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Nuclei

Nuclei different mass unit is used for expressing atomic masses. This unit is the atomic mass ... One also uses the term nucleon for a proton or a neutron. Thus the ... elements, the sizes of nuclei of various elements have been accurately measured. ... apart a nucleus in this way, the nuclear binding energy is still a convenient.

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Weak Interaction

3 .6 The V - A Theory and the Coupling Constants in Neutron Decay ... Let the mass of the particle be m, its momentum p and its energy be E. We should expect the de Broglie wave of the particle to b e ... expectation value of a given measurement is defined as the most probable ... Mathematical Physics, C .U .P ., 1956, pp .

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I found an answer from www.britannica.com

https://www.britannica.com/topic/argumentum-ad-misericordiam ...

A page from a first-grade workbook typical of “new math” might state: “Draw ... https://www.britannica.com/science/arithmetic-mean 2021-03-27 monthly 1.0 ... composition https://cdn.britannica.com/62/183862-050-06E2F117/World-Data- ... one electron (dark green) but differ in the number of neutrons (gray) in the nucleus.

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I found an answer from en.wikipedia.org

Neutron - Wikipedia

The neutron is a subatomic particle, symbol n or n ⁰ , which has a neutral (not positive or negative) charge, and a mass slightly greater than that of a proton. Protons and neutrons constitute the nuclei of atoms. Since protons and neutrons behave similarly within the nucleus, and each ... Neutrons are required for the stability of nuclei, with the exception of the ...

For more information, see Neutron - Wikipedia

Separation energy formula E = \Delta m c^2 where, \Delta m mass defect and c - speed of light

Step 1: Determine the energy required to separate the neutron from the nucleus ^{41}_{20} Ca

Nuclear reaction for Calcium

^{41}_{20} Ca \rightarrow ^{40}_{20} Ca + ^{1}_{0} n

Mass defect \Delta m = \text{ mass of } ^{40}_{20} Ca + \text{ mass of } ^{1}_{0} n - \text{ mass of } ^{40}_{20} Ca   

Given that

m (^{40}_{20} Ca) = 39.962591 u

m (^{41}_{20} Ca) = 40.962278 u

m (^{1}_{0} n) = 1.008665   

\Delta m = 39.962591 + 1.008665 - 40.962278

\Delta m = 0.008978 u

Separation energy E = \Delta m c^2 = 0.008978 c^2

E = 0.008978 c^2 * 931.5 MeV/c^2

E = 8.363007 MeV

Hence, energy required to separate the neutron E = 8.363007 MeV

Step 2: Determine the energy required to separate the neutron from the nucleus ^{27}_{13} Al

Nuclear reaction for Aluminium

^{27}_{13} Al \rightarrow ^{26}_{13} AI + ^{1}_{0} n

Mass defect \Delta m = \text{ mass of } ^{26}_{13} Al + \text{ mass of } ^{1}_{0} n - \text{ mass of } ^{27}_{13} Al

Given masses

m (^{26}_{13} Al) = 25.986895 u

m (^{27}_{13} Al) = 26.981541 u

\Delta m = 25.986895 + 1.008665 - 26.981541

\Delta m = 0.0140190

Separation energy E = \Delta m c^2 = 0.0140190 c^2

E = 0.0140190 c^2 * 931.5 MeV/c^2

E = 13.059 MeV  

Thus, energy required to separate the neutron E = 13.059 MeV