A stationary nucleus of mass 25 u emits a photon of energy 3.05 MeV . The recoil energy of the nucle

Question

A stationary nucleus of mass 25 u emits a photon of energy 3.05 MeV . The recoil energy of the nucleus is (Take $1 \mathrm{u}=931 \mathrm{MeV}$ )

Professor PC Thomas Classes · Answer

$\begin{gathered}p_{	ext {nucleus }}=p_{	ext {photon }}=\frac{E}{c} \ K=\frac{p^2}{2 M} \ K=\frac{(E / c)^2}{2 M}=\frac{E^2}{2 M c^2} \ M c^2=25 	imes 931 \mathrm{MeV} \ K=\frac{(3.05)^2}{2 	imes 25 	imes 931} \mathrm{MeV} \ K=\frac{9.3025}{46550} \mathrm{MeV} \ K \approx 0.0002 \mathrm{MeV}=0.2 \mathrm{keV}\end{gathered}$