Name the particle 'X' emitted in the nuclear reaction given below -
${_{11}^{22}Na} → {_{10}^{22}Ne}+v+X$
Answer & explanation
Correct answer: option 3
$^{22}_{11}\text{Na} \rightarrow \, ^{22}_{10}\text{Ne} + \nu + X$
$\text{Mass number: } 22 = 22 + 0 + 0$
$\text{Atomic number: } 11 = 10 + 0 + Z_X$
$Z_X = +1$
$X = \beta^+ \ (\text{positron})$
The particle $X$ is a positron $(\beta^+)$.