Question

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

• A. ${}_{12}^{22}Mg$
• B. ${}_{11}^{23}Na$
• C. ${}_{10}^{23}Ne$
• D. ${}_{10}^{22}Ne$

Answer 1

Answer: (D)

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