Physics - Modern Physics Question with Solution | TestHub

${{}_3\text{Li}}^7+{{}_1\text{H}}^1\to {{}_4\text{Be}}^8+{{}_{\mathit{\text{Z}}}\mathit{\text{X}}}^{\mathrm{A}}$

$\mathit{\text{Z}}$for the unknown$\mathit{\text{X}}$nucleus =$(3+1)-4=0$

$\mathit{\text{A}}$for the unknown$\mathit{\text{X}}$nucleus =$(7+1)-8=0$

Hence particle emitted has zero$\mathit{\text{Z}}$and zero$\mathit{\text{A}}$

It is a gamma photon.