The positron decay is given by
$p \to n + e^+ + \nu$
'Since the mass of the proton is less than a neutron, the process can only take place inside a nucleus.'
Can someone explain why this is so? And why exactly is a neutrino released in this case? While an antinutrino is released in beta decay
$n \to p + e^- + \bar \nu$
Note :if possible please explain the first part qualitatively(for my understanding).
'Since the mass of the proton is less than a neutron, the process can only take place inside a nucleus.'
Can someone explain why this is so? And why exactly is a neutrino released in this case? While an antineutrino is released in beta decay
This process cannot happen with a free proton. Energy has to be supplied for the reaction. This is possible if a lower energy nuclear state is be available, once the positron and e_neutrino are released.
A positron has to be emitted because of charge conservation for the proton to turn into a neutron. There is also lepton number conservation, which is balanced by the electron neutrino to zero, as the proton has zero lepton number.
The same is true for the neutron decay, the anti_electron_neutrino balancing the lepton number for the electron. It is the basic reason neutrino decay is a three body decay.
The reason Beta decay can only happen in the nucleus is because in the nucleus energy can be borrowed and taken between the nucleons(protons and neutrons). This borrowed energy can then be used to create the intermediate W+ boson and then this W+ boson decays to the positron and electron neutrino. The neutrino is created to conserve lepton number. An antineutrino is emitted in beta decay to negate the lepton number of the electron.
