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In the nuclear decay given below: A Z X → A−4 Z−2 Y → A−4 Z−1 B → A−4 Z−1 B⁻ The particles emitted in the sequence are

1. γ, β, α
2. β, γ, α
3. α, β, γ
4. β, α, γ

Correct answer: α, β, γ

Solution

The sequence of nuclear decay shows a decrease in mass number by 4 and atomic number by 2 in the first step, indicating alpha (α) emission. The second step involves a change in atomic number by +1, indicating beta (β) emission. The final step involves no change in mass or atomic number, indicating gamma (γ) emission.

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