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).

Step-by-step solution

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