JEE Advanced 2016 Paper 1

Section: Physics

Question : 16 of 54

Marks: +1, -0

The isotope 512B having a mass 12.014 u undergoes β-decay to612C,612C has an excited state of the nucleus (612C*) at 4.041MeV above ground state.If 512B decays to (612C*) ,the maximum kinectic energy of the β-particles in units of MeV is (1u=931.5MeV/c2,where c is the speed of the light in vacuum ),

[JEE Adv 2016 P1]

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