₈₈ R _ a ²²⁶ is converted into ₈₂ P _ b ²⁰⁶ by emission of alpha ( ) and beta ( ) particles. The number of alpha and beta particles emitted are respectively

Step-by-step solution

The nuclear decay from ₈₈ Ra ²²⁶ to ₈₂ Pb ²⁰⁶ involves emission of x alpha particles (₂ He ⁴ ) and y beta particles (_ -1 e ⁰ ). Mass number conservation gives 226 = 206 + 4x , reducing to 4x = 20 and x = 5 . Atomic number conservation yields 88 = 82 + 2x - y . Substituting x = 5 gives 88 = 82 + 10 - y , so y = 92 - 88 = 4 . The final answer is A .