0
$\begingroup$

Any nucleus will stable if it has neutron number= proton number. Or Even mass number=even atomic number or proton number.(maximum stability ) Or neutron number /proton number less than 1.6(not sure). If i apply first two conditions then boron 11 and 12 then For boron 11 . . . . . .
Proton number=5; . Neutron number =6; For boron 12 Proton number =5;neutron number =7; From that boron 11 should be unstable as for odd odd nucleus.

$\endgroup$

closed as unclear what you're asking by Ben Crowell, GiorgioP, John Rennie, Kyle Kanos, ZeroTheHero Apr 21 at 0:14

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What A=11 nucleus would you prefer to be stable, and why? $\endgroup$ – Jon Custer Apr 18 at 20:25
  • 2
    $\begingroup$ The writing is impossible to decode. There is strange punctuation and capitalization, and sentences without verbs. Any nucleus will stable if it has neutron number= proton number. This is not true. Odd-odd nuclei are almost never stable, and heavy nuclei with N=Z are unstable. $\endgroup$ – Ben Crowell Apr 18 at 21:46
  • $\begingroup$ Here i never said that odd odd nuclei is stable. please read it carefully $\endgroup$ – Sneha Banerjee Apr 19 at 2:56
1
$\begingroup$

Nuclear physics is more complicated than a mixture of statements just on the number of protons and neutrons, otherwise there would be no need for nuclear models .

Light nuclei lie close to the N=Z line, but the stable isotopes of heavier nuclei require an excess of neutrons to be stable. The neutrons contribute to the attractive strong interaction and moderate the strong electrostatic repulsion of the protons.

Modeling boron with a nuclear model would show that the (5,6) isotope (together with the (5,5) is stable, i.e. the orbitals of the neutrons are within the binding of the strong force, whereas an extra neutron has a quantum mechanical probability of decay through the weak force.

Please keep in mind that nuclear physics is more complicated than numerical countings. See this for example.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.