A

Step 1: Begin by identifying the initial and final nuclei. The initial uranium nucleus is typically $^{238}U$ (Uranium-238), and the final product after decay is usually $^{206}Pb$ (Lead-206).
Step 2: Evaluate the changes in atomic number (Z) and mass number (A).
For uranium: Z = 92, A = 238.
For lead: Z = 82, A = 206.
Step 3: Calculate the changes:
Change in mass number (A) = 238 - 206 = 32.
Change in atomic number (Z) = 92 - 82 = 10.
Step 4: Since each alpha decay reduces the mass number by 4 and the atomic number by 2, determine how many alpha particles are emitted:
Let the number of alpha particles be $n_a$, then:
4$n_a$ = 32 ◄ mass number
$ ightarrow n_a = 32/4 = 8$ alpha particles.
Step 5: For beta decays:
Let the number of beta particles be $n_b$, then:
2$n_a$ + $n_b$ = 10 ◄ atomic number
$n_b = 10 - 2*n_a = 10 - 2*8 = 2$ beta particles.
Final Answer: Therefore, the number of alpha particles emitted is 8 and the number of beta particles emitted is 2.