Engineering
Physics
Nuclear Fission and Fusion New

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.  np+e+v¯e , around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino  (v¯e) 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.

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Linked Question 1

What is the maximum energy of the anti-neutrino?

Zero

Nearly 0.8 × 106 eV

Much larger than 0.8 × 106 eV

Much less than 0.8 × 106 eV

Solution

Q value of reaction slightly decreases because mass of the product increases. So the Q value of the reaction is less than 0.8 × 106 eV.

The energy is now shared between the anti-neutrino and the electron. For the maximum of electron, the energy of anti-neutrino is zero and for minimum case, energy of electron is zero.

Linked Question 2

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 × 106 eV

0 K < 0.8 × 106 eV

3.0 eV K 0.8 × 106 eV

3.0 eV K < 0.8 × 106 eV

Solution

Q value of reaction slightly decreases because mass of the product increases. So the Q value of the reaction is less than 0.8 × 106 eV.

The energy is now shared between the anti-neutrino and the electron. For the maximum of electron, the energy of anti-neutrino is zero and for minimum case, energy of electron is zero.