AQA物理 粒子与辐射 夸克 光电效应 能级
粒子与辐射是 AQA A-Level Physics 7408 的第 3.2 章，按官方 spec（2015 年起施教，A-level 首考 2017 年）安排在第 1 卷，与第 1 到 5 章和 6.1 一起考。这一章的内容分成两半：前半是原子核与粒子物理，后半是量子现象。它看起来零碎，但只要把几条守恒律和两个公式吃透，题目几乎都是同一套路。第 1 卷共 85 分，其中 60 分是短答与长答，25 分是选择题，粒子与辐射通常占十几分。下面的写法按解题顺序走，从比荷和核素记号开始，一直写到德布罗意波长。每一节都配一个能直接套用的计算例。
Particles and radiation is section 3.2 of AQA A-Level Physics 7408, examined in Paper 1 alongside sections 1 to 5 and 6.1. The material splits into two halves that behave very differently: nuclear and particle physics, then quantum phenomena. The particle half rewards careful bookkeeping rather than insight, because charge, baryon number and lepton number have to balance in every equation you write. The quantum half is built on two formulas, so most of the marks come from substituting correctly and converting units before you reach for the calculator. What follows is the order I would work through the questions in, with one worked calculation per section so you can check your own method against it.
比荷与核素记号：代入公式前先确定电荷和质量
比荷是电荷量与质量的比值，电子的比荷是 1.76 × 10^11 C/kg，这个值在 AQA 的数据表里直接给出。质子的比荷小得多，大约是 9.58 × 10^7 C/kg，因为质子质量约是电子的 1836 倍。算原子核的比荷时，电荷是 Z × 1.60 × 10^-19 C，质量是 A × 1.66 × 10^-27 kg，Z 和 A 分别来自核素记号的下标和上标。很多学生以为比荷是电子的专属概念，其实任何带电体都能算，离子还要在电荷上加减它失去的电子数。常见的错误是把核的质量写成单个质子质量而不乘 A，或者忘记电荷那一项要乘 Z。举个数：氧 16 的核，8 × 1.60 × 10^-19 除以 16 × 1.66 × 10^-27，得到 4.82 × 10^7 C/kg。同位素的比荷互不相同，因为 A 变了而 Z 没变，这一点常在选择题里出现。
Specific charge is charge divided by mass, and the electron value of 1.76 × 10^11 C/kg is printed on the AQA data sheet. A proton is roughly 1836 times heavier, which pulls its own value down to about 9.58 × 10^7 C/kg. For a nucleus the charge is Z × 1.60 × 10^-19 C and the mass is A × 1.66 × 10^-27 kg, both read straight from the nuclide notation; an ion simply keeps that mass and changes the charge. Isotopes of one element therefore differ in specific charge, since the proton number is unchanged while the nucleon number is not. Oxygen-16 works out at 4.82 × 10^7 C/kg once the numbers are substituted.
强核力的作用范围与衰变方程配平
强核力负责把核子困在核内，它的行为分成两段：距离超过约 3 fm 基本没有吸引力，小于约 0.5 fm 时反而变成斥力。这两个数字写在 AQA spec 的 3.2.1.2 里，考简答题时可以直接引用。原子核能稳定存在靠的是吸引与斥力之间的平衡，核子数越多，需要额外提供的中子就越多。α 衰变写成 A 减 4、Z 减 2，β 负衰变里中子变成质子、电子和反中微子，A 不变而 Z 加 1。中微子最初是泡利在 1930 年为解释 β 衰变的能量守恒而假设的，这一点常以简答题形式出现。配平方程只需检查两条：两边质量数之和相等，电荷数之和也相等。错得最多的是 β 衰变只写一个电子，漏掉反中微子，而反中微子正是同时满足能量守恒与轻子数守恒的那一项。
The strong nuclear force holds nucleons together and behaves in two regimes, with attraction up to roughly 3 fm and repulsion inside about 0.5 fm; both figures appear in section 3.2.1.2 of the AQA spec; heavier nuclei need proportionally more neutrons to remain stable. Alpha decay removes two protons and two neutrons at once, so the nucleon number falls by four; in beta minus decay a neutron turns into a proton, an electron and an antineutrino, which raises the proton number by one and leaves the nucleon number untouched. Pauli proposed the neutrino in 1930 to rescue energy conservation in beta decay, and examiners expect that name and year to appear. Balancing these equations needs only two tallies, mass numbers and charges.
反粒子、湮灭与电子对产生的能量账
每种粒子都有对应的反粒子，正电子、反质子、反中子和反中微子分别对应电子、质子、中子和中微子。它们质量相同、电荷相反，相遇时发生湮灭，静质量全部转成能量。慢的正负电子对湮灭产生两个 γ 光子，每个光子带走 0.511 MeV，也就是 8.18 × 10^-14 J，两光子能量之和正好等于 2 m c^2。为什么一定是两个光子而不是一个？因为动量也要守恒，静止状态下单个光子无法带走动量。反向过程叫电子对产生，一个高能光子在核附近转成电子与正电子，光子能量至少要 1.02 MeV，低于这个门槛，光束再强也不会发生。PET 扫描仪就是湮灭的应用，spec 里点名的例子，两个方向相反的光子被探测器同时接收，就能反推病灶位置。
Every particle has a matching antiparticle with equal mass and opposite charge, and when the two meet they annihilate into pure energy. A slow electron and a slow positron give two gamma photons of 0.511 MeV each, so the pair releases 1.02 MeV in total. Two photons rather than one is forced by momentum conservation, since a single photon cannot carry the momentum of a stationary pair. Run the process backwards and you get pair production, where a photon of at least 1.02 MeV near a nucleus creates an electron and a positron; below that threshold intensity makes no difference at all. PET scanners exploit the signature, because the two back-to-back photons reach the detectors almost simultaneously.
