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JEE Advanced · Physics · 28. Nuclear Physics

The isotope B512 having a mass 12.014 u undergoes β - decay to C . 612C  612 has an excited state of the nucleus C*612 at 4.041 MeV above its ground state. If B  512 decays to C*,612  the maximum kinetic energy of the β - particle in units of MeV is ( 1μ=931.5 MeVc2, where c is the speed of light in vacuum).

  1. A 3
  2. B 6
  3. C 9
  4. D 12
Verified Solution

Answer & Solution

Correct Answer

(C) 9

Step-by-step Solution

Detailed explanation

B512 C612+ e-10+ v-
Mass defect =(12.014 - 12) u
Released energy = 13.041 MeV
Energy used for excitation of C612=4.041 MeV
Energy converted to KE of electron
=13.041-4.041=9 MeV
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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