AQA A-Level Nuclear Physics: Decay, Binding Energy, Fission and Fusion — 核物理：衰变、结合能、裂变与聚变

1. What Makes a Nucleus Radioactive? Proton-Neutron Balance and Stability | 什么让原子核具有放射性？质子-中子平衡与稳定性

原子核由质子和中子（统称核子）构成，质子带正电，彼此之间会产生强烈的静电排斥。按照常理，这么多带正电的质子挤在半径只有几飞米（1 fm = 10⁻¹⁵ m）的空间里，原子核早就应该四分五裂了。原子核之所以能稳定存在，靠的是一种比电磁力强得多、但作用距离极短的力 – 强核力（strong nuclear force）。它只在相邻核子之间起作用，把核子牢牢地”粘”在一起，同时抵消了质子之间的库仑排斥。

The nucleus is made of protons and neutrons, collectively called nucleons. Protons carry positive charge, so they repel each other electrostatically. In principle, so many positively charged protons squeezed into a region only a few femtometres across (1 fm = 10⁻¹⁵ m) should blow the nucleus apart. The nucleus survives because of the strong nuclear force, an attraction far stronger than electromagnetism but with an extremely short range. It acts only between neighbouring nucleons, gluing them together and cancelling the Coulomb repulsion between protons.

是否稳定，取决于质子数与中子数之间的平衡。轻核（质子数 Z 较小）在中子数 N 大致等于质子数 Z 时最稳定，即 N ≈ Z。随着 Z 增大，为了把更多质子”拉”在一起并抵消不断增长的静电排斥，稳定核需要越来越多的中子，于是稳定核落在一条 N 略大于 Z 的曲线上，这条线被称为”稳定线”（line of stability）。凡是偏离这条线太远的核都会不稳定，通过发射粒子或电磁辐射来重新回到平衡，这个过程就是放射性衰变。

Stability depends on the balance between protons and neutrons. Light nuclei (small proton number Z) are most stable when the neutron number N is roughly equal to Z, that is N ≈ Z. As Z grows, more and more neutrons are needed to bind the extra protons together and counteract the growing electrostatic repulsion, so stable nuclei follow a curve where N is slightly larger than Z, known as the line of stability. Any nucleus too far from this line is unstable and moves back towards balance by emitting particles or electromagnetic radiation, a process we call radioactive decay.

不稳定的原因可以归结为三类：核子数过多、质子数过多，或者核内能量过高。中子过多时，一个中子会转变成质子并发射 β⁻ 粒子；质子过多时，一个质子会转变成中子并发射 β⁺ 粒子（或通过电子俘获）；而当核内能量过高时，原子核会通过发射 γ 光子释放多余能量。理解”为什么衰变”，比单纯记住”会发生衰变”更重要，这也是 AQA 考试中反复考察的核心观念。

Instability arises for three main reasons: too many nucleons, too many protons, or too much internal energy. When there are too many neutrons, a neutron converts into a proton and emits a β⁻ particle. When there are too many protons, a proton converts into a neutron and emits a β⁺ particle (or captures an orbital electron). When the nucleus simply carries too much energy, it releases the surplus by emitting a gamma photon. Understanding why decay happens matters more than memorising that it happens, and this is a recurring core idea in AQA examinations.

2. Three Types of Decay: Alpha, Beta and Gamma Radiation Compared | 三种衰变类型：α、β、γ辐射对比

放射性衰变主要产生三种辐射：α（阿尔法）、β（贝塔）和 γ（伽马）。α 粒子本质是一个氦-4 原子核，由 2 个质子和 2 个中子组成，带 +2e 的电荷，质量相对较大。β⁻ 粒子是高速电子（电荷 -e），β⁺ 粒子是正电子（电荷 +e）。γ 辐射则不是粒子，而是一种高能电磁波，不带电荷、没有质量。

Radioactive decay produces three main types of radiation: alpha (α), beta (β) and gamma (γ). An alpha particle is essentially a helium-4 nucleus, made of two protons and two neutrons, carrying a charge of +2e and a relatively large mass. A β⁻ particle is a fast-moving electron (charge -e), while a β⁺ particle is a positron (charge +e). Gamma radiation is not a particle at all but a high-energy electromagnetic wave with no charge and no mass.

