📚 Nuclear Energy: Deriving Binding Energy and Reaction Q-Values | 核能：结合能与反应能公式推导
Nuclear energy is one of the most profound applications of Einstein’s mass–energy equivalence. In A-Level Physics, you are expected not only to recall E = mc² but to use it to derive binding energy, explain fission and fusion, and calculate the energy released in nuclear reactions. This article walks through every key formula step by step, always pairing English explanations with Chinese translations, and supplies worked examples that mirror the style of Oxford AQA International A-Level exam questions.
核能是爱因斯坦质能等价关系最深刻的应用之一。在 A-Level 物理中，你不仅要记住 E = mc²，还要运用它推导结合能、解释裂变与聚变、计算核反应释放的能量。本文逐步推导每一个关键公式，始终以英文解释与中文翻译配对呈现，并提供符合 Oxford AQA International A-Level 考试风格的例题。
1. Mass–Energy Equivalence | 质能等价
Einstein’s famous equation relates mass and energy: a mass m has an equivalent rest energy E₀ given by
爱因斯坦著名的方程联系了质量和能量：质量 m 具有等效的静止能量 E₀，表达式为
E₀ = m c²
where c = 3.00 × 10⁸ m s⁻¹ is the speed of light in vacuum. Because c² is huge, a tiny amount of mass corresponds to an enormous amount of energy. In nuclear physics we often use the unified atomic mass unit u (1 u = 1.6605 × 10⁻²⁷ kg). Converting 1 u into energy gives a very handy factor:
其中 c = 3.00 × 10⁸ m s⁻¹ 是真空中光速。由于 c² 极大，微小质量对应巨大的能量。核物理中常常使用原子质量单位 u（1 u = 1.6605 × 10⁻²⁷ kg）。将 1 u 转换为能量，得到极其方便的换算因子：
1 u c² ≈ 931.5 MeV
This conversion is the backbone of all binding energy and reaction energy calculations.
这一换算是所有结合能与反应能计算的基础。
2. Mass Defect: The Source of Binding Energy | 质量亏损：结合能的来源
The mass of a stable nucleus is always less than the sum of the masses of its separate nucleons. This difference is called the mass defect Δm. For a nucleus with Z protons and N neutrons (mass number A = Z + N),
稳定原子核的质量总是小于其单独核子质量之和。这个差值称为质量亏损 Δm。对于含有 Z 个质子、N 个中子（质量数 A = Z + N）的原子核，
Δm = Z mₚ + N mₙ – Mnucleus
where mₚ is the proton mass, mₙ is the neutron mass, and Mnucleus is the actual mass of the nucleus. Often we use neutral atomic masses to avoid electron binding energy complications; then the mass defect is calculated from atomic masses M(atom) and the hydrogen atom mass M(¹H) together with the neutron mass.
其中 mₚ 为质子质量，mₙ 为中子质量，Mnucleus 为原子核的实际质量。为避免电子结合能带来的麻烦，常使用中性原子质量；此时质量亏损由原子质量 M(atom)、氢原子质量 M(¹H) 及中子质量共同算出。
The mass defect appears because when nucleons bind together, some of their mass is converted into binding energy that holds the nucleus together. The greater the mass defect, the more stable the nucleus – more energy would be required to pull it apart.
出现质量亏损的原因是核子结合时，部分质量转化为维持原子核稳定的结合能。质量亏损越大，原子核越稳定——要将它拆散需要越多的能量。
3. Nuclear Binding Energy Formula | 核结合能公式
The total binding energy Eb of a nucleus is simply the energy equivalent of the mass defect:
原子核的总结合能 Eb 就是质量亏损对应的能量：
Eb = Δm c²
Substituting the mass defect in kilograms gives Eb in joules; using atomic mass units and the conversion 1 u = 931.5 MeV/c² yields Eb directly in MeV:
代入以千克为单位的质量亏损得到焦耳数；使用原子质量单位并利用 1 u = 931.5 MeV/c²，可直接得到以 MeV 为单位的 Eb：
Eb (MeV) = Δm (u) × 931.5
For example, the helium-4 nucleus (²⁴He) has a mass defect of about 0.0304 u, giving a total binding energy of roughly 28.3 MeV. This is the energy that would be released if the nucleus were assembled from free nucleons, or the energy needed to break it apart completely.
