Is it because a neutron decays into a proton and electron rather than a positron. Which type of nucleus emits positron and which emits electrons . Is it something to do with beta plus and beta minus decay .
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Beta decay occurs, approximately, in nuclei where the Fermi energy of one species of nucleon is higher than the first unoccupied orbital for the other species. In these nuclei energy can be liberated by turning one type of nucleon into another --- the new nucleon moves into the available, lower-energy orbital.
By definition, a nucleus with neutron excess has "too many" neutrons already, and it would cost energy to turn a proton into another neutron by positron emission. Such transitions don't happen spontaneously due to conservation of energy.
I have a nice illustration to complement this word salad but I will have to upload it later --- sorry for now.
There are a small number of nuclei, mostly where the number of protons and the number of neutrons are both odd numbers, where the nucleus can release energy by either electron or positron decay. Potassium-40 is one example: most decays are electron decays to calcium-40, but a small fraction are positron decays to argon-40. This effect some some others in nuclear structure suggest the protons like to pair up with protons, and neutrons like to pair up with neutrons.
_$p$_is converted into a_$n$_inside the nucleus" -- (That's by plain charge conservation; a gimmee.) "but in a neutron rich nucleus adding a_$n$_takes more energy than you get from removing a_$p$_so [...]" -- So the presumed logic holds. But why does converting $p$ to $n$ in an already $n$-rich nucleous take more energy than the reverse?? (Why "valley of stability" rather than "ridge of instability"??) And: Does the OP ask this question? $\endgroup$ – user12262 Apr 17 '16 at 6:52