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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^(bar)` are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. ` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is `0.8 xx 10^(6) eV` The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino ?A. The nucleus `._3^6Li` can emit an alpha particleB. The nucleus `._(84)^(120) P0` cam emit a proton.C. Deuteron and alpha particle can undergo complete fusion.D. The nuclei `._(30)^(70) Zn` and `._(34)^(82) Se` can undergo complete fusion.

Answer» Correct Answer - C
`KE_max of beta^(bar)`
`Q=0.8 xx 10^(6) eV`
`KE_p +KE_(beta^(bar))+KE_(bar(V))=Q`
`KE_p` is almost zero
When `KE_(beta(bar)) =0`
Then `KE_(bat(barv))=Q-KE~=Q`


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