Eduprobe
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 0.8×10⁶eV. The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino?
Much less than 0.8×10⁶eV.
Much larger than 0.8×10⁶eV.
Nearly 0.8×10⁶eV.
Zero.
Solution:
Anti-neutrino will have maximum energy when electron will have minimum energy or nearly zero energy. In a three-body decay like n → p + e⁻ + νe, energy and momentum must be conserved. If the electron has minimum kinetic energy (approximately zero), then the maximum kinetic energy is shared between the proton and the anti-neutrino. Since the proton's kinetic energy is relatively small (recoil energy), the anti-neutrino carries away most of the remaining energy. Therefore, the maximum energy of the anti-neutrino will be nearly 0.8 × 10⁶ eV.