Calculate the binding energy and binding energy per nucleon \[\left( {in{\text{ }}MeV} \right)\] of a nitrogen nucleus ${(^7}{N_{14}})$ from the following data
Mass of proton \[ = 1.00783u\]
Mass of neutron \[ = 1.00867u\]
Mass of nitrogen nucleus \[ = 14.00307u\]

Hint: Nuclear binding energies are usually expressed in terms of \[kJ/mole\] if nuclei or \[MeV/nucleon\]. Calculation of the nuclear binding energy involves the following three steps:
(A) Determining the mass defect between Reactants and Products
(B) Conversion of mass defect into energy
(C) Expressing nuclear binding energy as energy per mole or energy per nucleon.

Formula used:
$E = \Delta M \times 931.5MeV$
To convert to MeV per nucleon, simply divide by the number of nucleons

Complete step by step answer:
So we have all the data to get a mass defect. First will calculate the mass defect then we will find energy It is given that-
\[Mp = 1.00783u\]
\[Mn = 1.00867u\]
Mass of nitrogen nucleus \[ = 14.00307u\]
First we will write a balanced nuclear reaction-
Nuclear reaction : $7p + 7n{ \to ^7}{N_{14}}$
So the mass defect will be equals to $\Delta M = 7Mp + 7Mn - $ M of nucleus
$\therefore \Delta M = 7(1.00783) + 7(1.00867) - 14.00307amu$
$ \Rightarrow \Delta M = 0.11243amu$
Now we will find binding energy so,
Binding energy $E = \Delta M \times 931.5MeV$
$\therefore E = 0.11243 \times 931.5 = 104.73MeV$
Now we will find binding energy per nucleon,
Binding energy per nucleon $ = \dfrac{{104.73}}{{14}}$
Binding energy per nucleon \[ = 7.48MeV\]

So the answers are \[104.73Mev\] and \[7.48{\text{ }}MeV\] respectively.

Additional Information:
Nuclear binding energy is that the minimum energy that might be required to disassemble the nucleus of an atom into its component parts. These component parts are neutrons and protons, which are collectively called nucleons. The binding energy is usually a positive number, as we'd like to spend energy in moving these nucleons, interested in one another by the strong nuclear force, far away from one another . The mass of an atomic nucleus is a smaller amount than the sum of the individual masses of the free constituent protons and neutrons, consistent with Einstein's equation \[E = M{C^2}\]. This 'missing mass' is understood because of the mass deficiency , and represents the energy that was released when the nucleus was formed.

Note:
We should take mass in kg if we are using equation \[E = M{C^2}\] joule.
But in above question it is already in the form of $E = M \times 931.5$MeV, so we can use mass in amu.