A nucleus of mass \( 20 \mathrm{u} \) emits a \( \mathrm{y} \) photon of energy \( 6 \mathrm{MeV} \). If the emission assume to occur
when nucleus is free and rest, then the nucleus will have kinetic energy nearest to
(take \( 1 \mathrm{u}=1.6 \times 10^{-27} \mathrm{~kg} \) )

Given, nucleus of mass \( 20 \mathrm{u} \) emits a \( \mathrm{y} \) photon of energy \( 6 \mathrm{MeV} \)
Now, Momentum of photon
\( p_{p}=\frac{E}{C}=\frac{6 \times 1.6 \times 10^{-19} \times 10^{6}}{3 \times 10^{8}}=3.2 \times 10^{-21} \mathrm{kgms}^{-1} \)
Kinetic energy of nucleus
\( =\frac{\text { Momentum of photon }}{2 \times m}=\frac{3.2 \times 10^{-21}}{2 \times 20 \times 1.6 \times 10^{-27}} \)
\( =1.6 \times 10^{-16} \mathrm{~J} \)
\( =\frac{1.6 \times 10^{-16}}{1.6 \times 10^{-19}} \mathrm{eV}=10^{3} \mathrm{eV}=1 \mathrm{keV} \)
Therefore, nucleus will have kinetic energy nearest to \( 1 \mathrm{keV} \)