Question

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?

Answer 1

Step-by-step explanation:

From the given information, we are to compare the minimum energies required to remove a neutron from
`^(41)_(20)Ca`,
`^(42)_(20)Ca` and
`^(43)_(20)Ca`

To start with
`^(41)_(20)Ca`; the minimum energy required to remove a neutron from
`^(41)_(20)Ca` is :

=
`E ( ^(41)_(20) Ca) - E ( ^(40)_(20) Ca) - E ( ^(1)_(0) n)`

= (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
`^(42)_(20)Ca` is :

=
`E ( ^(42)_(20) Ca) - E ( ^(41)_(20) Ca) - E ( ^(1)_(0) n)`

= ( 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
`^(43)_(20)Ca` is :

=
`E ( ^(43)_(20) Ca) - E ( ^(42)_(20) Ca) - E ( ^(1)_(0) n)`

= (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
`^(42)_(20)Ca` ?

This is as a result of electronic configuration of
`^(42)_(20)Ca` which posses the same number of proton and neutron, as such,
`^(42)_(20)Ca` tends to acquire more stability. For this reason, it will be difficult to remove a neutron from
`^(42)_(20)Ca` .