A=202;Z=80
A=200;Z=81
A=208;Z=82
A=208;Z=80
The initial nucleus is 23290Th.
6 α-particles are emitted, and each α-particle has a mass number of 4 and an atomic number of 2.
4 β-particles are emitted, and each β-particle has a mass number of 0 and an atomic number of -1.
Let the final nucleus be AZX.
The mass number A of the final nucleus is given by:
A = 232 - 6(4) - 4(0) = 232 - 24 = 208
The atomic number Z of the final nucleus is given by:
Z = 90 - 6(2) - 4(-1) = 90 - 12 + 4 = 82
Therefore, the final nucleus is 20882X, where X is the element with atomic number 82, which is lead (Pb).
The correct option is A=208; Z=82