The nuclear binding energy per nucleon of a nucleus is the binding energy divided by the total number of nucleons
^{56}Fe has the highest nuclear binding energy per nucleon and is therefore the most stable isotope (see graph above). The negative gradient of the curve beyond ^{56}Fe implies that electrostatic forces of repulsion increase per nucleon for isotopes with mass number greater than 56.
Question
With reference to the data below and using the concept of nuclear binding energy per nucleon, explain why isotopes with atomic numbers lower than that of ^{12}C have relative masses that are slightly higher than their respective mass numbers, while those with atomic numbers that are higher than ^{12}C have relative masses that are slightly lower than their respective mass numbers.
Atomic no. (Z) 
Mass no. (A)  Symbol 
Relative isotopic mass* 
1 
1  ^{1}H 
1.007825 
2 
^{2}H 
2.014104 

2 
3  ^{3}He 
3.016029 
4 
^{4}He 
4.002603 

3 
6  ^{6}Li 
6.015122 
4 
9  ^{9}Be 
9.012182 
5 
10 
^{10}B 
10.012937 
11 
^{11}B 
11.009305 

6 
12  ^{12}C 
12.000000 
8 
16 
^{16}O 
15.994915 
17  ^{17}O 
16.999132 

18 
^{18}O 
17.999160 

9 
19  ^{19}F 
18.998403 
10 
20 
^{20}Ne 
19.992440 
21  ^{21}Ne 
20.993847 

22 
^{22}Ne 
21.991386 
Answer
Let’s rewrite eq9 of the previous article as:
Dividing the above equation throughout by A and rearranging,
where is the binding energy per nucleon.
Using data in a previous article, m_{proton} + m_{electron} – m_{neutron} = 0.000839869u. Furthermore, with the exception of ^{1}H, . Therefore, the RHS of eq11 approximately equals to m_{neutron}:
For ^{12}C, , and so . If , then . By the same logic, if , then .