Question

A certain radioactive material \(_x^AX\) starts emitting \(\alpha\) and \(\beta\) particles successively such that the end product is \(_{Z-3}^{A-8}\)A. The number of \(\alpha\) and \(\beta\) particles emitted are respectively

• A. 4 and 3 respectively
• B. 2 and 1 respectively
• C. 3 and 4 respectively
• D. 3 and 8 respectively

Hint:

\(\alpha\)-decay reduces \(A\) by 4 and \(Z\) by 2, while \(\beta^-\)-decay keeps \(A\) same but increases \(Z\) by 1.

Answer 1

The Correct Option is B

Step 1: Understand changes due to \(\alpha\)-decay.
Each \(\alpha\)-particle emission decreases mass number by 4 and atomic number by 2.
So if \(n\) alpha particles are emitted:
\[ A \to A - 4n \quad \text{and} \quad Z \to Z - 2n \]
Step 2: Compare with given final mass number.
Final mass number is \(A-8\).
So:
\[ A - 4n = A - 8 \Rightarrow 4n = 8 \Rightarrow n = 2 \]
Step 3: Now compare atomic number change.
After 2 alpha decays:
\[ Z \to Z - 2(2) = Z - 4 \]
But final atomic number is given as \(Z-3\).
Step 4: Effect of \(\beta^-\)-decay.
Each \(\beta^-\) emission increases atomic number by 1 (mass number unchanged).
So if \(m\) beta particles emitted:
\[ Z - 4 + m = Z - 3 \Rightarrow m = 1 \]
Step 5: Final conclusion.
Number of \(\alpha\) particles = 2 and number of \(\beta\) particles = 1.
Final Answer:
\[ \boxed{\text{2 and 1 respectively}} \]