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The isotope boron-12 (atomic mass 12.014 u) undergoes beta-minus decay to carbon-12. Carbon-12 has a nuclear excited state (C-12*) at 4.041 MeV above its ground state. If boron-12 decays directly to this excited state C-12*, what is the maximum kinetic energy of the emitted beta particle in MeV? (Use 1 u = 931.5 MeV/c².)

1. 9.000 MeV
2. 4.041 MeV
3. 13.041 MeV
4. 0.541 MeV

Correct answer: 9.000 MeV

Solution

Q-value = (12.014 - 12.000) * 931.5 = 13.041 MeV. Decaying to C-12* at 4.041 MeV above ground leaves KE_max = 13.041 - 4.041 = 9.000 MeV for the beta particle (maximum, when neutrino gets zero KE).

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