#### Question

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei `""_20^41Ca` and `""_13^27 Al` from the following data:

`m(""_20^40Ca) = 39.962591 u`

`m(""_20^41Ca) = 40.962278 u

`m(""_13^26Al) = 25.986895 u`

`m(""_13^27Al) = 26.981541 u`

`

#### Solution

For `""_20^41 Ca:` Separation energy = 8.363007 MeV

For `""_13^27Al:` Separation energy = 13.059 MeV

A neutron `(""_0n^1)` is removed from a `""_20^41 Ca` nucleus. The corresponding nuclear reaction can be written as:

`""_20^41Ca -> ""_20^40Ca + _0^1n`

It is given that:

Mass `m(""_20^40 Ca)`= 39.962591 u

Mass `m(""_20^41 Ca)` = 40.962278 u

Mass m(`""_0n^1`) = 1.008665 u

The mass defect of this reaction is given as:

Δ*m* = `m(""_20^40Ca) + (""_0^1n) - m(""_20^41 Ca)`

`= 39.962591 + 1.008665 - 40.962278 = 0.008978 u`

But `1 u = 931.5 "MeV/c^2"`

∴Δ*m* = 0.008978 × 931.5 MeV/*c*^{2}

Hence, the energy required for neutron removal is calculated as:

`E = trianglemc^2`

= 0.008978 xx 931.5 = 8.363007 MeV

For `""_13^27 Al` the neutron removal reaction can be written as:

`""_13^27 Al -> _13^26 Al + _0^1 n`

it us given that.

Mass `m(""_13^27 Al)` = 26.981541 u

Mass `m(""_13^26 Al)` = 25.986895 u

The mass defect of this reaction is given as:

`trianglem = m(""13^26 Al) + m(""_0^1 n) - m(""_13^27 Al)`

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

`= 0.014019 xx 931.5 "MeV/c"^2`

Hence, the energy required for neutron removal is calculated as:

`E = trianglemc^2`

= 0.014019 x 931.5 = 13.059 MeV