Nuclear Fusion Reactions and Energy Release | 核聚变反应与能量释放

📚 Nuclear Fusion Reactions and Energy Release | 核聚变反应与能量释放

Nuclear fusion is the process by which two light atomic nuclei combine to form a heavier nucleus, releasing a tremendous amount of energy. This process powers the Sun and other stars, and understanding it is essential for both astrophysics and the pursuit of clean energy on Earth.

核聚变是两个轻原子核结合形成一个较重原子核的过程，在此过程中释放出巨大的能量。这一过程为太阳及其他恒星提供能量，理解核聚变对天体物理学以及在地球上追求清洁能源都至关重要。

1. What Is Nuclear Fusion? | 什么是核聚变？

Nuclear fusion is a reaction in which two light nuclei, typically isotopes of hydrogen such as deuterium (²H) and tritium (³H), merge to form a heavier nucleus such as helium-4 (⁴He), along with a neutron and a substantial release of energy.

核聚变是指两个轻原子核（通常是氢的同位素，如氘（²H）和氚（³H））合并形成一个较重的原子核（如氦-4（⁴He）），同时释放出一个中子和大量能量的反应。

The fundamental condition for fusion is that the reacting nuclei must come within the range of the strong nuclear force, approximately 10⁻¹⁵ m. To overcome the electrostatic repulsion between positively charged nuclei, extremely high temperatures and pressures are required.

发生聚变的基本条件是反应原子核必须进入强核力的作用范围，约为10⁻¹⁵ m。为了克服带正电原子核之间的静电斥力，需要极高的温度和压力。

2. Mass-Energy Equivalence | 质能等效

Albert Einstein’s famous equation E = mc² is the theoretical foundation for understanding energy release in nuclear reactions. The mass of the products of a fusion reaction is slightly less than the mass of the reactants; this mass difference, known as the mass defect, is converted into kinetic energy.

爱因斯坦著名的方程E = mc²是理解核反应中能量释放的理论基础。聚变反应产物的质量略小于反应物的质量；这个质量差称为质量亏损，它被转化为动能。

E = Δm × c²

Here, E is the energy released, Δm is the decrease in mass (mass defect), and c is the speed of light in a vacuum (3.00 × 10⁸ m s⁻¹). Because c² is an enormous number, even a tiny mass defect corresponds to a very large energy release.

其中，E是释放的能量，Δm是质量的减少量（质量亏损），c是真空中的光速（3.00 × 10⁸ m s⁻¹）。由于c²是一个巨大的数字，即使是微小的质量亏损也对应着非常大的能量释放。

3. The Most Important Fusion Reaction: D-T Fusion | 最重要的聚变反应：氘-氚聚变

The deuterium-tritium (D-T) fusion reaction is considered the most promising for controlled fusion power on Earth because it has the largest cross-section at relatively achievable temperatures.

氘-氚（D-T）聚变反应被认为是地球上受控聚变能源最有前景的反应，因为它在相对可达到的温度下具有最大的反应截面。

²H + ³H → ⁴He (3.5 MeV) + n (14.1 MeV)

The total energy released in this reaction is 17.6 MeV. The helium-4 nucleus carries 3.5 MeV of kinetic energy, while the neutron carries 14.1 MeV. The neutron energy is more difficult to harness, but it can be captured in a blanket of lithium to produce more tritium, achieving breeding.

该反应释放的总能量为17.6 MeV。氦-4原子核携带3.5 MeV的动能，而中子携带14.1 MeV。中子的能量较难利用，但它可以被锂层捕获以生产更多的氚，实现增殖。

Let us verify the energy release using atomic masses:

让我们用原子质量来验证能量释放：

Mass of ²H: 2.014102 u

氘的质量：2.014102 u
Mass of ³H: 3.016049 u

氚的质量：3.016049 u
Mass of ⁴He: 4.002603 u

氦-4的质量：4.002603 u
Mass of n: 1.008665 u

中子的质量：1.008665 u

Δm = (2.014102 + 3.016049) − (4.002603 + 1.008665) = 0.018883 u

Using the conversion 1 u = 931.5 MeV/c², we find E = 0.018883 × 931.5 ≈ 17.6 MeV. This confirms the energy release calculated from the mass defect.

利用换算1 u = 931.5 MeV/c²，得到E = 0.018883 × 931.5 ≈ 17.6 MeV。这证实了由质量亏损计算出的能量释放。

4. Fusion in the Sun: The Proton-Proton Chain | 太阳中的聚变：质子-质子链

In the Sun and other main-sequence stars of similar mass, the dominant fusion process is the proton-proton (p-p) chain. This series of reactions converts four protons into one helium-4 nucleus, releasing energy in several steps.

