When ${}_3{\text{Li}}^7$ nuclei are bombarded by protons, and the resultant nuclei are ${}_4{\text{Be}}^8$, the emitted particles will be:

The reaction is:
$${}_3{\text{Li}}^7+{}_1{\text{H}}^1\to {}_4{\text{Be}}^8+X$$
Checking mass number: $7+1=8+0⟹$ mass number of $X$ is $0$.Checking atomic number: $3+1=4+0⟹$ atomic number of $X$ is $0$.The particle with $0$ mass and $0$ charge is a gamma photon ($\gamma$).