Question

Refer to illustration `5.10`, the energy released by the `alpha` - decay of `._(92)^(238) U` is found to be` 4.3 MeV`. Since this energy is carried away as kinetic energy of the recoiling `._(90)^(234)Th`. nucleus and the `alpha`- particles, it follows that `KE_(Th) +KE_(alpha )=4.3 MeV`. However, `KE_Th ` and `KE_alpha` are not equal. Which particle carries away more kinetic energy, the `overset(234)(90)Th` nucleus or the `alpha`- particle? .

Answer 1

Kinetic energy depends on the mass m and speed v of a particle, since `KE =(1)/(2) mnu^(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, it is tempting to conclude that the `._(90)^(234)Th` nucleus has a greater kinetic energy. This conclusion is not correct, however, since it does not take into account the fact that the `._(90)^(234)Th` nucleus has greater kinetic energy. This conclusion is not correct, however, since it does not take into account the fact that the `._(90)^(234)Th` nucleus and `alpha` - particle have diiferent speeds after the decay. In fact, we wxpect the thorium nucleus to recoil with smaller speed precisely because it has the greater mass. The decaying of `._(92)^(238)U` is like a father and his young daughter on ice skates, pushing off against one another. The more massive father recoils with much less speed than the daughter. We can use the pricnc iple of conservation. As we know, the conservation princliple states that the total linear momentum of an isolated sysyem is one for which the vector sum of the external forces acting on the system is zero, and the decaying `._(92)^(238)U` nucleus fits this description. It is stationary inititally, and since momentum is mass times velocity, its initial momentum is zero. In its final from, the system consists of the `._(90)^(234)Th` nucleus and the `alpha `- particle and has a final total momentum of `m_(Th) nu_(Th) +m_(alpha) v_(alpha)` . According to momentum conservation, the initial and final values of the total momentum of the system must 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, the kinetic energy depends on the square of the speed and only the first power of the mass. As a result of its much greater speed, the `alpha `- particle has the greater kinetic energy.