How Is Kinetic Energy Distributed in a Thorium to Radon Decay Process?

[MAPgirl23]

A 232 Th (thorium) nucleus at rest decays to a 228 Ra (radon) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is 6.54 x 10^-3 J. An alpha particle has 1.76% of the mass of a 228 Ra nucleus.

a) Calculate the kinetic energy of the recoiling 228 Ra nucleus.

b) Calculate the kinetic energy of the alpha particle.

** Here again I would use the formula k_f = 0.5(m_1 + m_2)v_(particle)^2 now I know we don't have any velocities to work with but the problem gives me the total kinetic energy = 6.54*10^-3 Therefore I should solve maybe for velocities first in order to plug in and get the kinetic energy of Ra and do the same for the alpha particle. Assuming m_1 = 228 and the alpha particle m_2 = 4.0 I need help seeing if I'm going on the right track with finding the velocities of the particles or is that not necessary ?

 

[Doc Al]

Hint: The speeds of the two fragments can be related using conservation of momentum.

 

[Doc Al]

** Here again I would use the formula k_f = 0.5(m_1 + m_2)v_(particle)^2 now I know we don't have any velocities to work

Again, that equation would apply if the fragments remained together, but they don't!