The isotope B 5 12 having a mass 12.014 u undergoes β - decay to C . 6 12 C 6 12 has an excited state of the nucleus C * 6 12 at 4.041 MeV above its ground state. If B 5 12 decays to C * , 6 12 the ma

Question

The isotope B 5 12 having a mass 12.014 u undergoes β - decay to C . 6 12 C 6 12 has an excited state of the nucleus C * 6 12 at 4.041 MeV above its ground state. If B 5 12 decays to C * , 6 12 the maximum kinetic energy of the β - particle in units of MeV is ( 1 μ = 931.5 M e V c 2 , where c is the speed of light in vacuum).

TestHub · Accepted Answer

B → 5 12 C 6 12 + e - 1 0 + v - Mass defect =(12.014 - 12) u ∴ Released energy = 13.041 MeV Energy used for excitation of C 6 12 = 4.041 M e V ∴ Energy converted to KE of electron = 13.041 - 4.041 = 9 M e V 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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