A stationary `Pb^(200)` nucleus emits an alpha-particle with kinetic energy `T_(alpha)= 5.77MeV`. Find the recoil velocityof a daughter nucleus. What friction of the total energy liberated in this decay is accountd for by the recoil energy of the daughter nucleus?

Explanation

The momentum of the `alpha`-particle is
`sqrt(2M_(alpha)T)`. This is also the recoil momentum of the daughter nuclear in opposite direction.
The recoil velocity of the daughter nucleus is
`(sqrt(2M_(alpha)T))/(M_(d))=(2)/(196)sqrt((2T)/(M_(p)))= 3.39xx10^(5)m//s`
The energy of the daughter nucleus is `(M_(alpha))/(M_(p))T` and this represents a fraction
`(M_(alpha)//M_(d))/(1+(M_(alpha))/(M_(d)))=(M_(alpha))/(M_(alpha+M_(d)))=(4)/(200)=(1)/(50)=0.02`
of total energy. Here `M_(d)`is the mass of the daughter nucleus.