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- Homework Statement
- The polonium isotope 211 Po is radioactive and undergoes alpha decay. In the decay process, a 211 Po nucleus at rest explodes into an alpha particle (a 4 He nucleus) and a 207 Pb lead nucleus. The lead nucleus is found to have 0.12 MeV of kinetic energy. The energy released in a nuclear decay is the total kinetic energy of all the decay products.
How much energy is released, in MeV, in a 211 Po decay?
- Relevant Equations
- Ei = Ef
K total = K alpha + K Pb
Since we are looking for K total, I summed the given kinetic energy for lead and the typical kinetic energy of an alpha particle: 0.12 + 5 = 5.12 MeV. My answer is definitely wrong, but I don't how I should approach the problem.