A nucleus of mass 220 amu in free state decays to emit an alpha particle. Kinetic energy of the alpha particle emitted is 5.4 MeV. The recoil energy of the daughter nucleus is
220 amu decay into alpha particle to give 216 amu particle.
now given Kinetic energy of alpha particle= 5.4 Mev
Let the velocity of alpha particle be "v"
so 1/2 m v2 = 5.4
v comes out to be 1.634
Now applying conservation of momentum:
m1v1 + m2v2 = 0
v2= -0.0304 ( velocity of daughter nucleus)
so its Kinetic energy = 1/2 m v2 = 0.1 Mev
now given Kinetic energy of alpha particle= 5.4 Mev
Let the velocity of alpha particle be "v"
so 1/2 m v2 = 5.4
v comes out to be 1.634
Now applying conservation of momentum:
m1v1 + m2v2 = 0
v2= -0.0304 ( velocity of daughter nucleus)
so its Kinetic energy = 1/2 m v2 = 0.1 Mev