Let say lead has mass number A and atomic number Z after emission of 6α and 4β-particles. Nuclear reaction for given situation is given as ‌90232Th→‌Z1Pb+6‌24He+410R Since, atomic number of both sides are equal hence 90=Z+6×2+4×(−1) 90=Z+12−4 or Z=82 Similarly, comparing the mass number of both sides, 232‌‌=A+6×4+4×0 232‌‌=A+24 A‌‌=208