Answer
(a) $Bi_{83}^{209} + \alpha_{2}{4} -> At_{85}^{211} + 2n_{0}{1}$
(b) $Bi_{83}^{209} (\alpha, 2n) At_{85}^{211}$
Work Step by Step
We find:
(a) $Bi_{83}^{209} + \alpha_{2}{4} -> At_{85}^{211} + 2n_{0}{1}$
(b) $Bi_{83}^{209} (\alpha, 2n) At_{85}^{211}$
