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?

A stationary `Pb^(200)` nucleus emits an alpha-particle with kinetic e

1.	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?
Answer» 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.

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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?

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