JEE (Advanced) 2012 Paper-2 Question 9

Instructions

The $$\beta$$- decay process, discovered around 1900, is basically the decay of a neutron (n), In the laboratory, a proton (p) and an electron $$(e^{-})$$ are observed as the decay products of the neutron. therefore, considering the decay of a neutron as a tro-body dcay process, it was predicted theoretically that thekinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $$n \rightarrow p + e^{-} + \overline{v_e}$$, around 1930, pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $$(\overline{v_e})$$ to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, themaximum kinetic energy of the lectron is $$0.8 \times 10^6$$ eV. The kinetic energy carried by the proton is only the recoil energy.

Question 9

What is the maximum energy of the anti-neutrino ?



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