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

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The decay of a 232 Th nucleus to a 228 Ra nucleus involves the emission of an alpha particle, resulting in a total kinetic energy of 6.54 x 10^-3 J. To calculate the kinetic energy of the recoiling 228 Ra nucleus and the alpha particle, one can use conservation of momentum and the given mass ratios. The alpha particle constitutes 1.76% of the mass of the 228 Ra nucleus, which is approximately 4.0 units. While velocities are not directly provided, they can be inferred through the conservation of momentum principle. Understanding the relationship between the velocities of the decay fragments is crucial for determining their respective kinetic energies.
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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 ?
 
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Hint: The speeds of the two fragments can be related using conservation of momentum.
 
MAPgirl23 said:
** 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!
 

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