Answer
$^{14}_{6}C$ $\longrightarrow$ $^{14}_{7}N+^{0}_{-1}e$
Work Step by Step
(a) $\beta^{-}$ deacy of $^{14}_{6}C$
in $\beta^{-}$ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A}_{Z+1}D+^{0}_{-1}e$
Here $P$ is parent nucleus, $D$ is daughter nucleus and $^{0}_{-1}e$ is positron or $\beta^{-}$ particle
$^{14}_{6}C$ $\longrightarrow$ $^{14}_{7}N+^{0}_{-1}e$
