Nuclear binding energy per nucleon

The nuclear binding energy per nucleon of a nucleus is the total binding energy of the nucleus divided by the total number of nucleons.

In general, the heavier the atom — that is, the more protons and neutrons it contains — the larger the total binding energy required to hold the nucleons together against the electrostatic repulsion between the positively charged protons. However, the quantity that determines nuclear stability is the binding energy per nucleon, not the total binding energy.

The binding energy per nucleon shows an overall increasing trend from very light nuclei to medium-mass nuclei, reaching a maximum near ⁵⁶Fe and ⁶²Ni, before gradually decreasing for heavier nuclei (see diagram above). However, this trend is not perfectly smooth; individual nuclei may deviate because nuclear stability also depends on factors such as proton–neutron pairing. For example, ¹²C has 6 protons and 6 neutrons, so both are even and fully paired, making it particularly stable. In contrast, ¹⁴N has 7 protons and 7 neutrons, leaving unpaired nucleons and slightly reducing its binding energy per nucleon despite being heavier.

Nuclei near the peak of the curve are therefore among the most stable nuclei. The negative gradient of the curve beyond ⁵⁶Fe indicates that, for very heavy nuclei, the increasing electrostatic repulsion between protons outweighs the additional binding contributed by the short-ranged strong nuclear force, causing the binding energy per nucleon to decrease.

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 ¹²C have relative masses that are slightly higher than their respective mass numbers, while those with atomic numbers that are higher than ¹²C have relative masses that are slightly lower than their respective mass numbers.

Atomic no. (Z)	Mass no. (A)	Symbol	Relative isotopic mass
1	1	¹H	1.007825
1	2	²H	2.014104
2	3	³He	3.016029
2	4	⁴He	4.002603
3	6	⁶Li	6.015122
4	9	⁹Be	9.012182
5	10	¹⁰B	10.012937
5	11	¹¹B	11.009305
6	12	¹²C	12.000000
8	16	¹⁶O	15.994915
	17	¹⁷O	16.999132
	18	¹⁸O	17.999160
9	19	¹⁹F	18.998403
10	20	²⁰Ne	19.992440
	21	²¹Ne	20.993847
	22	²²Ne	21.991386

Answer

Let’s rewrite eq9 of the previous article as:

$m_{isotope}+m_{binding}=Z\left ( m_{proton}+m_{electron} \right )+(A-Z)m_{neutron}$

Dividing the above equation throughout by A and rearranging,

$\frac{m_{isotope}}{A}+\frac{m_{binding}}{A}=\frac{Z}{A}\left ( m_{proton}+m_{electron}-m_{neutron}\right )+m_{neutron}\; \; \; \; \; \; \; \; 11$

where $\frac{m_{binding}}{A}$ 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 ¹H, $\frac{Z}{A}<1$ . Therefore, the RHS of eq11 approximately equals to m_neutron:

$\frac{m_{binding}}{A}\approx m_{neutron}-\frac{m_{isotope}}{A}$

For ¹²C, $m_{isotope}=A=12u$ , and so $\biggr$\frac{m_{binding}}{A}\biggr$_{^{12}C}\approx m_{neutron}-1$ . If $\left (\frac{m_{binding}}{A}\right )_{isotope}<\left (\frac{m_{binding}}{A}\right )_{^{12}C}$ , then $m_{neutron}-\frac{m_{isotope}}{A}<m_{neutron}-1$ , or equivalently, $m_{isotope}>A$ . By the same logic, if $\left (\frac{m_{binding}}{A}\right )_{isotope}>\left (\frac{m_{binding}}{A}\right )_{^{12}C}$ , then $m_{isotope}<A$ .