explain the relevance of binding energy per nucleon to nuclear reactions, including nuclear fusion and nuclear fission

Mass Defect and Nuclear Binding Energy

1. Mass Defect

The mass defect of a nucleus is the difference between the sum of the masses of its constituent nucleons and the actual mass of the nucleus.

\$\Delta m = Z\,mp + N\,mn - m_{\text{nucleus}}\$

where:

• \$Z\$ = number of protons
• \$N\$ = number of neutrons
• \$m_p\$ = mass of a proton
• \$m_n\$ = mass of a neutron
• \$m_{\text{nucleus}}\$ = measured mass of the nucleus

2. Nuclear Binding Energy

The binding energy is the energy required to separate a nucleus into its individual nucleons. By Einstein’s mass‑energy equivalence:

\$E_{\text{b}} = \Delta m\,c^{2}\$

where \$c\$ is the speed of light in vacuum (\$c = 3.00\times10^{8}\,\text{m s}^{-1}\$).

3. Binding Energy per Nucleon

To compare nuclei of different sizes we use the binding energy per nucleon:

\$\frac{E{\text{b}}}{A} = \frac{E{\text{b}}}{Z+N}\$

This quantity indicates how tightly, on average, each nucleon is bound within the nucleus.

4. Relevance to Nuclear Reactions

The shape of the binding‑energy curve explains why certain nuclear processes release energy:

• For light nuclei (\$A \lesssim 56\$), moving to a higher \$A\$ (fusion) increases \$E_{\text{b}}/A\$ → energy is released.
• For heavy nuclei (\$A \gtrsim 56\$), moving to a lower \$A\$ (fission) also increases \$E_{\text{b}}/A\$ → energy is released.

Thus, the difference in binding energy per nucleon before and after a reaction directly gives the energy output:

\$Q = \left(\frac{E{\text{b}}}{A}\right){\text{final}}A{\text{final}} - \left(\frac{E{\text{b}}}{A}\right){\text{initial}}A{\text{initial}}\$

5. Nuclear Fusion

Fusion combines two light nuclei to form a heavier nucleus. Example: the fusion of deuterium (\$^{2}\!\$H) and tritium (\$^{3}\!\$H) to produce helium‑4 and a neutron.

\$^{2}\!{\rm H} + ^{3}\!{\rm H} \rightarrow ^{4}\!{\rm He} + n + Q\$

Using binding‑energy data:

1. Calculate \$E_{\text{b}}\$ for each reactant and product.
2. Find the total binding energy before and after the reaction.
3. The released energy \$Q\$ equals the increase in total binding energy.

Because \$E_{\text{b}}/A\$ for \$^{4}\!\$He (≈7.07 MeV) is greater than the average for \$^{2}\!\$H and \$^{3}\!\$H (≈1–2 MeV), the reaction releases about \$17.6\,\$MeV.

6. Nuclear Fission

Fission splits a heavy nucleus into lighter fragments, each with a higher \$E_{\text{b}}/A\$ than the original heavy nucleus.

Typical example: neutron‑induced fission of uranium‑235.

\$^{235}\!{\rm U} + n \rightarrow ^{92}\!{\rm Kr} + ^{141}\!{\rm Ba} + 3n + Q\$

Using the table values (or more detailed data):

• Binding energy per nucleon of \$^{235}\$U ≈ 7.6 MeV.
• Binding energy per nucleon of the fission fragments (≈8.5 Me \cdot on average).

The increase in total binding energy corresponds to an energy release of roughly \$200\,\$Me \cdot per fission event.

7. Summary of Key Points

• Mass defect \$\Delta m\$ quantifies the “missing” mass that appears as binding energy.
• Binding energy \$E_{\text{b}} = \Delta m c^{2}\$ measures nuclear stability.
• The binding energy per nucleon peaks near iron‑56; nuclei far from this peak have the greatest potential to release energy.
• Fusion releases energy for light nuclei moving up the curve; fission releases energy for heavy nuclei moving down the curve.
• Calculating \$Q\$ from differences in total binding energy provides the quantitative energy yield of a nuclear reaction.