在太阳及其他类似质量的主序星中，主导的聚变过程是质子-质子（p-p）链。这一系列反应将四个质子转化成一个氦-4原子核，分几个步骤释放能量。

The first step involves two protons fusing to form a deuterium nucleus:

第一步是两个质子聚变形成氘核：

¹H + ¹H → ²H + e⁺ + νₑ

This step requires the weak nuclear force and converts a proton into a neutron, emitting a positron (e⁺) and an electron neutrino (νₑ). It occurs very slowly, which is why the Sun has a long lifetime of about 10 billion years.

这一步需要弱核力的参与，将一个质子转化为中子，同时发射一个正电子（e⁺）和一个电子中微子（νₑ）。这一步骤发生得非常缓慢，这就是太阳拥有约100亿年长寿命的原因。

The subsequent steps are:

后续步骤如下：

²H + ¹H → ³He + γ

³He + ³He → ⁴He + ¹H + ¹H

The overall effect of the p-p chain is summarized as:

质子-质子链的净效果总结为：

4¹H → ⁴He + 2e⁺ + 2νₑ + 26.7 MeV

Thus, each helium-4 nucleus produced in the Sun’s core releases 26.7 MeV of energy, sustaining the Sun’s luminosity.

因此，太阳核心每产生一个氦-4原子核，就释放26.7 MeV的能量，维持着太阳的光度。

5. Nuclear Fusion vs Nuclear Fission | 核聚变与核裂变对比

While both nuclear fusion and nuclear fission involve changes in atomic nuclei and release energy through mass-energy equivalence, they differ significantly in several important aspects.

虽然核聚变和核裂变都涉及原子核的变化并通过质能等效释放能量，但它们在几个重要方面存在显著差异。

Aspect	Fusion	Fission
Process	Light nuclei combine	Heavy nuclei split
Fuel	Deuterium, tritium	Uranium-235, Plutonium-239
Energy per kg	≈ 4 × 10¹⁴ J	≈ 8 × 10¹³ J
Waste	Helium (non-radioactive)	Long-lived radioactive waste
Reactor conditions	≥ 10⁸ K, low density	Room temperature, chain reaction
Safety	No meltdown risk; inherently self-limiting	Meltdown and runaway risk

Fusion offers significant advantages: abundant fuel supply from seawater, minimal radioactive waste, and no risk of a runaway chain reaction. However, achieving and sustaining the extreme conditions required for fusion remains a major engineering challenge.

聚变具有显著优势：燃料来源丰富（可从海水中提取）、放射性废物极少、不存在失控链式反应的风险。然而，实现并维持聚变所需的极端条件仍然是一项重大的工程挑战。

6. Energy Barrier: Coulomb Repulsion and Quantum Tunnelling | 能量势垒：库仑斥力与量子隧穿

Two positively charged nuclei experience a Coulomb repulsion that acts as an energy barrier to fusion. To fuse, the nuclei must approach each other closely enough for the strong nuclear force to dominate, which requires overcoming this electrostatic barrier.

两个带正电的原子核之间存在着库仑斥力，它构成了聚变的能量势垒。要实现聚变，原子核必须靠近到强核力主导的距离，这需要克服这一静电势垒。

The height of the Coulomb barrier for two hydrogen isotopes is approximately 0.4 MeV. At temperatures of 10⁷ K (the core of the Sun), the average kinetic energy of particles is only about 1 keV, far below the barrier height.

对于两个氢同位素，库仑势垒的高度约为0.4 MeV。在10⁷ K的温度下（太阳核心），粒子的平均动能仅为约1 keV，远低于势垒高度。

So how does fusion occur in the Sun? The answer lies in two quantum mechanical and statistical effects:

那么聚变如何在太阳中发生？答案在于两个量子力学和统计效应：

Maxwell-Boltzmann distribution: Some particles in the high-energy tail of the distribution have enough kinetic energy to overcome the barrier.

麦克斯韦-玻尔兹曼分布：分布的高能尾部中的一些粒子拥有足够的动能来克服势垒。
Quantum tunnelling: Even particles with energy below the barrier have a finite probability of tunnelling through it due to their wave-like nature.

量子隧穿：即使能量低于势垒的粒子，由于其波动性，也有一定概率隧穿通过势垒。

The Gamow peak is the energy range where fusion reactions are most likely, representing the product of the Maxwell-Boltzmann distribution (which decreases with energy) and the tunnelling probability (which increases with energy).

