The beta spectrum for ¹²B decay is shown in the graph. The kinetic energy of an emitted $\beta$ particle is 6.00 MeV. The energy of the associated neutrino is approximately A) 6.00 MeV B) 13.4 MeV C) 7.40 MeV D) 11.3 MeV E) 2.54 MeV

Question

Quizplus · Accepted Answer

According to the law of conservation of energy in Beta decay, the sum of kinetic energy (KE) of the emitted particle and the energy of the associated neutrino (Eν) is equal to the maximum energy released in the decay (Q). 
Q = KE + Eν

From the graph, we see that the maximum energy released in the decay (Q) is approximately 13.4 MeV. We also know that the KE of the emitted particle (F1F1F1S1F1F1F10) is 6.00 MeV. Substituting these values in the above equation, we get:

13.4 MeV = 6.00 MeV + Eν 
Eν = 13.4 MeV - 6.00 MeV 
Eν ≈ 7.40 MeV

Therefore, the energy of the associated neutrino is approximately 7.40 MeV. The correct choice is (C).

The Beta Spectrum for 12B Decay Is Shown in the Graph

The Beta Spectrum for ¹²B Decay Is Shown in the Graph