question_answer

(a) The observed decay products of a free neutron are a proton and an electron. The emitted electrons are found to have a continuous distribution of kinetic energy with a maximum of \[({{m}_{n}}-{{m}_{p}}-{{m}_{e}}){{c}^{2}}\]. Explain clearly why the presence of a continuous distribution of energy is a pointer to the existence of other unobserved product(s) in the decay. (b) If a neutron is unstable with a half life of about 1000 s, why don't all the neutrons of a nucleus decay eventually into protons? How can a nucleus of \[\mathbf{Z}\] protons and (A-Z) neutrons ever remain stable if the neutrons themselves are unstable?

Answer:

                (a) Consider the decay of a free neutron at rest. \[_{0}^{1}n\to _{1}^{1}p+_{-1}^{0}e\] By momentum conservation, if the momentum of electron is \[{{p}_{e}},\] then the momentum of proton is - \[{{p}_{e}}\] .By energy conservation, \[{{m}_{n}}{{c}^{2}}={{[p_{e}^{2}{{c}^{2}}+m_{e}^{2}{{c}^{4}}]}^{1/2}}+{{[p_{e}^{2}{{c}^{2}}+m_{p}^{2}{{c}^{4}}]}^{1/2}}\] Thus a definite momentum \[{{p}_{e}}\] is given to the electron. This means the energy of the electron in the above decay is fixed. An electron in the above decay cannot have a continuous distribution of energies. The presence of an additional particle, however, allows this possibility. The available energy can now be shared by the electron and the third particle, and the electron's energy is no longer fixed. This led Pauli to postulate the existence of a new particle till then unobserved. We now know that the correct equation for \[\beta \]-decay is: \[_{0}^{1}n\to _{1}^{1}p+_{-1}^{0}e+\overset{-}{\mathop{v}}\,\]  where the particle denoted by \[\overset{-}{\mathop{v}}\,\] is called antineutrino. It is a neutral particle of negligibly small rest mass and intrinsic spin 1/2. (b) A free neutron has rest mass greater than that of a proton. Thus \[{{\beta }^{-}}\]- decay is energetically allowed, but the \[{{\beta }^{+}}\]- decay of a proton into a neutron is not allowed. In a nucleus, individual neutrons and protons are not free. Thus the \[{{\beta }^{+}}\] -decay of a proton \[(p\to n+{{e}^{+}}+{{v}_{e}})\] is possible when the proton is bound in a nucleus. The energy needed for the decay comes from the difference in binding energies of a proton and a neutron in the nucleus. In a stable nucleus with \[\text{Z}\] protons and \[\left( \text{A }-\text{ Z} \right)\] neutrons, the two reciprocal processes (neutron decay and proton decay) are in dynamic equilibrium.