An artificial radioactive decay series begins with unstable94Pu241 . The stable nuclide obtained after eight α -decays and five β -decays is
83Br209
82Pb209
82Se205
82Mg201
Decrease in mass no.=8×4+5×0=32
Decrease in charge no.=8×2−5×1=11
Therefore (a) is the right choice.