A nucleus of mass number 220 decays by decay. The energy released in the reaction is 5 MeV . The kinetic energy of an -particle is

Step-by-step solution

Let the reaction be represented as _ Z ²²⁰ X Z -2 216 gathered 4 Y + H e 2 gathered energy released in the reaction is 5 MeV 1 2 m_Y v_Y^2+ 1 2 m_ v_ ^2=5 MeV Also using conservation of linear momentum. aligned & m_Y v_Y=-m_ v_ v_Y= -m_ m_Y v_ & v_Y= -4 216 v_ = -1 54 v_ aligned Putting in eqn. (i) aligned & 1 2 (216) ( v_ 54 )^2+ 1 2 4 (v_ )^2=5 MeV & 1 2 (216) ( v_ 54 )^2+ K _ E E _ =5 MeV & 1 2 4 v_ ^2 54 + K E _ =5 MeV & 1 54 ( ~K _ E _ )+ K _ E _ =5 MeV & K.E. = 5 54 55 = 54 11 MeV aligned