With the addition of an $\alpha$ particle, atomic mass increases by 4 and atomic number increases by 2. With the addition of a $\beta$ particle, atomic number decreases by 1, and with the release of a $\beta^{+}$ particle, atomic number decreases by 1.Atomic mass of Y $= 234+4$ (due to addition of $\alpha$) $=238$Atomic number of Y $= 92-1$ (due to addition of $\beta^{-}$) $-2$ (due to release of $\beta^{+}$) $+2$ (due to addition of $\alpha$) $=91$Thus, the resulting nucleus is $^{238}_{91}\mathrm{Y}$.