三者的穿透能力与电离能力恰好相反。α 粒子电离能力最强，但在空气中只能前进几厘米，一张纸或几厘米空气就能把它挡住。β 粒子电离能力中等，在空气中能前进约 1 米，需要几毫米的铝板才能阻挡。γ 射线电离能力最弱，穿透能力却最强，需要几厘米厚的铅或很厚的混凝土才能显著削弱。记住这条规律：电离能力越强，穿透能力越弱。

The three types have opposite trends in penetrating power and ionising power. Alpha particles ionise most strongly but travel only a few centimetres in air, stopped by a sheet of paper or a few centimetres of air. Beta particles ionise moderately and travel about one metre in air, requiring a few millimetres of aluminium to stop them. Gamma rays ionise least but penetrate most, needing several centimetres of lead or thick concrete to attenuate them significantly. Remember the rule: the more strongly a radiation ionises, the less deeply it penetrates.

下面的表格总结了三种辐射的关键属性，考试中经常要求你根据这些性质选择或解释某种辐射的用途。

The table below summarises the key properties of the three types of radiation, which exam questions frequently ask you to use when choosing or explaining a particular application.

性质 Property	α 粒子	β 粒子	γ 射线
本质 Nature	氦-4 核 He-4 nucleus	电子/正电子 electron/positron	电磁波 EM wave
电荷 Charge	+2e	-e 或 +e	0
穿透力 Penetration	几张纸几厘米空气 stopped by paper	几毫米铝 a few mm of Al	几厘米铅 several cm of Pb
电离力 Ionising power	最强 Strongest	中等 Moderate	最弱 Weakest

在磁场或电场中的偏转行为也是常考点。α 粒子带正电，β⁻ 带负电，二者在磁场中会向相反方向偏转；由于 β 粒子质量远小于 α 粒子，其偏转半径更小、偏转更明显。γ 射线不带电，穿过磁场时完全不偏转。利用这一差异可以区分三种辐射。

Deflection in magnetic or electric fields is another common exam point. Alpha particles are positively charged and β⁻ negatively charged, so they deflect in opposite directions in a magnetic field. Because beta particles are far lighter than alpha particles, they deflect more sharply along a smaller radius. Gamma rays carry no charge and pass straight through a magnetic field without any deflection. This difference is used to distinguish the three types.

3. Writing Nuclear Decay Equations: Balancing Mass and Atomic Numbers | 书写核衰变方程：质量数与原子序数守恒

书写核衰变方程有两条铁律：质量数（上标）在反应前后必须守恒，原子序数（下标，即质子数）也必须守恒。这两条守恒定律让你即使忘记某个产物的具体符号，也能把它推导出来。以最常见的 α 衰变为例，铀-238 发射一个 α 粒子后，质量数减少 4、原子序数减少 2，因此产物必然是钍-234。

Writing nuclear decay equations follows two iron rules: the mass number (superscript) must be conserved across the reaction, and the atomic number (subscript, the proton number) must also be conserved. These two conservation laws let you deduce any product even if you forget its symbol. In the most common example, alpha decay, uranium-238 emits an alpha particle, losing 4 from its mass number and 2 from its atomic number, so the product must be thorium-234.

β⁻ 衰变的规律略有不同：中子转变为质子并发射一个电子（和一个反中微子），因此质量数不变，而原子序数增加 1。例如碳-14 衰变成氮-14。β⁺ 衰变则相反，质子转变为中子，原子序数减少 1，质量数不变。理解”质量数不变、原子序数 ±1″是 β 衰变的关键，也是学生最容易出错的地方。

Beta-minus decay follows a different rule: a neutron turns into a proton and emits an electron (plus an antineutrino), so the mass number stays the same while the atomic number increases by 1. Carbon-14, for example, decays into nitrogen-14. Beta-plus decay is the reverse: a proton turns into a neutron, so the atomic number decreases by 1 with the mass number unchanged. Understanding that the mass number is constant while the atomic number changes by ±1 is the key to beta decay, and the point where students most often slip.