伽莫夫峰是聚变反应发生概率最高的能量区间，它反映了麦克斯韦-玻尔兹曼分布（随能量增加而下降）与隧穿概率（随能量增加而上升）的乘积。

7. Conditions for Controlled Fusion: The Lawson Criterion | 受控聚变的条件：劳森判据

For a fusion reactor to produce net energy output, the confinement must be good enough to keep the plasma hot and dense for a sufficient time. This is quantified by the Lawson criterion.

要使聚变反应堆产生净能量输出，约束条件必须足以在足够长的时间内保持等离子体的高温和高密度。这可以用劳森判据来量化。

n τ_E > 10²⁰ m⁻³ s

Here, n is the particle density of the plasma, and τ_E is the energy confinement time. For a D-T plasma at a temperature of approximately 10 keV, the triple product of density (n), temperature (T), and confinement time (τ) must satisfy nTτ > 3 × 10²¹ m⁻³ keV s.

其中，n是等离子体的粒子密度，τ_E是能量约束时间。对于温度约10 keV的氘-氚等离子体，密度（n）、温度（T）和约束时间（τ）的三乘积必须满足nTτ > 3 × 10²¹ m⁻³ keV s。

There are two main approaches to achieving controlled fusion:

实现受控聚变主要有两种方法：

Magnetic confinement fusion (MCF): Using strong magnetic fields (tokamaks, stellarators) to confine the plasma at low density (n ≈ 10²⁰ m⁻³) for a long time (τ ≈ several seconds).

磁约束聚变（MCF）：利用强磁场（托卡马克、仿星器）在低密度（n ≈ 10²⁰ m⁻³）下长时间（τ ≈ 几秒）约束等离子体。
Inertial confinement fusion (ICF): Using intense laser or ion beams to compress a small fuel pellet to extremely high density (n ≈ 10³² m⁻³) for a very short time (τ ≈ 10⁻¹¹ s).

惯性约束聚变（ICF）：利用强激光或离子束将小燃料靶丸压缩到极高密度（n ≈ 10³² m⁻³），持续极短时间（τ ≈ 10⁻¹¹ s）。

8. Energy Release in the D-T Reaction: A Worked Example | 氘-氚反应的能量释放：计算示例

Let us calculate the number of D-T reactions needed to produce 1 J of energy, and then estimate the fuel mass required.

让我们计算产生1 J能量所需的氘-氚反应次数，然后估算所需燃料质量。

First, convert 17.6 MeV to joules:

首先，将17.6 MeV转换为焦耳：

E_per_reaction = 17.6 × 10⁶ × 1.602 × 10⁻¹⁹ J = 2.82 × 10⁻¹² J

Number of reactions per joule:

每焦耳的反应次数：

N = 1 / (2.82 × 10⁻¹²) ≈ 3.55 × 10¹¹ reactions

Each D-T pair has a combined mass of (2.014102 + 3.016049) u = 5.030151 u. The total mass consumed per joule is:

每对氘-氚的总质量为(2.014102 + 3.016049) u = 5.030151 u。消耗的总质量每焦耳为：

m = 3.55 × 10¹¹ × 5.030151 × 1.661 × 10⁻²⁷ kg ≈ 2.96 × 10⁻¹⁴ kg

In practice, the mass actually converted to energy is only Δm = 0.018883 u per reaction, which equals about 0.375% of the total fuel mass. This means that 1 kg of D-T fuel mixture can produce:

实际上，每个反应真正转化为能量的质量仅为Δm = 0.018883 u，约占总燃料质量的0.375%。这意味着1 kg的氘-氚混合燃料可以产生：

E = 1 × 0.00375 × c² = 1 × 0.00375 × 9 × 10¹⁶ ≈ 3.4 × 10¹⁴ J

This is equivalent to burning approximately 2,000 tonnes of coal. For comparison, 1 kg of uranium-235 in fission releases about 8 × 10¹³ J, so D-T fusion releases about four times more energy per kilogram of fuel.

这相当于燃烧约2000吨煤。作为比较，1 kg铀-235通过裂变释放约8 × 10¹³ J，因此氘-氚聚变每千克燃料释放的能量约为裂变的四倍。

9. The Binding Energy Perspective | 从结合能角度理解核聚变

Nuclear fusion releases energy because the binding energy per nucleon increases up to iron-56 (⁵⁶Fe) at the peak of the binding energy curve. When light nuclei fuse into heavier ones, the products have higher binding energy per nucleon, meaning they are more stable and have lower total mass.

