The neutron separation energy is defined to be the energy required to remove a neutron form nucleus. Obtain the neutron separation energy of the nuclei `._(20)Ca^(41)` and `._(13)Al^(27)` from the following data : `m(._20Ca^(40))=39.962591u` and `m(._(20)Ca^(41))=40.962278u`
`m(._(13)Al^(26))=25.986895u` and `m(._(13)Al^(27))=26.981541u`

Explanation

To find the neutron separation energy (NSE) for the nuclei \( _{20}^{41}\text{Ca} \) and \( _{13}^{27}\text{Al} \), we will follow these steps:

### Step 1: Calculate the mass defect for \( _{20}^{41}\text{Ca} \)

The mass defect (\( \Delta m \)) is calculated using the formula:
\[ \Delta m = m(\text{Ca}^{41}) - \left( m(\text{Ca}^{40}) + m(\text{neutron}) \right) \]
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