JEE Advanced 2016 Paper 1 Physics Q15

Physics · Nuclei & Radioactivity

Problem

The isotope 512B{}^{12}_{5}\mathrm{B} having a mass 12.014 u undergoes β\beta-decay to 612C{}^{12}_{6}\mathrm{C}. 612C{}^{12}_{6}\mathrm{C} has an excited state of the nucleus (612C∗{ }^{12}_{6}\mathrm{C}^{*}) at 4.041 MeV above its ground state. If 512B{}^{12}_{5}\mathrm{B} decays to 612C∗{}^{12}_{6}\mathrm{C}^{*}, the maximum kinetic energy of the β\beta-particle in units of MeV is (1 u=931.5 MeV/c21\,\mathrm{u}=931.5\,\mathrm{MeV}/c^2, where cc is the speed of light in vacuum).

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