A positron is emitted from _11Na^22. The ratio of the atomic mass and atomic number in the resulting — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Positron Emission ($\beta^+$ Decay) In $\beta^+$ decay, a proton converts into a neutron inside the nucleus, emitting a positron and neutrino: $$\ce{^1_1p -> ^1_0n + ^0_{+1}e + \nu_e}$$ Conservation rules: mass number $A$ is unchanged; atomic number $Z$ decreases by 1. Step 2 - Apply to Sodium-22 $$\ce{^{22}_{11}Na -> ^{A'}_{Z'}Y + ^0_{+1}e + \nu_e}$$ Conservation of $A$: $22 = A' + 0 \implies A' = 22$ Conservation of $Z$: $11 = Z' + 1 \implies Z' = 10$ (Neon, \ce{Ne}) Resulting nuclide: $\ce{^{22}_{10}Ne}$ Step 3 - Calculate the Ratio $$\frac{\text{Atomic mass}}{\text{Atomic number}} = \frac{A'}{Z'} = \boxed{\frac{22}{10}}$$ Step 4 - Evaluate Options - **(A) 22/10** — Correct. $\ce{^{22}_{10}Ne}$. - **(B) 22/11** — Incorrect. Assumes $Z$ unchanged (only true for $\beta^-$ isotopic change). - **(C) 23/10** — Incorrect. Would apply to $\ce{^{23}_{11}Na}$ parent. - **(D) 23/12** — Incorrect. Wrong both in $A$ and $Z$.