Joe's Uranium Problem: Alpha Particle Emission & Final Nucleus Mass

[josephcollins]

Here I have a problem:

A 232/92 Uranium nucleus emits an alpha particle with kinetic energy=5.32MeV. What is the final nucleus and what is the approximate mass(in units) of the final atom.

I can write the equation for the reaction, this will give 4/2 He and 228/90 Th. If the alpha particle has KE=5.32MeV then is this the energy emitted in the reaction? If this is then I can obtain the change in mass. Is this correct? Any help is welcomed, thx. Joe

 

[Doc Al]

https://www.physicsforums.com/showthread.php?t=71746

Please do not double (or triple) post!

 

[sir-pinski]

Hi Joe,

Yes, your approach is correct. The equation for this reaction is:

232/92 U -> 4/2 He + 228/90 Th + energy

The alpha particle has a kinetic energy of 5.32 MeV, which is the energy emitted in the reaction. This energy is equal to the difference in mass between the initial and final nuclei, as given by Einstein's famous equation E=mc^2.

To find the final nucleus, simply subtract the mass numbers and atomic numbers of the alpha particle and thorium nucleus from the uranium nucleus. This gives us the final nucleus of 228/90 Th.

As for the approximate mass of the final atom, we can use the atomic mass unit (u) as the unit of measurement. The mass of a 4/2 He atom is approximately 4.0015 u and the mass of a 228/90 Th atom is approximately 228.028 u. Therefore, the approximate mass of the final atom is 232.0295 u.

I hope this helps clarify the problem for you. Good luck!