核聚变释放能量是因为每个核子的平均结合能在铁-56（⁵⁶Fe）处达到结合能曲线的峰值。当轻核聚变成较重的核时，产物每个核子的平均结合能更高，意味着它们更稳定，总质量更低。

The graph of binding energy per nucleon against nucleon number shows a steep rise from hydrogen (²H) to helium-4 (⁴He). This sharp increase is responsible for the large energy release in fusion reactions. Iron-56 has the highest binding energy per nucleon of approximately 8.8 MeV/nucleon. Fusion releases energy when moving toward iron from the left side of the curve, while fission releases energy when moving toward iron from the right side.

每个核子平均结合能对核子数的图表显示，从氢（²H）到氦-4（⁴He）急剧上升。这一急剧上升导致了聚变反应中巨大的能量释放。铁-56具有最高的每个核子平均结合能，约为8.8 MeV/核子。当从曲线左侧向铁方向移动时，聚变释放能量；当从曲线右侧向铁方向移动时，裂变释放能量。

For the D-T fusion reaction specifically, we can calculate the binding energy difference:

具体到氘-氚聚变反应，我们可以计算结合能差异：

²H: binding energy = 2.22 MeV → 1.11 MeV/nucleon

²H：结合能 = 2.22 MeV → 1.11 MeV/核子
³H: binding energy = 8.48 MeV → 2.83 MeV/nucleon

³H：结合能 = 8.48 MeV → 2.83 MeV/核子
⁴He: binding energy = 28.30 MeV → 7.07 MeV/nucleon

⁴He：结合能 = 28.30 MeV → 7.07 MeV/核子

The dramatic increase in stability from the reactants to the product accounts for the 17.6 MeV released.

从反应物到产物稳定性的显著提高正是17.6 MeV能量释放的原因。

10. Challenges of Achieving Fusion on Earth | 在地球上实现聚变面临的挑战

Although the physics of fusion is well understood, engineering a device that produces more energy than it consumes has proven extremely difficult. Key challenges include:

尽管聚变的物理原理已被充分理解，但要制造一个产生的能量多于消耗能量的装置已被证明是极其困难的。关键挑战包括：

Achieving extreme temperatures: The fuel must be heated to temperatures above 100 million °C, hotter than the core of the Sun. At these temperatures, matter exists as a plasma and no physical container can survive direct contact.

实现极端温度：燃料必须被加热到超过1亿°C的温度，比太阳核心还要热。在这样的温度下，物质以等离子体形式存在，任何物理容器都无法承受直接接触。
Plasma confinement: Confining a plasma at 10⁸ K using magnetic fields requires extremely precise control. Instabilities in the plasma can cause it to escape the magnetic trap.

等离子体约束：使用磁场约束10⁸ K的等离子体需要极其精确的控制。等离子体中的不稳定性可能导致其逃逸出磁阱。
Neutron damage: The 14.1 MeV neutrons produced in D-T fusion carry most of the energy but also damage the reactor walls over time through displacement damage and nuclear transmutation.

中子损伤：氘-氚聚变产生的14.1 MeV中子携带大部分能量，但也会通过位移损伤和核嬗变随时间损坏反应堆壁。
Tritium breeding: Tritium is radioactive with a half-life of 12.3 years and does not occur naturally in significant quantities. It must be bred by surrounding the reactor with a lithium blanket: ⁶Li + n → ³H + ⁴He.

氚增殖：氚具有放射性，半衰期为12.3年，自然界中不存在大量天然氚。必须通过在反应堆周围设置锂层来增殖：⁶Li + n → ³H + ⁴He。
Net energy gain: In addition to the energy carried by the fusion products, a significant amount of energy is required to heat the plasma and power the confinement magnets and auxiliary systems. Reaching Q > 1 (where Q = output/input) is essential for a viable power plant.

净能量增益：除了聚变产物携带的能量外，还需要大量能量来加热等离子体并为约束磁体和辅助系统供电。实现Q > 1（其中Q = 输出/输入）是可行电厂的关键。

11. Energy Balance: Q Value and Ignition | 能量平衡：Q值与点火

The energy gain factor Q is defined as the ratio of the fusion power output to the input heating power:

能量增益因子Q的定义是聚变功率输出与输入加热功率之比：

Q = P_fusion / P_heating

Modern experiments such as JET and JT-60SA have achieved Q values of up to 0.67 and 1.25 respectively. The International Thermonuclear Experimental Reactor (ITER) is designed to achieve Q = 10.

