In the nuclear transmutation 4 9 Be + X ⟶ 4 8 Be + Y , identify X ,   Y .

By charge and mass conservation, we can complete the nuclear transformation.

$\begin{array}{c}9\\ 4\end{array}\text{Be}+\mathrm{\gamma }⟶\begin{array}{c}8\\ 4\end{array}\text{Be}+\begin{array}{c}1\\ 0\end{array}\text{n}$
$\begin{array}{c}9\\ 4\end{array}\text{Be}+{}_1{\text{H}}^1⟶\begin{array}{c}8\\ 4\end{array}\text{Be}+\begin{array}{c}2\\ 1\end{array}\text{D}$
$\begin{array}{c}9\\ 4\end{array}\text{Be}+{}_0{\text{n}}^1⟶\begin{array}{c}8\\ 4\end{array}\text{Be}+2\begin{array}{c}1\\ 0\end{array}\text{n}$
As with $\mathrm{\gamma }$-rays, the emission is neutron and not proton.