1.

94Pu239 is undergoing α− decay according to the equation94Pu239→ 92U235+ 2He4The energy released in the process is mostly kinetic energy of the α− particle. However, a part of the energy is released as γ− rays. What is the speed of the emitted α− particle if the γ−rays radiated out have energy of 0.90 MeV?Given:Mass of 94Pu239=239.05122 uMass of 92U235=235.04299 u Mass of2He4=4.002602 u(1uc2=931 MeV)

Answer» 94Pu239 is undergoing α decay according to the equation



94Pu239 92U235+ 2He4



The energy released in the process is mostly kinetic energy of the α particle. However, a part of the energy is released as γ rays. What is the speed of the emitted α particle if the γrays radiated out have energy of 0.90 MeV?



Given:

Mass of 94Pu239=239.05122 u

Mass of 92U235=235.04299 u

Mass of2He4=4.002602 u

(1uc2=931 MeV)


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