If the atomic number of 'H' is 88, what is the atomic number of 'A'? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Decay Sequence The transition series from the passage: $$A \xrightarrow{-\alpha} B \xrightarrow{-\alpha} C \xrightarrow{-\beta} D \xrightarrow{-\beta} E \xrightarrow{-\alpha} F \xrightarrow{-\beta} G \xrightarrow{-\alpha} H$$ Count: **4 alpha** emissions and **3 beta** emissions from $A$ to $H$. Step 2 - Change in Atomic Number - Each $\alpha$ emission: $\Delta Z = -2$ - Each $\beta^-$ emission: $\Delta Z = +1$ $$Z_H = Z_A + (4 \times (-2)) + (3 \times (+1)) = Z_A - 8 + 3 = Z_A - 5$$ Step 3 - Solve for $Z_A$ Given $Z_H = 88$: $$88 = Z_A - 5 \implies Z_A = 93$$ Step 4 - Evaluate Options - **(A) 91**: $Z_H = 91 - 5 = 86$. Incorrect. - **(B) 92**: $Z_H = 92 - 5 = 87$. Incorrect. - **(C) 93**: $Z_H = 93 - 5 = 88$ ✓. **Correct.** - **(D) 94**: $Z_H = 94 - 5 = 89$. Incorrect. Note: $Z_A = 93$ is Neptunium (Np); $Z_H = 88$ is Radium (Ra). $$\boxed{\text{Answer: C — Atomic number of A} = 93}$$