The correct answer is 8 and 8. Hence the correct option is a.
The nuclear reaction ₈₈A¹⁹⁶ → ₇₈B¹⁹⁴ involves the emission of both alpha (α) and beta (β) particles.
An alpha particle consists of two protons and two neutrons (4^He), while a beta particle is an electron (e− ). In this decay, the parent nucleus 198^A undergoes alpha decay, emitting an alpha particle, and transforms into the daughter nucleus 194 B. This process releases energy and changes the atomic and mass numbers of the nucleus.
The balanced equation for the reaction is ₈₈A¹⁹⁶ → ₇₈B¹⁹⁴4^He. Therefore, the number of alpha particles (α) emitted is 8, as eight alpha particles are released in the decay of the parent nucleus. There is no mention of beta particles in this particular reaction, so the number of beta particles (β) emitted is 0.
Hence, the correct answer is (a) 8 and 8, representing the emission of 8 alpha particles and 0 beta particles in the given nuclear reaction. Hence the correct option is a.