Kinetic energy depends on the mass m and speed v of a particle, since `KE=(1//2)mn^(2)`, The `._(90)^(234)Th` nucleus has a much greater mass than the `alpha`-particle, and since the kinetic energy is proportional to the mass, we may conclude that the `._(90)^(234)Th` nucleus has the greater kinetic energy. This conclusion is not correct, since it does not take into account the fact that the `._9^(234)Th` nucleus and tha `alpha`-particle have different speeds after the decay. The conservation principle states that the total linear momentum of an isolated system remains costant. An isolated system is one for which the vecotor sum of the external forces acting on the system is zero and the decaying .`_(92)^(238)U` nucleus is staionery initially , and since momnetum is mass times velocity, its initial momentum is zero. In its final form, the system consists of the `._(90)^(234)Th` nucleus and the `alpha`- particles and has a final total momentum of `m_(Th) v_(th) + m_(alpha) v_(alpha)` According to momentum of the system must to be the same, so that `m_(Th) v_(Th) +m_(alpha) v_(alpha) =0`. Solving this equation for the velocity of the thorium nucleus, we find that `v_(Th) = -m_(alpha) v_(alpha)//m_(Th)`. Since `m_(Th)` is much greater than `m_(alpha)`, we can see that the speed of the thorium nucleus is less than the speed of the `alpha`-particle.
Moreover, `(KE_(Th))/(KE_(alpha))=((1)/(2)m_(Th)v_(Th)^(2))/((1)/(2)m_(alpha)v_(alpha^(2)))`
`=(m_(Th)v_(Th)^(2))/(m_(Th))xx(m_(alphav_(alpha)))^(2)=(m_(alpha))/(m_(Th))`
As `|m_(Th) v_(Th)|=|m_(alpha) v_(alpha)|`, thus we say that kinetic energy is shared in inverse ratio of mass.
