Question

A radioactive decay forms an isotope of the original nucleus with the emission of the following particles

• A. one $\alpha$- and four $\beta^-$-particles
• B. one $\alpha$- and one $\beta^-$-particles
• C. one $\alpha$- and two $\beta^-$-particles
• D. four $\alpha$- and one $\beta^-$-particles

Hint:

Decay Chains and Isotopes

• $\alpha$ decay: $A - 4$, $Z - 2$
• $\beta^-$ decay: $A$ unchanged, $Z + 1$
• Isotopes: Same $Z$, different $A$
• To restore $Z$ after one $\alpha$ decay, emit two $\beta^-$ particles.

Answer 1

The Correct Option is C

Let the initial nucleus be $^A_ZX$. Emission of:

• One $\alpha$-particle decreases $A$ by 4 and $Z$ by 2.
• Each $\beta^-$-particle increases $Z$ by 1.

For the final nucleus to be an \textit{isotope}, $Z$ must remain unchanged. Let one $\alpha$ and $x$ $\beta^-$ particles be emitted. Change in $Z$: \[ Z_\text{final} = Z - 2 + x = Z \Rightarrow x = 2 \] So, one $\alpha$ and two $\beta^-$ emissions restore $Z$, making the final nucleus an isotope (same $Z$, different $A$). Hence, option (3) is correct.