例如，氦-4 核（²⁴He）质量亏损约 0.0304 u，总结合能约为 28.3 MeV。若由自由核子组装成该核，便会释放这些能量；反之，将其完全打散也需要这么多能量。
4. Binding Energy per Nucleon and the Stability Curve | 平均结合能与稳定曲线
Dividing the total binding energy by the number of nucleons A gives the average binding energy per nucleon:
总结合能除以核子数 A 得到平均结合能（每核子结合能）：
B.E. per nucleon = Eb / A
This quantity measures the relative stability of a nucleus. A plot of binding energy per nucleon against mass number A produces the famous ‘binding energy curve’. The curve rises steeply for light nuclei, reaches a broad maximum around iron-56 (about 8.8 MeV per nucleon), and then falls slowly for heavy nuclei.
这一量度反映了原子核的相对稳定性。平均结合能对质量数 A 作图，便得到著名的“结合能曲线”。该曲线对轻核急速上升，在铁-56 附近达到峰值（约 8.8 MeV/核子），随后对重核缓慢下降。
The shape of the curve explains why energy can be released either by fusing light nuclei (fusion) or by splitting heavy nuclei (fission). In both processes, the products are closer to the iron peak, so the total binding energy increases and the corresponding mass defect is converted into kinetic energy of the products.
曲线的形状解释了为何既可以通过轻核聚变，也可以通过重核裂变释放能量。两种过程都使产物更靠近铁峰值，总结合能增大，相应的质量亏损转化为产物的动能。
5. Energy Release in Fission – A Derivation | 核裂变能量释放——推导
A typical fission reaction is the neutron-induced fission of uranium-235:
一个典型的裂变反应是中子引发的铀-235 裂变：
¹₀n + ²³⁵₉₂U → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n
The energy released Q is given by the difference in total mass before and after the reaction:
释放能量 Q 由反应前后总质量之差给出：
Q = (minitial – mfinal) c²
Using atomic masses in atomic mass units, the calculation becomes:
采用原子质量（单位 u），计算为：
minitial = m(²³⁵U) + m(n) = 235.0439 u + 1.0087 u = 236.0526 u
minitial = m(²³⁵U) + m(n) = 235.0439 u + 1.0087 u = 236.0526 u
mfinal = m(¹⁴¹Ba) + m(⁹²Kr) + 3 m(n) = 140.9144 u + 91.9262 u + 3×1.0087 u = 235.8667 u
mfinal = m(¹⁴¹Ba) + m(⁹²Kr) + 3 m(n) = 140.9144 u + 91.9262 u + 3×1.0087 u = 235.8667 u
The mass defect is Δm = 0.1859 u, so
质量亏损为 Δm = 0.1859 u，因此
Q = 0.1859 u × 931.5 MeV/u ≈ 173 MeV
This energy is shared among the fission fragments and neutrons as kinetic energy. The derivation directly uses Einstein’s relation and mass–energy conservation.
该能量以动能形式在裂变碎片与中子之间分配。推导直接使用了爱因斯坦关系及质能守恒。
6. Energy Release in Fusion – A Derivation | 核聚变能量释放——推导
Consider the deuterium–tritium fusion reaction, one of the most promising for controlled fusion:
考虑氘-氚聚变反应，这是最有希望实现受控聚变的反应之一：
²₁H + ³₁H → ⁴₂He + ¹₀n
Mass before reaction: m(²H) + m(³H) = 2.0141 u + 3.0160 u = 5.0301 u
反应前质量：m(²H) + m(³H) = 2.0141 u + 3.0160 u = 5.0301 u
Mass after reaction: m(⁴He) + m(n) = 4.0026 u + 1.0087 u = 5.0113 u
反应后质量：m(⁴He) + m(n) = 4.0026 u + 1.0087 u = 5.0113 u
Mass defect Δm = 5.0301 u – 5.0113 u = 0.0188 u. The energy released is:
质量亏损 Δm = 5.0301 u – 5.0113 u = 0.0188 u。释放能量为：
Q = 0.0188 × 931.5 MeV ≈ 17.5 MeV
Although the energy per reaction is smaller than in fission, the energy per unit mass of fuel is much larger because the reacting nuclei are so light. This derivation highlights why fusion is the power source of stars.
