Question

neutron breaks into proton and electron. the energy released during this process.

Answer 1

Neutron breaks into proton and electron. The energy released during this process.

Free neutrons decay by emission of an electron and an electron antineutrino to become a proton, a process known as beta decay:

$$n \rightarrow p + e + \bar {v _ {e}}$$ $$n - \text{neutron}$$ $$p - \text{proton}$$ $$e - \text{electron}$$ $$\bar {v _ {e}} - \text{electron antineutrino}$$

The law conservation of energy for this process:

$$E _ {in} = E _ {fin}$$ $$E _ {in} - \text{energy in initial state}$$ $$E _ {fin} - \text{energy in final state}$$ $$E _ {in} = m _ {n} c ^ {2}$$ $$E _ {fin} = m _ {p} c ^ {2} + m _ {e} c ^ {2} + m _ {v} c ^ {2} + Q$$ $$m _ {n}, m _ {p}, m _ {e}, m _ {v} - \text{masses of neutron, proton, electron and electron antineutrino}$$ $$m _ {v} \ll m _ {e} \quad \text{so} \quad m _ {v} \approx 0$$ $$m _ {n} c ^ {2} = m _ {p} c ^ {2} + m _ {e} c ^ {2} + m _ {v} c ^ {2} + Q$$

energy released during this process:

$$Q = m _ {n} c ^ {2} - \left(m _ {p} c ^ {2} + m _ {e} c ^ {2}\right) = 0.782343 \mathrm{MeV}$$

Answer: $Q = 0.782343$ MeV