1)Neutron emission is not a possible mode of decay for K(40,19).Why?
I think it is because the neutron to proton ratio for potassium(40,19) is almost unity(more precisely N/P=1.1). So, neutron emission is not a mode of decay to gain stability. Is it right?
At wiki (potassium/isotopes) I found:
"Three isotopes occur naturally: 39K (93.3%), 40K (0.012%) and 41K (6.7%). Naturally occurring 40K decays to stable 40Ar (11.2%) by electron capture and by positron emission, and decays to stable 40Ca (88.8%) by beta decay; 40K has a half-life of 1.250×109 years."
Note that 93.3% of potassium is the stable isotope with one less neutron, so there are even less neutrons to overcome the mutual repulsion between the protons. So I would say your guess is completely wrong.
I guess, but know little of this field, that the two decay modes are aided by the electrostatics. I.e. an external electron is attracted by the protons to form argon, 40Ar, (converting a proton into into a neutron and thus improving the binding together of the nucleus) or 40K creating a positron (from a proton converting* it to a neutron) and immediately expelling the positron (as the short range nuclear forces only act between barrions) So there is nothing to keep the positive charge of the nucleus from tossing it out.
If 40K emits an electron, (converts a neutron into a proton) it forms 40Ca. I am too lazy to look up and not sure from memory, but think the neutron converted was more massive than the proton. If true, then the energy needed for the beta to escape the attraction of the now 20 protons comes from the mass difference. If the proton is the more massive, I do not understand why forming Ca is the more probably.
To throw a neutron out of the nucleus is very hard. It is attracted by strong short range nuclear forces. How can it escape? If you want to throw out barions from the nucleus, you need to throw out some protons also in a bound unit so the mutual repulsion between those thrown out and those remaining can supply the departure energy. For example toss out an alpha particle.
In payment for my answer (from above wiki and by thinking only) please post the which is heaver; proton or neutron.
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*Although this will not significantly change the nuclear binding, it will reduce the mutual repulsion between the 18 protons that remain in the 40Ar, which has the same net effect on the nuclear stability.
Later by edit:
MT's "fits very nicely with a graph made to represent spheres of a range of sizes. using surface area as proton charge, and volume as mass. in such a graph... we see that as spheres get bigger... the volume always grows faster than the surface area." is NONSENSE. -The reason why the stable ratio of N/P increase as you move up thru the periodic table is due to fact that the Coulumb force extends far but the strong force (barion/barion attraction) is of such short range that in the larger nuclei, it can not even "reach" across the nucleus.
