence, $Na_{11}^{22} ightarrow Ne_{10}^{22} + e^{+} + 
u$

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

Solution