A stationary nitrogen (
\(^{14}_7N\)) nucleus is bombarded with Ξ±- particle (
\(^{4}_2He\)) and the following nuclear reaction takes place:
If the kinetic energies of
\(^4_2\) and
\(^4_2H\) are 5.314MeV and 4.012MeV, respectively, then the kinetic energy of
\(^{17}_80\) is MeV. (Rounded off to one decimal place)
(Masses are given in units of π’ = 931.5MeV/c
2 )