Question

If a radioactive parent nucleus \({}^{236}_{94}X\) emits two alpha particles and two \(\beta\) particles successively to reach the daughter nucleus \({}^{a}_{b}Y\), then the values of \(a\) and \(b\) are

• A. $224$ and $90$
• B. $220$ and $94$
• C. $228$ and $92$
• D. $230$ and $92$
• E. $226$ and $92$

Hint:

Decay rules: - $\alpha$: $A-4$, $Z-2$ - $\beta^-$: $Z+1$

Answer 1

The Correct Option is C

Concept: - $\alpha$ decay: $A \rightarrow A-4$, $Z \rightarrow Z-2$
- $\beta^-$ decay: $Z \rightarrow Z+1$, $A$ unchanged

Step 1: After two $\alpha$ decays.
\[ A = 236 - 8 = 228, \quad Z = 94 - 4 = 90 \]

Step 2: After two $\beta$ decays.
\[ Z = 90 + 2 = 92 \]

Step 3: Final nucleus.
\[ A = 228, \quad Z = 92 \]