γ 辐射通常伴随 α 或 β 衰变出现，是原子核在衰变后仍处于激发态时释放的能量。γ 发射不改变质量数，也不改变原子序数，所以在衰变方程中它只是作为产物被加上去。写出完整、配平的方程（包括 α、β、γ 以及中微子）是 AQA 试卷中每年必考的基本技能。

Gamma radiation usually accompanies alpha or beta decay, released when the daughter nucleus is left in an excited state. Gamma emission changes neither the mass number nor the atomic number, so it is simply added to the equation as a product. Writing complete, balanced equations, including the α, β, γ particles and neutrinos, is a basic skill that appears in AQA papers every year.

4. Half-Life and the Decay Constant: Exponential Decay Mathematics | 半衰期与衰变常数：指数衰变的数学

放射性衰变是一个随机过程：你无法预测某一个特定的原子核会在什么时候衰变，但对于大量原子核的集合，其衰变却遵循精确的统计规律。原子核的数量随时间按指数规律减少，这一规律可以用公式 N = N₀e^(−λt) 描述，其中 λ 是衰变常数（decay constant），单位为 s⁻¹，表示单位时间内每个原子核发生衰变的概率。

Radioactive decay is a random process: you cannot predict when any particular nucleus will decay, yet for a large collection of nuclei the decay follows a precise statistical law. The number of nuclei decreases exponentially with time, described by N = N₀e^(−λt), where λ is the decay constant, measured in s⁻¹, representing the probability per unit time that a given nucleus will decay.

半衰期（half-life, T½）是理解衰变快慢最直观的量：它表示放射性核的数量（或活度）减少到原来一半所需的时间。半衰期与衰变常数由公式 T½ = ln 2 / λ 联系在一起，即 T½ = 0.693 / λ。半衰期越长，衰变常数越小，样品衰变得越慢。这两个量互为反比，是计算题中最常用的一组关系。

The half-life (T½) is the most intuitive measure of how fast a sample decays: it is the time taken for the number of radioactive nuclei (or the activity) to fall to half its original value. The half-life and the decay constant are linked by T½ = ln 2 / λ, or T½ = 0.693 / λ. The longer the half-life, the smaller the decay constant and the slower the decay. These two quantities are inversely related and form one of the most frequently used pairs in calculation questions.

半衰期的应用非常广泛。考古学家用碳-14（半衰期约 5730 年）来测定古代有机物的年代；医学上用锝-99m（半衰期约 6 小时）作为示踪剂，因为它衰变得足够快，不会让病人长期暴露在辐射中，又足够慢，能在检查完成前持续发出可探测的信号。选择同位素时，半衰期必须与用途相匹配，这也是常考的评估类问题。

Half-life has wide-ranging applications. Archaeologists use carbon-14 (half-life about 5730 years) to date ancient organic material. Medicine uses technetium-99m (half-life about 6 hours) as a tracer because it decays fast enough not to leave the patient exposed for long, yet slowly enough to keep emitting a detectable signal until the scan is complete. When choosing an isotope, the half-life must match the purpose, and this is a common evaluation-style exam question.

5. Activity and Count Rate: Measuring How Fast a Sample Decays | 活度与计数率：测量样品衰变的快慢

活度（activity, A）定义为每秒发生的衰变次数，单位是贝克勒尔（Bq），1 Bq = 每次衰变每秒。活度与尚未衰变的核数成正比，A = λN，因此活度同样随时间按指数规律衰减：A = A₀e^(−λt)。这是一个非常重要的结论，因为实验通常测量的是活度或计数率，而不是直接数原子核的个数。

Activity (A) is defined as the number of decays per second, measured in becquerels (Bq), where 1 Bq equals one decay per second. Activity is proportional to the number of undecayed nuclei, A = λN, so activity also decays exponentially with time: A = A₀e^(−λt). This is a crucial result because experiments usually measure activity or count rate rather than counting nuclei directly.

