1
Question
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
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
Open in App
Solution
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
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
