Write the nuclear reaction

The question states that ${}_3\mathrm{Li}^7$ is bombarded by protons and produces ${}_4\mathrm{Be}^8$.

So the reaction is:

where $X$ is the emitted particle.

Apply conservation of mass number

On the left-hand side, total mass number is:

On the right-hand side, ${}_4\mathrm{Be}^8$ already has mass number $8$.

Therefore, emitted particle must have mass number:

Apply conservation of atomic number

On the left-hand side, total atomic number is:

On the right-hand side, ${}_4\mathrm{Be}^8$ has atomic number $4$.

Therefore, emitted particle must have atomic number:

Identify the particle

So the emitted particle must have:

• mass number $0$
• atomic number $0$

Among the options:

• Alpha particle: mass number $4$, atomic number $2$ ❌
• Beta particle: mass number $0$, atomic number $-1$ or $+1$ depending on type ❌
• Gamma photon: mass number $0$, atomic number $0$ ✅
• Neutron: mass number $1$, atomic number $0$ ❌

Hence, the emitted particle is a gamma photon.

Final answer

Therefore, the correct option is C.