在实际实验中，盖革-米勒计数器记录到的”计数率”（count rate）并不等于活度，因为探测器只能捕获到一部分衰变（几何因素、探测效率、以及样品到探测器的距离都会影响结果），同时还存在环境本底辐射。处理这类实验数据时，必须先减去本底计数率，再对结果进行分析。忽略本底是实验题中最常见的失分原因之一。

In practice, the count rate recorded by a Geiger-Müller counter is not equal to the activity, because the detector captures only a fraction of the decays (geometry, detector efficiency and the sample-to-detector distance all matter), and there is also background radiation from the environment. When analysing such data, you must first subtract the background count rate before drawing conclusions. Forgetting to subtract background is one of the most common reasons for losing marks in experimental questions.

当样品含有半衰期很短的同位素，或测量时间跨度远小于半衰期时，计数率在一小段时间内可近似看作不变。反之，测量半衰期本身时，可以通过记录计数率随时间的变化，绘出计数率对时间的图像，再从中读取半衰期：每过半个半衰期，计数率就减半。能从图像中准确读出半衰期是一项明确的考试技能。

When a sample contains a very short-lived isotope, or when the measurement time span is much smaller than the half-life, the count rate can be treated as roughly constant over a short interval. Conversely, to measure a half-life itself, you record how the count rate changes with time, plot count rate against time, and read the half-life from the graph: every half-life, the count rate halves. Reading a half-life accurately from a graph is a specific exam skill.

6. Mass Defect and Binding Energy: Where Nuclear Energy Comes From | 质量亏损与结合能：核能量从何而来

核物理中最反直觉的事实之一，是原子核的质量总是小于组成它的各个核子质量之和。这个差值被称为质量亏损（mass defect, Δm）。根据爱因斯坦的质能方程 E = mc²，这一”消失”的质量其实转化成了把核子束缚在一起的能量，也就是结合能（binding energy）。质量亏损越大，核子被束缚得越牢固。

One of the most counterintuitive facts in nuclear physics is that the mass of a nucleus is always less than the sum of the masses of its individual nucleons. This difference is called the mass defect (Δm). According to Einstein’s mass-energy equation E = mc², this missing mass has actually been converted into the energy that binds the nucleons together, namely the binding energy. The larger the mass defect, the more tightly the nucleons are held.

计算结合能通常分三步：先求出质量亏损 Δm（用核子总质量减去核质量，单位统一成 kg 或 u），再用 E = Δmc² 算出能量，最后换算成 MeV 或 J。计算中要特别注意单位：原子质量单位 1 u ≈ 931.5 MeV/c²，这个换算因子是考试计算题的基石。答题时务必先写出质量亏损的表达式，再代入能量公式，步骤分往往比最终答案更值钱。

Calculating binding energy usually involves three steps: find the mass defect Δm (total nucleon mass minus the nuclear mass, converting units consistently to kg or u), then use E = Δmc² to find the energy, and finally convert to MeV or J. Pay close attention to units: one atomic mass unit is 1 u ≈ 931.5 MeV/c², a conversion factor that is the bedrock of exam calculations. Always write out the mass-defect expression before substituting into the energy formula, as method marks often outweigh the final answer.

更有用的是”每个核子的结合能”（binding energy per nucleon），即总结合能除以核子数。把它对质量数作图，会得到一条先升后降的曲线，峰值大约出现在铁-56 附近。位于峰值附近的核最稳定；质量数比铁小得多的轻核（如氢、氦）以及比铁大得多的重核（如铀）结合能都较低。这条曲线解释了裂变与聚变为何都能释放能量：两者都是向更稳定的中间区域”移动”。

More useful is the binding energy per nucleon, the total binding energy divided by the number of nucleons. Plotting this against mass number gives a curve that rises then falls, peaking near iron-56. Nuclei near the peak are the most stable; light nuclei well below iron (such as hydrogen and helium) and heavy nuclei well above it (such as uranium) both have lower binding energy per nucleon. This curve explains why both fission and fusion release energy: each moves towards the more stable middle region.

