# Why positron emission is unlikely to occur for nuclei with an excess of neutrons?

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 .

• Possible duplicate? physics.stackexchange.com/questions/249036/… Apr 16, 2016 at 6:59
• @Farcher: "Possible duplicate? Why beta+- decay occurs?" -- Not really: at least the OP question title implies an inquiry to justify, roughly, "why there is a line/valley of stability in the isotope chart (rather than, say, a line/ridge of instability)". However, admittedly, this request is not (yet) spelled out in the OP question text. Apr 16, 2016 at 7:57
• Follow the logic. "Positron emission can only occur when a ___ is converted into a ___ inside the nucleus, but in a neutron rich nucleus adding a ___ takes more energy than you get from removing a ___ so the event results in a net energy gain to the nucleus." Apr 16, 2016 at 15:28
• @dmckee: "Follow the logic. [...]" -- Logic allows at least to fill in the blanks you left: "Positron emission can only occur when a _$p$_ 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? Apr 17, 2016 at 6:52