Compare the minimum energies required to remove a neutron from Ca, zzCa and 23Ca, using m40=37225.15 MeV/c2, m41=38156.36 MeV/c?, m42=39084.46 MeV/c?, m43=40016.10 MeV/c and mn=939.57 MeV/c2. Why there is a distinct increase of this energy for Ca?

Explanation:

From the given information, we are to compare the minimum energies required to remove a neutron from  ,    and  

To start with   ; the minimum  energy required to remove a neutron from is :

=

= (38156.36 - 37225.15 -939.57) MeV

= -8.36 MeV   Since energy is being given out

Thus, E = 8.36 MeV

the minimum  energy required to remove a neutron from is :

=

= ( 39084.46 - 38156.36 - 939.57) MeV

= -11.47 MeV     Since energy is being given out

Thus, E = 11.47 MeV  

the minimum  energy required to remove a neutron from is :

=

= (40016.10 -39084.46 -939.57) MeV

=  - 7.93 MeV     Since energy is being given out

Thus, E =7.93  MeV

Why there is a distinct increase of this energy for ?

This is as a result of electronic configuration of which posses the same number of proton and neutron, as such, tends to acquire more stability. For this reason, it will be difficult to remove a neutron from .

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

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the answer would be 32*f