7. Nuclear Fission: Splitting Heavy Nuclei and Chain Reactions | 核裂变：分裂重核与链式反应

核裂变（nuclear fission）是指一个重核（如铀-235 或钚-239）吸收一个慢中子后，分裂成两个较轻的裂变碎片，同时释放出能量和两到三个中子的过程。释放的能量来自产物碎片比原来的重核具有更高的”每核子结合能”，两者之差就是裂变释放的能量。铀-235 裂变时，每个核释放的能量约为 200 MeV，远大于任何化学反应。

Nuclear fission is the process in which a heavy nucleus such as uranium-235 or plutonium-239 absorbs a slow neutron and splits into two lighter fission fragments, releasing energy and two or three further neutrons. The energy released comes from the products having a higher binding energy per nucleon than the original heavy nucleus; the difference is the energy liberated. When uranium-235 fissions, each nucleus releases roughly 200 MeV, vastly more than any chemical reaction.

裂变释放的中子可以继续轰击其他铀-235 核，引发更多裂变，形成链式反应（chain reaction）。要让链式反应持续，必须满足两个条件：中子的速度要足够慢（所以反应堆中使用慢化剂，如石墨或水），以及裂变材料的质量要超过临界质量。若中子数量失控增长，反应会爆炸式加速；核反应堆的核心任务就是通过控制棒（吸收中子）把反应控制在稳定的速率。

The neutrons released by fission can go on to strike other uranium-235 nuclei, triggering further fissions and creating a chain reaction. For the chain reaction to sustain itself, two conditions must be met: the neutrons must be slowed down (which is why reactors use moderators such as graphite or water), and the mass of fissile material must exceed the critical mass. If the neutron population grows out of control, the reaction accelerates explosively; the core task of a nuclear reactor is to hold the reaction at a steady rate using control rods that absorb neutrons.

核反应堆的各个部件各司其职，考试经常要求你逐一说明它们的作用：燃料棒提供铀-235；慢化剂减慢中子速度以提高裂变概率；控制棒吸收多余中子以调节反应速率；冷却剂带走热量用于发电；屏蔽层阻挡逃逸的辐射。能够把每个部件与它的功能一一对应，是拿到这道”解释反应堆如何工作”题满分的关键。

Each component of a nuclear reactor has a specific job, and exams frequently ask you to explain them one by one: the fuel rods supply uranium-235; the moderator slows neutrons to increase the fission probability; the control rods absorb excess neutrons to regulate the rate; the coolant carries heat away for electricity generation; and the shielding blocks escaping radiation. Being able to match each component to its function is the key to full marks on the explain-how-a-reactor-works question.

8. Nuclear Fusion: Joining Light Nuclei in the Stars | 核聚变：恒星中轻核的融合

核聚变（nuclear fusion）是裂变的反过程：两个轻核（通常是氢的同位素氘和氚）结合成一个更重的核（氦），并释放出巨大的能量。轻核在聚合成靠近铁-56 的核时，每核子结合能上升，因此同样有能量释放。太阳及所有恒星的能量就来自聚变 – 太阳内部每秒钟都在把大约 6 亿吨氢转化成氦。

Nuclear fusion is the reverse of fission: two light nuclei, typically the hydrogen isotopes deuterium and tritium, combine to form a heavier nucleus (helium), releasing enormous energy. When light nuclei fuse into a nucleus closer to iron-56, the binding energy per nucleon rises, so energy is again released. The energy of the Sun and all stars comes from fusion, with the Sun converting roughly 600 million tonnes of hydrogen into helium every second.

聚变要发生，两个原子核必须靠得足够近，让强核力压过它们之间的静电排斥。这要求极高的温度和压强，因此聚变被称为”热核”反应。在地球上，科学家用磁约束（托卡马克装置）或惯性约束来把高温等离子体约束住。为什么聚变如此吸引人？因为它所需的燃料氘可以从海水中大量提取，产物基本无长寿命放射性废料，而且单次反应释放的能量远高于裂变。

For fusion to occur, the two nuclei must come close enough for the strong nuclear force to overcome their electrostatic repulsion. This demands extremely high temperatures and pressures, which is why fusion is described as thermonuclear. On Earth, scientists confine the hot plasma using magnetic confinement (tokamak devices) or inertial confinement. Why is fusion so attractive? Because its fuel, deuterium, can be extracted in abundance from seawater, the products leave almost no long-lived radioactive waste, and a single reaction releases far more energy than fission.

