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Calculate the energy required to remove...

Calculate the energy required to remove the least tighly bond neutron in `""._(20)Ca^(40)` .Given that
Mass of `""._(20)Ca^(40)=39.962589 am u`
Mass of `""._(20)Ca^(39)=38.970691 am u`
Mass fo neutron `=1.008665`

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Total mass (neutron plus product nucleus) after the removal is expected to be more than the initial nucleus mass: This mass difference when expressed in energy units (MeV) is the binding energy of the least bound neutron. The mass after break-up
`M(._(30)^(90)Ca)=38.970719 u`
`m_(n) =1.08665 u`
Total mass `=39.979384 u`
The mass difference is
`(Delta m)=[M(.^(39)Ca)+m_(n)]-M(.^(40)Ca)`
` =(39.964314 +1.008665)-39.962383 u`
`=0.010596 u=(0.010596)xx(931.5)MeV =9.870 MeV`
So, the binding energy of least bound neutron is `15.64 MeV`.
Similarly, for `._(20)^(39)Ar`,
`Delta m=[(._(20)^(39) Ar)+-M(._(18)^(40)Ar)]`
`=(38.964314 +1. 008665)-39.9623 u`
`=0010596 u=-(0.010596) xxx (931.5) MeV`
`=9.870 MeV`
The inert elements are relatively unreactive beause their outer shells of electrons are full. Large energies are involved in gaining or losing electrons which is therefore unlikely. An analogus behaviour takes place in the nucleus. Experimental evidence does indicate the existence of 'closed nuclear shell when the number of protons or neutrons is `2,8,20,28,50,82or 126,` althought the concepts of individual nucleon 'orbits' and filed shells inside the nucleus is hard to belive. The above mentioned number of protons or neutron are called magic numbers. The elemnets with magic number of protons have unsually high number of stable isotope. For example, aluminiunm with `13` protons has just one stable isotope `.^(27)A1`, but tin with `Z=5` (a magic number) has` 10` stable isotopes ranging from `N=62` to `N=74`, whereas neighboring indium with `Z =49` and antimony with `Z=51` have only two.
Thus, the magic numbers are associated with extra binding energies, implying higher stability. Thus, in the given example it takes about `6 MeV`more energy to remove a paired neutron from a pair that completes a closed nuclear shell (a magic number) than from a pair that dose not complete a shell.
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Calculate the energy required to remove the least tightly neutron form .^20(Ca^(40)) . Given that Mass of .^20(Ca^(40)) = 39.962589 amu Mass of .^20(Ca^(39)) = 38.970691 amu Mass of neutron = 1.008665 amu

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