If atomic mass of   Th  is 232 and its atomic number is 90. If it forces 4 β -particles and some α-particles. The mass number of final stable element is 208 and atomic number 82. How many α-particles are emitted during this process?

90Th232→82Pb208+x 2He4+4 −1e0  Balancing mass no,  232 =208 + 4x + 0 ⇒x=6 Or, Balancing atomic no, 90=82 + 2x -4 ⇒x=6when a  beta particle is emitted by a nucleus  mass number of the nucleus remains same but its atomic number will be increased  by one , when an alpha particle is emitted  by the nucleus its mass number will be decreased by 4 and atomic number will be decreased by 2.