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 `""_20^41"Ca"` and `""_13^27 "Al"` from the following data:

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

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

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

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

#### Solution

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

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

A neutron `(""_0"n"^1)` is removed from a `""_20^41 "Ca"` nucleus. The corresponding nuclear reaction can be written as:

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

It is given that:

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

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

Mass m(`""_0"n"^1`) = 1.008665 u

The mass defect of this reaction is given as:

Δ m = `"m"(""_20^40"Ca") + (""_0^1"n") - "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" = triangle"mc"^2`

= 0.008978 xx 931.5 = 8.363007 MeV

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

\[\ce{^27_13Al -> ^26_13Al + ^1_0n}\]

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:

`triangle"m" = "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 = triangle"mc"^2`

= 0.014019 x 931.5 = 13.059 MeV