尽管每次反应释放的能量小于裂变，但由于反应核极轻，单位质量燃料释放的能量要大得多。此推导体现了为何聚变是恒星的能源。
7. General Q-Value Equation for Nuclear Reactions | 核反应的通用 Q 值方程
For any nuclear reaction of the form a + X → Y + b, the Q-value is defined as the energy released (positive for exothermic, negative for endothermic). It is calculated from the rest masses:
对于任何 a + X → Y + b 形式的核反应，Q 值定义为释放的能量（放热为正，吸热为负）。由静止质量计算：
Q = [(ma + mX) – (mY + mb)] c²
When the masses are given in atomic mass units, the result is
当质量用原子质量单位给出时，结果为
Q (MeV) = [Minitial(u) – Mfinal(u)] × 931.5
This equation encapsulates conservation of mass–energy: the decrease in total rest mass appears as kinetic energy of the products.
此方程体现了质能守恒：总静止质量的减少表现为产物的动能。
8. Q-Value in Terms of Binding Energies | 用结合能表示的 Q 值
A very useful alternative form expresses Q as the difference between the total binding energies of products and reactants. Since the mass of a nucleus is the sum of its nucleon masses minus its binding energy equivalent, the Q-value becomes:
另一种非常有用的形式将 Q 表示为产物与反应物总结合能之差。因为原子核质量等于其核子质量之和减去结合能对应的质量，Q 值变为：
Q = Σ Eb(products) – Σ Eb(reactants)
Notice that the free nucleon masses cancel out. This expression explains the energy release on the binding energy curve: if the products are more tightly bound than the reactants, the total binding energy increases, so Q is positive. This is the fundamental reason fission and fusion release energy – the fragment nuclei sit closer to the iron peak.
注意，自由核子的质量项会相互抵消。该表达式在结合能曲线上解释了能量释放：若产物比反应物结合得更紧，总结合能增加，Q 为正。这就是裂变和聚变释放能量的根本原因——碎片核更靠近铁峰。
9. Converting Atomic Mass Units to Energy | 原子质量单位与能量换算
In exam problems, you need to convert between atomic mass units and MeV quickly. The key relationship is derived from the definition of 1 u = 1/12 of the mass of a carbon‑12 atom and the speed of light. The conversion is
在考试题目中，你需要快速在原子质量单位和 MeV 之间换算。关键关系来自 1 u 的定义（碳‑12 原子质量的 1/12）和光速。换算为
1 u c² = 931.494 MeV ≈ 931.5 MeV
Thus, to find the energy equivalent of a mass difference Δm in u, multiply by 931.5. If Δm is given in kg, use E = Δm c² directly with c = 3.00×10⁸ m/s to obtain joules, then convert to eV (1 eV = 1.602×10⁻¹⁹ J).
因此，若质量差 Δm 以 u 为单位，乘以 931.5 即得能量当量。若 Δm 以 kg 给出，直接用 E = Δm c²（c = 3.00×10⁸ m/s）得到焦耳，再转化为 eV（1 eV = 1.602×10⁻¹⁹ J）。
Many students lose marks by forgetting to square c or by using wrong units – always check whether the question expects an answer in J, MeV, or kWh.
许多学生因忘记将 c 平方或单位使用错误而失分——务必确认题目要求答案的单位是 J、MeV 还是 kWh。
10. Worked Example: Fission of Uranium-235 | 计算范例：铀-235 裂变
Question: Calculate the energy released in the fission of one ²³⁵U nucleus when it splits into ¹⁴⁰Xe and ⁹⁴Sr, with two neutrons emitted. Given masses: m(²³⁵U) = 235.0439 u, m(¹⁴⁰Xe) = 139.9216 u, m(⁹⁴Sr) = 93.9154 u, m(n) = 1.0087 u.
题目：计算一个 ²³⁵U 核裂变成 ¹⁴⁰Xe 和 ⁹⁴Sr 并释放两个中子时释放的能量。已知质量：m(²³⁵U) = 235.0439 u，m(¹⁴⁰Xe) = 139.9216 u，m(⁹⁴Sr) = 93.9154 u，m(n) = 1.0087 u。
Solution: Total initial mass = 235.0439 + 1.0087 = 236.0526 u. Total final mass = 139.9216 + 93.9154 + 2×1.0087 = 235.8544 u. Δm = 0.1982 u. Q = 0.1982 × 931.5 ≈ 184.6 MeV.
