How many alpha and beta particles are emitted when Uranium $$_{92}U^{238}$$ decays to lead $$_{82}Pb^{206}$$?

We need to find the number of alpha and beta particles emitted when $$_{92}U^{238}$$ decays to $$_{82}Pb^{206}$$.

The number of alpha particles emitted can be determined by noting that each alpha particle ($$_2He^4$$) reduces the mass number by 4 and the atomic number by 2. Since the change in mass number is $$238 - 206 = 32$$, the number of alpha particles is $$\frac{32}{4} = 8$$.

In turn, each beta particle ($$_{-1}e^0$$) increases the atomic number by 1 without changing the mass number. After eight alpha emissions, the atomic number decreases by $$8 \times 2 = 16$$ to $$92 - 16 = 76$$, whereas the final lead nucleus has atomic number 82. Therefore, beta emissions must raise the atomic number from 76 to 82, giving $$82 - 76 = 6$$ beta particles.

Verifying this result shows that the mass number becomes $$238 - 8(4) = 238 - 32 = 206$$ and the atomic number becomes $$92 - 8(2) + 6(1) = 92 - 16 + 6 = 82$$, confirming that 8 alpha particles and 6 beta particles are emitted.

The correct answer is Option D: 8 alpha particles and 6 beta particles.