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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Physics
Chapter
Nuclear Physics

JEE Advanced 2016 · Paper 1 · Physics · Nuclear Physics

The isotope B 512 having a mass 12.014 u undergoes β - decay to C ....

The isotope B512B 512 having a mass 12.014 u undergoes β\beta - decay to C.612C612C . 612 C 612 has an excited state of the nucleus (C∗612)\left({C}^{*} 612\right) at 4.041 MeV above its ground state. If B512B 512 decays to C∗,612{C}^{*}, 612 the maximum kinetic energy of the β\beta - particle in units of MeV is ( 1μ=931.5MeVc2,1\mu = 931.5Me\dfrac{V}{{c}^{2}}, where c is the speed of light in vacuum).

Correct answer

9

Step-by-step solution

B→512C612+e−10+v−B \to 512 C 612 + e - 10 + \overset{-}{v}
Mass defect =(12.014 - 12) u
∴∴ Released energy = 13.041 MeV
Energy used for excitation of C612=4.041MeVC 612 = 4.041MeV
∴∴ Energy converted to KE of electron
=13.041−4.041=9MeV= 13.041 - 4.041 = 9MeV 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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