KCET 2016 Physics Solved Paper

Given, nucleus of mass 20 u emits a γ photon of energy 6 MeV
Now, Momentum of photon
Kinetic energy of nucleus
=1.6×10−16J
=‌

1.6×10−16

1.6×10−19

eV=103eV=1keV
Therefore, nucleus will have kinetic energy nearest to 1 keV