JEE Advanced · Physics · 28. Nuclear Physics
The isotope having a mass 12.014 u undergoes - decay to has an excited state of the nucleus at 4.041 MeV above its ground state. If decays to the maximum kinetic energy of the - particle in units of MeV is ( where c is the speed of light in vacuum).
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
Mass defect =(12.014 - 12) u
Released energy = 13.041 MeV
Energy used for excitation of
Energy converted to KE of electron
The β particle will have maximum K.E when antineutrino will have minimum K.E i.e 0.
Therefore maximum K.E of the β particle is 9 MeV
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