Answer:
Explanation:
From the given information, we are to compare the minimum energies required to remove a neutron from [tex]^{41}_{20}Ca[/tex], [tex]^{42}_{20}Ca[/tex] and [tex]^{43}_{20}Ca[/tex]
To start with [tex]^{41}_{20}Ca[/tex]; the minimum energy required to remove a neutron from[tex]^{41}_{20}Ca[/tex] is :
= [tex]E ( ^{41}_{20} Ca) - E ( ^{40}_{20} Ca) - E ( ^{1}_{0} n)[/tex]
= (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 [tex]^{42}_{20}Ca[/tex] is :
= [tex]E ( ^{42}_{20} Ca) - E ( ^{41}_{20} Ca) - E ( ^{1}_{0} n)[/tex]
= ( 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 [tex]^{43}_{20}Ca[/tex] is :
= [tex]E ( ^{43}_{20} Ca) - E ( ^{42}_{20} Ca) - E ( ^{1}_{0} n)[/tex]
= (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 [tex]^{42}_{20}Ca[/tex] ?
This is as a result of electronic configuration of [tex]^{42}_{20}Ca[/tex] which posses the same number of proton and neutron, as such, [tex]^{42}_{20}Ca[/tex] tends to acquire more stability. For this reason, it will be difficult to remove a neutron from [tex]^{42}_{20}Ca[/tex] .
