No, your answer is not correct.
When U238 decays into Pb206, it undergoes a series of alpha and beta decays. Let's break down the process:
1) U238 undergoes alpha decay, which means it emits an alpha particle (consisting of two protons and two neutrons) and transforms into Th234.
2) Th234 then undergoes beta decay, emitting a beta particle (an electron) and transforming into Pa234.
3) Pa234 also undergoes beta decay, emitting another beta particle and transforming into U234.
4) U234 undergoes alpha decay, emitting an alpha particle and transforming into Th230.
5) This process continues until Pb206 is produced, with a series of alpha and beta decays.
Therefore, the correct answer is B) 8 alpha particles, 6 electrons, 6 neutrinos, and energy.