The nucleus of 214Po decays radioactively by emitting an alpha particle (mass 6.65 × 10?27 kg) with kinetic energy 1.23 × 10?12 J, as measured in the laboratory reference frame. Assuming that the Po was initially at rest in this frame, find the recoil velocity of the nucleus that remains after the decay.

Solution 75P Step 1: Mass of Alpha particle m = 6He x10 -27Kg Mass of Po m = Po x 1.67 x10 -27Kg = 357.38 x10 -27Kg Where 1.67 x10 -27Kg is mass per nucleon. -12 Kinetic Energy of Alpha particle K = 1.Hex 10 J Step 2: Kinetic energy of alpha particles is given by K = 1 m v 2 He 2 He He Where v iHehe speed of alpha particles Step 3: Rearranging the above equation 2K He vHe m He 2*1.2* 1012 vHe 6.6* 1027 v = 1.92 x 10 m/s He 7 Speed of Alpha particles is 1.92 x 10 m/s Step 4: To find the recoil velocity. When an alpha particle is ejected from a Po nucleus, Po experiences an opposite force. Initially the Po nucleus is at rest, hence initial momentum is zero. After emission alpha particle moves in a particular direction and Po nucleus moves in the opposite direction.