现代实验装置如JET和JT-60SA分别实现了最高约0.67和1.25的Q值。国际热核实验反应堆（ITER）的设计目标是实现Q = 10。

When Q becomes infinite, the plasma is said to reach ignition: the energy deposited by the helium-4 particles (alpha particles) is sufficient to sustain the fusion reactions without any external heating. At this point, the reaction is self-sustaining, analogous to a flame that can sustain itself with its own combustion energy.

当Q变为无穷大时，等离子体被称为达到点火状态：氦-4粒子（α粒子）沉积的能量足以维持聚变反应而无需任何外部加热。此时，反应是自持的，类似于火焰在自身燃烧能量的作用下可以自行维持。

12. Exam Tips: Fusion Calculations and Concepts | 考点提示：聚变计算与概念

For examinations, the following key skills and concepts are frequently tested:

在考试中，以下关键技能和概念经常被考查：

Be able to define mass defect and calculate energy release using E = Δmc², converting between atomic mass units (u) and MeV using 1 u = 931.5 MeV/c².

能够定义质量亏损并使用E = Δmc²计算能量释放，在原子质量单位（u）和MeV之间换算，使用1 u = 931.5 MeV/c²。
Distinguish between nuclear fission and fusion in terms of reactants, products, energy release per nucleon, and the binding energy curve.

从反应物、产物、每个核子的能量释放以及结合能曲线等方面区分核裂变和核聚变。
Explain why extremely high temperatures are required for fusion, referencing the Coulomb barrier and quantum tunnelling.

解释为什么聚变需要极高的温度，涉及库仑势垒和量子隧穿。
Calculate the number of fusion reactions per second needed for a given power output: N/s = P / E_per_reaction.

计算给定功率输出所需的每秒聚变反应次数：N/s = P / E_per_reaction。
Describe the conditions required for controlled fusion (n, T, τ) using the Lawson criterion.

使用劳森判据描述受控聚变所需的条件（n、T、τ）。
Calculate the total energy released by the p-p chain and explain the role of neutrinos in carrying away energy.

计算质子-质子链释放的总能量，并解释中微子在带走能量中的作用。

Sample problem: A fusion power plant produces 1 GW of electrical power with 40% efficiency. If the energy comes from D-T fusion (17.6 MeV per reaction), how many reactions occur per second, and what mass of fuel is consumed per day?

示例问题：一座聚变电站以40%的效率产生1 GW的电功率。如果能量来自氘-氚聚变（每个反应17.6 MeV），每秒发生多少次反应，每天消耗多少质量的燃料？

The fusion power must be 1 GW / 0.40 = 2.5 GW. The number of reactions per second is:

聚变功率必须为1 GW / 0.40 = 2.5 GW。每秒的反应次数为：

N = P / E = 2.5 × 10⁹ / (2.82 × 10⁻¹²) ≈ 8.87 × 10²⁰ reactions s⁻¹

The mass of fuel consumed per day is N × m_per_reaction × time:

每天消耗的燃料质量为N × 每个反应质量 × 时间：

m = 8.87 × 10²⁰ × 5.03 × 1.661 × 10⁻²⁷ × 86,400 ≈ 64 kg day⁻¹

However, only about 0.375% of this mass is converted to energy; the rest becomes the helium-4 product. A conventional 1 GW coal plant consumes about 10,000 tonnes of coal per day, showing the dramatic advantage of fusion in fuel economy.

然而，这些质量中只有约0.375%转化为能量；其余部分成为氦-4产物。一座传统的1 GW燃煤电厂每天消耗约10,000吨煤，这显示了聚变在燃料经济性方面的巨大优势。

In conclusion, nuclear fusion represents one of the most promising yet challenging energy frontiers in physics. The principles of mass-energy equivalence, the binding energy curve, and the quantum mechanical phenomena of tunnelling form the theoretical bedrock of fusion science. From the natural fusion engine of our Sun to the ambitious engineering projects like ITER, understanding fusion energy release is essential for any physics student aiming to master nuclear physics topics.

总而言之，核聚变代表了物理学中最有前景但也最具挑战性的能源前沿领域之一。质能等效原理、结合能曲线以及量子隧穿现象构成了聚变科学的理论基础。从我们太阳的天然聚变引擎到ITER这样的雄心勃勃的工程项目，理解聚变能量释放对于任何希望掌握核物理主题的物理学生来说都是必不可少的。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Nuclear Fusion Reactions and Energy Release | 核聚变反应与能量释放