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

Question

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

Accepted Answer

1. **Identify the Decay Process:** The reaction shows the emission of a positron ($e^+$) and a neutrino ($
u$). This is characteristic of **$\beta^+$ decay** (positron emission) .
2. **Apply Conservation Laws:**
   - **Mass Number ($A$):** The total mass number is conserved. The positron and neutrino have zero mass number.
     $$ 22 = A_X + 0 \Rightarrow A_X = 22 $$
   - **Atomic Number ($Z$):** The total electric charge (atomic number) is conserved. The positron has a charge of $+1$.
     $$ 11 = Z_X + 1 \Rightarrow Z_X = 10 $$
3. **Identify the Element:** The element with atomic number $Z=10$ is Neon (Ne). Thus, the daughter nucleus $X$ is ${}^{22}_{10}\mathrm{Ne}$.

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