尽管聚变原理清晰，实现可控聚变仍是世界性难题：等离子体温度超过 1 亿摄氏度，任何容器都会被瞬间熔化，只能用磁场来”悬浮”它；同时，维持反应所需的能量目前常常超过反应释放的能量。考试中对聚变的考察通常聚焦于三点：为什么需要高温、为什么目前难以商用，以及它与裂变在能量来源和产物上的区别。

Although the principle is clear, achieving controlled fusion remains a global challenge: the plasma exceeds 100 million degrees Celsius, which would instantly melt any container, so it must be suspended by magnetic fields; meanwhile, the energy needed to sustain the reaction currently often exceeds the energy it releases. Exam questions on fusion typically focus on three points: why high temperatures are needed, why commercial fusion is still difficult, and how it differs from fission in energy source and products.

9. Radiation Hazards, Uses and Safety | 辐射的危害、应用与安全

电离辐射对人体有害，因为它能电离细胞中的原子，破坏 DNA 和细胞结构。短期大剂量照射会导致辐射病，长期低剂量照射则会增加患癌风险。辐射防护遵循三条基本原则：尽量减少受照时间、尽量远离辐射源、并在必要时使用屏蔽。辐射源的处理、使用和废弃都必须严格遵守规范。

Ionising radiation is harmful because it ionises atoms inside cells, damaging DNA and cell structures. A large short-term dose causes radiation sickness, while long-term low-dose exposure raises the risk of cancer. Radiation protection follows three basic principles: minimise exposure time, maximise distance from the source, and use shielding when necessary. Radioactive sources must be handled, used and disposed of in strict accordance with regulations.

然而，辐射在受控条件下有着广泛的正面用途。医学上，γ 射线用于对癌细胞进行放射治疗和杀灭医疗器具上的细菌；示踪剂（如碘-131）用于追踪甲状腺功能；α 粒子则被用于烟雾探测器。工业上，γ 射线用于检测金属焊缝和管道中的裂纹（无损探伤），以及测量材料的厚度。农业上，辐射还被用来延长食品保质期和培育抗病作物新品种。

Yet radiation has many beneficial uses when properly controlled. In medicine, gamma rays are used in radiotherapy to destroy cancer cells and to sterilise medical equipment; tracers such as iodine-131 track thyroid function; and alpha particles power smoke detectors. In industry, gamma rays detect cracks in metal welds and pipes (non-destructive testing) and measure material thickness. In agriculture, radiation extends food shelf life and helps breed disease-resistant crop varieties.

回答”某种用途为什么选择这种辐射”的问题时，要把辐射的性质与用途的需求对应起来：放射治疗需要穿透人体到达肿瘤，所以选 γ；示踪剂需要能被体外探测器跟踪，所以选发射 γ 的短半衰期同位素；烟雾探测器需要强电离能力来让空气导电，所以选 α。性质、用途、理由三者的对应，是 AQA 评价类问题的标准答题结构。

When answering why a particular use selects a particular radiation, match the radiation’s properties to the needs of the application: radiotherapy needs to penetrate the body to reach a tumour, so gamma is chosen; tracers need to be tracked by an external detector, so a short-half-life gamma emitter is chosen; smoke detectors need strong ionisation to make air conductive, so alpha is chosen. Matching property, use and reason is the standard answer structure for AQA evaluation questions.

