A positron is emitted from _11Na^23. The ratio of the atomic mass and atomic number in the resulting — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Positron Emission Rules In positron ($\beta^+$) emission: mass number $A$ is conserved; atomic number $Z$ decreases by 1. $$\ce{^A_Z X -> ^A_{Z-1}Y + ^0_{+1}e + \nu_e}$$ Step 2 - Apply to Sodium-23 $$\ce{^{23}_{11}Na -> ^{23}_{Z'}Y + ^0_{+1}e + \nu_e}$$ Conservation of mass number: $A' = 23$ Conservation of atomic number: $Z' = 11 - 1 = 10$ → element with $Z=10$ is Neon Resulting nuclide: $\ce{^{23}_{10}Ne}$ Step 3 - Calculate the Ratio $$\text{Ratio} = \frac{\text{atomic mass}}{\text{atomic number}} = \frac{A'}{Z'} = \frac{23}{10}$$ Step 4 - Option Analysis - (A) 22/10: incorrect — mass number does not change in $\beta^+$ decay - (B) 22/11: incorrect — both mass and atomic number assumptions are wrong - (C) 23/10: correct ✓ - (D) 23/12: incorrect — $Z$ increases in $\beta^-$ decay, not $\beta^+$ $$\boxed{C}$$