A nucleus with mass number 220, initially at rest, emits an α − particle. If the Q - value of the reaction is 5 .5 MeV , the kinetic energy of the α - particle is

Step-by-step solution

Let K 1   &   K 2 and P 1   &   P 2 are the K . E . and momentum of the α - particle and remaining nucleus, then K 1 + K 2 = 5 .5   MeV ......(i) From the conservation of linear momentum P 1 = P 2 ⇒ 2 K 1 × 4 m ⇒ K 1 = 54 K 2 ......(ii) From (i) and (ii) K 1 = 5.5 × 54 55 = 5 .4   MeV