Answer:
A free neutron has rest mass energy greater than that of a proton. Thus\[\beta \] -decay \[_{0}^{1}n\to _{1}^{1}p+_{-1}^{0}e+\overset{-}{\mathop{v}}\,\] is energetically allowed, but the (3-decay of a free proton into a neutron : \[_{1}^{1}p\to _{0}^{1}n+_{1}^{0}e+v\] is not allowed energetically. Inside a nucleus, individual neutrons and protons are not free. Thus the \[{{\beta }^{+}}\] - decay of a proton is possible when the proton is bound in a nucleus. The energy needed for the decay can come from the appropriate difference in binding energies of a proton and a neutron in the nucleus.
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