Determine the mass defect, binding energy and specific energy of the 14N7 nitrogen nucleus

The binding energy of the nucleus is equal to the minimum energy that must be applied for the complete splitting of the nucleus into particles.
E = Δm * c², where Δm is the difference between the mass of the nucleus and its components (mass defect).
Δm = Z * mp + N * mn-Mя, where Z is the number of protons in the nucleus, mp is the proton mass mp = 1.00728 amu, N is the number of neutrons in the nucleus, mn is the neutron mass mn = 1 , 00866, My is the mass of the nucleus.
m poison = m at – Z * me
m poison = 14.0031 – 7 * 0.000549 = 13.999257 a. eat.
Substituting the numbers, we find the mass defect:
Δm = Z * mp + N * mn-My = 7 * 1.00728 + 7 * 1.00866-13.999257 = 0.112323 amu
With this in mind, the binding energy:
E = Δm * c² = (Z * mp + N * mn-My) * c².
The nucleus of the nitrogen isotope contains 7 protons, 7 neutrons, My = 14.00307 amu.
Δm = Z * mp + N * mn-My = 7 * 1.00728 + 7 * 1.00866-14.00307 = 0.00855 amu
Let’s translate into kilograms:
Δm = 0.00855 * 1.6606 * 10 ^ -27 = 0.18 * 10 ^ -27 kg.
Communication energy:
E = Δm * c² = 0.18 * 10 ^ -27 * (3 * 10 ^ 8) ² = 1.678 * 10 ^ -11 J.
Specific binding energy is the binding energy per 1 nucleon, we have 3 of them:
Esp = E / A = 1.678 * 10 ^ -11 / 7 = 0.239 * 10 ^ -11 J.
Answer: Δm = 0.18 * 10 ^ -27 kg, E = 1.678 * 10 ^ -11 J., Esp = 0.239 * 10 ^ -11 J.