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Atoms and Nuclei question

2016 · Shift 1 · Q52
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Atoms and Nuclei question

2016 · Shift 1 · Q52

JEE AdvancedPhysicsAtoms and NucleiNumerical+3 / −1
The isotope 512B_5^{12}B512​B having a mass 12.014 u undergoes β\betaβ-decay to 612C_6^{12}C612​C. 612C_6^{12}C612​C has an excited state of the nucleus (612C_6^{12}C612​C*) at 4.041 MeV above its ground state. If 512B_5^{12}B512​B decays to 612C_6^{12}C612​C*, the maximum kinetic energy of the β\betaβ-particle in units of MeV is (1u = 931.5 MeV/c2, where c is the speed of light in vacuum).
Numerical answer
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