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 observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)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, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino has a mass of `3eV//c^(2)` (where `c` is the speed of light) instead of zero mass, what should be the range of the kinetic energy, `K` of the electron?A. `0 leK le0.8xx10^(6)eV`B. `3.0 eV le K le 0.8xx10^(6) eV`C. `3.0 eV le K lt 0.8xx10^(6)eV`D. `0 le K lt 0.8xx10^(6) eV`

The `beta`-decay process, discovered around `1900`, is basically the d

1.	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 observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)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, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino has a mass of `3eV//c^(2)` (where `c` is the speed of light) instead of zero mass, what should be the range of the kinetic energy, `K` of the electron?A. `0 leK le0.8xx10^(6)eV`B. `3.0 eV le K le 0.8xx10^(6) eV`C. `3.0 eV le K lt 0.8xx10^(6)eV`D. `0 le K lt 0.8xx10^(6) eV`
Answer» Correct Answer - D `0 le KE_(beta^(-))leQ-KE_(P)-KE_(overset(-)v)` `0 le KE_(beta^(-)) ltQ`

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