Answer
$He: U_{92}^{238} -> Pb_{82}^{206} + 8\alpha_{2}^{4} + 6\beta_{-1}^{0}$
The $\alpha$-particles produced are eventually converted to Helium atoms.
$Ne: Na_{11}^{22} -> Ne_{10}^{22} + \beta_{1}^{0}$
$Ar: K_{31}^{40} + e_{-1}^{0} -> Ar_{18}^{40}$
$Kr: U_{92}^{235} + n_{0}^{1} -> Kr_{36}^{85} + Ba{56}^{148} + 3n_{0}^{1}$
$Xe: U_{92}^{235} + n_{0}^{1} -> Sr_{38}^{90} + Xe{54}^{143} + 3n_{0}^{1}$
$Rn: Ra_{88}^{226} -> Rn_{86}^{222} + \alpha_{2}^{4} $
Work Step by Step
$He: U_{92}^{238} -> Pb_{82}^{206} + 8\alpha_{2}^{4} + 6\beta_{-1}^{0}$
The $\alpha$-particles produced are eventually converted to Helium atoms.
$Ne: Na_{11}^{22} -> Ne_{10}^{22} + \beta_{1}^{0}$
$Ar: K_{31}^{40} + e_{-1}^{0} -> Ar_{18}^{40}$
$Kr: U_{92}^{235} + n_{0}^{1} -> Kr_{36}^{85} + Ba{56}^{148} + 3n_{0}^{1}$
$Xe: U_{92}^{235} + n_{0}^{1} -> Sr_{38}^{90} + Xe{54}^{143} + 3n_{0}^{1}$
$Rn: Ra_{88}^{226} -> Rn_{86}^{222} + \alpha_{2}^{4} $
