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Question:

The β-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 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 a continuous spectrum. Considering a three-body decay process, i.e., n→p+e⁻+νe, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (νe) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 10⁶ eV. The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino had a mass of 3 eV/c² (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

0≤K≤0.8 × 10⁶eV

3.0eV≤K≤0.8 × 10⁶cV

0≤K<0.8 × 10⁶eV

3.0eV≤K<0.8 × 10⁶eV

Solution:

K can't be equal to 0.8 × 10⁶ eV as anti-neutrino must have some energy. If the anti-neutrino has a mass of 3 eV/c², it will carry away some energy. Therefore, the maximum kinetic energy of the electron will be less than 0.8 × 10⁶ eV. The minimum kinetic energy of the electron will be greater than or equal to 0 eV. The minimum kinetic energy will be greater than 0 because some energy will be carried by the anti-neutrino due to its mass. Considering the anti-neutrino's rest mass energy, the range of the kinetic energy of the electron will be 3.0 eV ≤ K < 0.8 × 10⁶ eV. Therefore, the correct option is 3.0eV≤K<0.8 × 10⁶eV