Question

A nucleus \( _{Z}^A X \) emits an \( \alpha \)-particle. The resultant nucleus emits a \( \beta^- \)-particle. The respective atomic and mass numbers of final nucleus will be:

• A. \( _{Z-2}^{A-4} \)
• B. \( _{Z-1}^{A-4} \)
• C. \( _{Z-2}^{A-2} \)
• D. \( _{Z-1}^{A-2} \)

Hint:

The emission of an \( \alpha \)-particle reduces the atomic number by 2 and the mass number by 4.

Answer 1

The Correct Option is A

Step 1: Understand the decay process.
The emission of an \( \alpha \)-particle decreases the atomic number by 2 and the mass number by 4. After the \( \alpha \)-particle is emitted, the remaining nucleus will have an atomic number \( Z-2 \) and mass number \( A-4 \). Then, the emission of a \( \beta^- \)-particle does not change the mass number but increases the atomic number by 1. Hence, the final nucleus will have \( Z-2 \) and \( A-4 \).
Final Answer: \[ \boxed{_ {Z-2} ^{A-4}} \]