Nuclear binding energy per nucleon

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

⁵⁶Fe 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 ⁵⁶Fe 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 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, $m_{isotope}=A=12$ , and so $\frac{m_{isotope}}{A}=1$ . If $\left (\frac{m_{binding}}{A}\right )_{isotope}<\left (\frac{m_{binding}}{A}\right )_{^{12}C}$ , then $\frac{m_{isotope}}{A}>1$ . By the same logic, if $\left (\frac{m_{binding}}{A}\right )_{isotope}>\left (\frac{m_{binding}}{A}\right )_{^{12}C}$ , then $\frac{m_{isotope}}{A}<1$ .

Nuclear binding energy per nucleon

Question

Answer

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