Q. In the nuclear decay given below: (A, Z)X → (A, Z+1)Y → (A−4, Z−1)B* → (A−4, Z−1)B the particles emitted in the sequence are: [notation: (mass number, atomic number)]
A) γ, β, α
B) β, γ, α
C) α, β, γ
D) β, α, γ
Explanation
[Solution from source] X to Y keeps the mass number and raises the atomic number by 1, which is β⁻ emission. Y to B* lowers A by 4 and Z by 2, which is α emission. B* to B changes neither A nor Z, so the excited nucleus de-excites by γ emission, giving the order β, α, γ. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.