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

Let   K1 & K2 and  P1 & P2  are the K.E and momentum of the  α-particle and remaining nucleus, then   K1+K2=5.5MeV--- (i)From conservation of linear momentum  P1=P2 ⇒2K1×4m=2K2×216m⇒K1=54K2 --- (ii)From (i) and (ii)  K1=5.5×5455=5.4MeV