解答：反应前总质量 = 235.0439 + 1.0087 = 236.0526 u。反应后总质量 = 139.9216 + 93.9154 + 2×1.0087 = 235.8544 u。Δm = 0.1982 u。Q = 0.1982 × 931.5 ≈ 184.6 MeV。
This released energy is distributed among the fragments as kinetic energy and is a typical value for uranium fission.
该能量以动能形式分布在碎片中，是铀裂变的典型值。
11. Worked Example: Fusion of Deuterium and Tritium | 计算范例：氘氚聚变
Question: For the reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, find the energy released per reaction and per kilogram of fuel. Masses: m(²H) = 2.014102 u, m(³H) = 3.016049 u, m(⁴He) = 4.002603 u, m(n) = 1.008665 u.
题目：对反应 ²₁H + ³₁H → ⁴₂He + ¹₀n，求每次反应及每千克燃料释放的能量。质量：m(²H) = 2.014102 u，m(³H) = 3.016049 u，m(⁴He) = 4.002603 u，m(n) = 1.008665 u。
Solution: Δm = (2.014102 + 3.016049) – (4.002603 + 1.008665) = 0.018883 u. Q = 0.018883 × 931.5 ≈ 17.59 MeV. Per kilogram of fuel: the reacting mass is 5.030151 u per reaction. 1 kg contains N = 1 kg / (5.030151 × 1.6605×10⁻²⁷ kg) ≈ 1.196×10²⁶ pairs. Total energy ≈ 1.196×10²⁶ × 17.59 MeV × 1.602×10⁻¹³ J/MeV ≈ 3.37×10¹⁴ J. This is about 80 times the energy per kg from fission of ²³⁵U.
解答：Δm = (2.014102 + 3.016049) – (4.002603 + 1.008665) = 0.018883 u。Q = 0.018883 × 931.5 ≈ 17.59 MeV。每千克燃料：每次反应的反应物质量为 5.030151 u。1 kg 含有 N = 1 kg / (5.030151 × 1.6605×10⁻²⁷ kg) ≈ 1.196×10²⁶ 对。总能量 ≈ 1.196×10²⁶ × 17.59 MeV × 1.602×10⁻¹³ J/MeV ≈ 3.37×10¹⁴ J。这约为每千克 ²³⁵U 裂变能量的 80 倍。
12. Common Pitfalls and Key Takeaways | 常见错误与重点归纳
• Unit confusion: Always ensure Δm is in kg when using E = Δm c² to get joules, or in u when multiplying by 931.5 to get MeV. Mixing units is the most frequent mistake.
• 单位混淆：用 E = Δm c² 求焦耳时务必保证 Δm 单位为 kg，用 931.5 乘时确保 Δm 单位为 u。单位混用是最常见的错误。
• Forgetting the neutron count: In fission, the number of free neutrons must be included in the final mass sum. Missing a neutron can give a wildly wrong Q-value.
• 忘记中子数目：裂变中，产物总质量必须包含自由中子。漏掉一个中子会使 Q 值严重错误。
• Sign of Q: A positive Q means energy is released; the reaction is exothermic. If you obtain a negative Q, the reaction is endothermic and cannot occur spontaneously.
• Q 的符号：正 Q 表示释放能量，反应放热。若得到负 Q，则为吸热反应，不能自发进行。
• Binding energy per nucleon vs total binding energy: Do not confuse them. Binding energy per nucleon determines stability; total binding energy determines the absolute energy needed to dismantle a nucleus.
• 平均结合能与总结合能：切勿混淆。平均结合能决定稳定性；总结合能决定拆散原子核所需的绝对能量。
The derivations above all rest on two pillars: mass–energy conservation and the detailed balance of initial and final masses. Mastering these will enable you to handle any nuclear energy calculation on the A-Level specification.
以上推导皆基于两大支柱：质能守恒以及初末态质量的精细平衡。掌握它们，你就能应对 A-Level 考纲中的任何核能计算。
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