10. Exam Technique: The Four Question Types You Must Master | 考试技巧：必须掌握的四种题型

AQA 核物理部分的题目可以归纳为四类，掌握了它们就掌握了大部分分数。第一类是”配平方程题”：给出一个不完整的衰变方程，要求你补齐缺失的粒子或核素，核心是质量数和原子序数守恒。第二类是”半衰期计算题”：给定初值和半衰期，求若干时间后的剩余量，或反过来求经过的时间，关键是熟练运用 N = N₀e^(−λt) 以及”每过半个半衰期数量减半”的捷径。

Questions on nuclear physics in AQA papers can be grouped into four types, and mastering them means mastering most of the marks. The first is the balancing-equation question: given an incomplete decay equation, complete the missing particle or nuclide, relying on conservation of mass number and atomic number. The second is the half-life calculation: given an initial value and a half-life, find the remaining amount after some time, or work out the elapsed time in reverse, with the key being fluency in N = N₀e^(−λt) and the shortcut that every half-life halves the quantity.

第三类是”结合能计算题”：求质量亏损、再用 E = Δmc² 计算能量，注意单位换算（1 u ≈ 931.5 MeV/c²）。第四类是”解释与评价题”：解释反应堆部件的作用、比较裂变与聚变、或论证某种同位素适用于某种用途，这类题要求用物理原理组织答案，而不是堆砌术语。无论哪一类，都要先写出公式或守恒关系，再代入数据，最后给出带单位的答案。

The third is the binding-energy calculation: find the mass defect, then compute the energy using E = Δmc², taking care with unit conversion (1 u ≈ 931.5 MeV/c²). The fourth is the explain-and-evaluate question: explain the role of reactor components, compare fission and fusion, or justify why a particular isotope suits a particular use, requiring you to organise your answer around physical principles rather than piling up terminology. Whichever type you face, always write the formula or conservation relation first, substitute the data, and finish with an answer carrying its unit.

一个常被忽视的细节是有效数字。核物理计算中的数据往往只有两位有效数字（例如半衰期给到 5730 年），最终答案不应给出过高的精度。另一个要点是”估计数量级”的能力 – AQA 有时要求你先估算一个量的大小，再判断某个说法是否合理，这类题考察的是物理直觉而非精确计算。

One often-overlooked detail is significant figures. Nuclear-physics data frequently carry only two significant figures (for example a half-life given as 5730 years), so the final answer should not claim excessive precision. Another point is the ability to estimate order of magnitude: AQA sometimes asks you to estimate the size of a quantity first and then judge whether a claim is reasonable, testing physical intuition rather than exact calculation.

Summary | 总结

核物理是 AQA A-Level 物理中逻辑清晰、规律性强的一个板块。核心内容可以浓缩为几条主线：原子核因质子-中子比例失衡而不稳定，通过 α、β、γ 三种辐射衰变回到稳定线；衰变遵循指数规律，由半衰期与衰变常数描述；质量亏损通过 E = mc² 转化为结合能，每核子结合能曲线解释了裂变与聚变为何释放能量；裂变链式反应驱动核电站，聚变则点亮了恒星。

Nuclear physics is a logically clear, rule-governed section of AQA A-Level Physics. The core content condenses into a few threads: nuclei become unstable when the proton-neutron ratio is unbalanced and decay back towards the line of stability via alpha, beta and gamma radiation; decay follows an exponential law described by the half-life and decay constant; mass defect converts into binding energy through E = mc², and the binding-energy-per-nucleon curve explains why fission and fusion release energy; fission chain reactions power nuclear stations, while fusion lights up the stars.

掌握这门内容的关键在于把守恒定律、公式和”性质与用途的对应”三者结合起来。配平方程靠质量数与原子序数守恒；半衰期与活度靠指数公式；结合能靠质能方程与单位换算；解释题靠把物理性质与具体用途对应起来。多做这些结构化、带单位的计算，并在实验数据中记得扣除本底，就能在这部分稳拿高分。

The key to mastering this material is combining conservation laws, formulas, and the property-to-use correspondence. Balance equations using mass-number and atomic-number conservation; handle half-life and activity with the exponential formula; work out binding energy with the mass-energy equation and unit conversion; and answer explanation questions by matching physical properties to specific applications. Practise these structured, unit-bearing calculations, and remember to subtract background in experimental data, and you will score reliably well on this section.

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