Calculate nuclear binding energy in real time with the Bethe-Weizsäcker semi-empirical mass formula. Explore the BE/A curve and discover why both nuclear fission and fusion can release energy.
Professor, I understand that energy can be extracted from nuclear fusion, but energy is also released in nuclear fission, right? Why does energy come out from both? I thought it would be the opposite?
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That's a sharp question. The key to understanding this is 'binding energy.' It's the energy required to break a nucleus into individual protons and neutrons. The larger the BE/A (binding energy per nucleon), the more 'stable' the nucleus. Looking at the graph, it starts increasing around Be-4, peaks near iron (A=56), and then slowly decreases, forming a mountain shape.
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So iron is at the peak of the mountain. That means for elements lighter than iron, fusion moves them toward the peak and releases energy, while for elements heavier than iron, fission moves them toward the peak and releases energy, right?
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Perfect understanding! Exactly that. For example, in hydrogen fusion (H-2 + H-3 → He-4), BE/A increases from about 1 MeV/nucleon to about 7 MeV/nucleon. For 4 nucleons, that releases about 24 MeV of energy. In uranium fission (U-235 → Ba-141 + Kr-92), BE/A increases from about 7.6 to about 8.4 MeV/nucleon, and for 235 nucleons, that releases about 200 MeV.
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So fusion has a larger energy per nucleon, but fission involves many more nucleons at once, so the total energy is larger. What is the 'Coulomb term' in the Bethe-Weizsäcker formula?
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Protons all have positive charge, so they repel each other. That 'energy loss due to Coulomb repulsion' is the Coulomb term $-a_C Z(Z-1)/A^{1/3}$. It increases as Z increases. This is one reason why heavier elements become more unstable. Meanwhile, the 'symmetry term' reflects the tendency for stability when the number of protons and neutrons are equal. For light nuclei, N=Z is stable, but for heavier ones, nuclei with N>Z become stable due to the Coulomb term's influence.
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So I've also heard the term 'valley of stability'—does that mean stable nuclei line up like a valley on the N-Z plane?
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Yes! On an N-Z plot, you can see the 'valley of stability,' where stable nuclides line up in a narrow band. Nuclei that deviate from this valley (with extreme N/Z ratios) are radioactive and decay via beta decay, etc., falling toward the valley. Check the 'Valley of Stability' tab in this simulator. The BE/A magnitude is shown by color intensity, so you can see at a glance which regions are stable.
Frequently Asked Questions
What is nuclear binding energy?
It is the energy required to break a nucleus into its individual protons and neutrons. For example, to separate He-4 (alpha particle) into 2 protons and 2 neutrons, about 28 MeV is needed. The larger the binding energy per nucleon (BE/A), the more stable the nuclide. Iron (Fe-56) has the maximum BE/A at about 8.8 MeV/nucleon.
Why is iron the most stable element?
Fe-56 (actually Ni-62 has the highest) has the maximum binding energy per nucleon. Lighter nuclei release energy through fusion, while heavier ones release energy through fission as they approach iron. This is why iron is abundant in the universe—stellar fusion reactions stop at iron (fusion beyond iron consumes energy).
How accurate is the Bethe-Weizsäcker formula?
It is accurate to within a few percent for stable nuclei with medium mass numbers (A = 20 to 200). However, for very light nuclei (like He-4) and those with magic numbers (2, 8, 20, 28, 50, 82, 126)—which are especially stable due to shell effects—the liquid-drop model cannot explain them. More precise calculations require nuclear structure models or numerical methods.
Which releases more energy per nucleon: fusion or fission?
Per nucleon, fusion releases about 3 to 4 times more energy. D-T fusion (deuterium + tritium → helium + neutron) yields about 17.6 MeV / 5 nucleons ≈ 3.5 MeV/nucleon. U-235 fission yields about 200 MeV / 235 nucleons ≈ 0.85 MeV/nucleon. However, since fission involves more nucleons per reaction, the total energy per reaction is larger for fission.
What are "magic numbers"?
Nuclides with proton or neutron numbers equal to 2, 8, 20, 28, 50, 82, or 126 are especially stable. This is explained by the nuclear shell model. Just like electron shells in atoms (closed shells are stable), nucleons also become particularly stable when they form closed shells. Examples include He-4 (Z=2, N=2), O-16 (Z=8, N=8), and Pb-208 (Z=82, N=126, a doubly magic nucleus).
Set mass number A (total nucleons) using slider sl-A or input the input field; range 4–238 covers He-4 to U-238.
Set proton number Z using slider sl-Z or input the input field; ensure Z ≤ A and N = A − Z ≥ 0.
Read binding energy in MeV from the semi-empirical mass formula: BE = a_v·A − a_s·A^(2/3) − a_c·Z(Z−1)/A^(1/3) − a_a·(A−2Z)²/A, where a_v=15.68, a_s=18.56, a_c=0.717, a_a=28.1 are empirical coefficients.
Worked Example
Iron-56 (A=56, Z=26, N=30): Volume term 15.68×56=877.8 MeV; surface term 18.56×56^(2/3)≈123.4 MeV; Coulomb term 0.717×26×25/3.83≈122.3 MeV; asymmetry (56−52)²/56≈0.57 MeV × 28.1≈16.0 MeV. Net BE ≈ 877.8 − 123.4 − 122.3 − 16.0 ≈ 616 MeV, matching experimental 492 MeV (formula overestimates light nuclei; precision improves at A>40).
Practical Notes
Pairing effects omitted in Bethe-Weizsäcker; even-even nuclei (Fe-56, Ni-58) show 1–2% higher binding than formula predicts.
Coulomb repulsion dominates for Z>50; uranium (Z=92) requires stable N≈1.55Z to minimize asymmetry penalty and Coulomb cost.
Magic numbers (Z or N = 2, 8, 20, 50, 82, 126) cause sharp BE jumps; Pb-208 achieves 7.87 MeV/nucleon, near-maximum stability.