交换粒子与四种基本相互作用
四种基本相互作用是引力、电磁力、弱核力和强核力，前两种与日常生活直接相关，后两种只在核尺度上出现。按场论的说法，力靠交换粒子传递：电磁力交换虚光子，弱作用交换 W 与 Z 玻色子，强作用在夸克之间交换胶子，而核子之间实际交换的是 π 介子。AQA 明确不考胶子和引力子的细节，但要求知道名称以及与哪种力对应。费曼图把过程画成时间和空间的图，横轴是时间，斜线表示交换粒子，每个顶点处电荷必须守恒。β 负衰变的图是中子放出一个 W 负粒子，W 负再变成电子和反中微子，看上去分三步，物理上是一瞬间的事。阅卷时经常看到学生把交换粒子画成水平线，那样时间顺序就错了。
Gravity, electromagnetism, the weak nuclear force and the strong nuclear force are the four fundamental interactions, and each is carried by its own exchange particle; virtual photons mediate the electromagnetic force, W and Z bosons carry the weak interaction, gluons act between quarks, and nucleons exchange pions. AQA does not test gluon or graviton detail, but it does expect each force to be matched to the right carrier; on a Feynman diagram the horizontal axis is time, so an exchange particle appears as a slanted line, and charge must be conserved at every vertex. Beta minus decay shows a neutron emitting a W boson that then splits into an electron and an antineutrino; drawing that exchange particle horizontally is a common way to lose the mark.
强子分类与三个守恒量
参与强作用的粒子叫强子，分两类：重子由三个夸克组成，介子由一个夸克加一个反夸克组成。质子是唯一稳定的重子，其他重子最终都会衰变成质子，这句话在简答题里可以直接用。轻子不参与强作用，电子、μ 子和对应的中微子都属于这一类，μ 子会衰变成电子。守恒量要看三个：电荷、重子数、轻子数，其中轻子数还要分电子型与 μ 型分别守恒。奇异数在强作用里守恒，在弱作用里可以变 0、加 1 或减 1，奇异粒子总是成对产生、单独衰变，这正是它们看起来奇异的原因。检查选项时先数电荷，再数重子数，接着看轻子数，最后看奇异数，四步里总有一处能排除错误项。
Hadrons feel the strong interaction and come in two classes, baryons built from three quarks and mesons built from a quark plus an antiquark; only the proton is stable, and every other baryon eventually decays into one. Leptons ignore the strong force altogether, and the muon simply decays into an electron; three conservation rules do most of the work in these questions: charge, baryon number and lepton number, with electron and muon lepton numbers counted separately. Strangeness is conserved in strong interactions but changes by 0, +1 or -1 in weak ones, which is why strange particles are produced in pairs and decay alone. Testing charge first, then baryon number and strangeness, will eliminate a wrong multiple-choice option almost every time.
夸克组合与 β 衰变里的夸克变化
AQA 只考三个夸克：上夸克带电 +2/3，下夸克和奇夸克都是 -1/3，三者的重子数都是 1/3。质子是 uud，净电荷 +1，中子是 udd，净电荷刚好为 0。介子需要一个夸克加一个反夸克：π 的正电版本是 u 加反 d，负电版本是 d 加反 u，K 加是 u 加反 s。β 负衰变在夸克层面是 d 变成 u，同时放出一个 W 负粒子，β 正衰变和电子俘获则是 u 变成 d。这条规则顺带解释了为什么 β 衰变算弱作用：弱作用能改变夸克味，强作用做不到。题目要求写出中子衰变的夸克式时，写 udd 到 uud，再加电子与反中微子，这两项都不能漏。
AQA tests only three quarks, up at +2/3 and down and strange both at -1/3, each carrying baryon number 1/3; a proton is uud and a neutron is udd, which produces charges of +1 and 0 without any further assumptions. Mesons take one quark and one antiquark, so the positive pion is u plus anti-down and the negative pion is d plus anti-up; at quark level beta minus decay is a down quark turning into an up quark while emitting a W boson, and beta plus decay runs the other way. That single change of flavour explains why beta decay belongs to the weak interaction, since the strong force cannot alter quark flavour.
光电效应方程与图像的两条读出规则
光电效应方程是 hf 等于功函数 φ 加上光电子的最大动能，两边都用焦耳。图像题考两条：最大动能对频率作图的斜率是普朗克常数，横向截距是阈值频率，纵向截距是负的功函数。强度只影响飞出光电子的数量，不改变单个电子的最大动能，这是概念题上最容易丢掉的分。遏止电压与最大动能的关系是 e 乘 V 等于最大动能，所以把遏止电压乘以 1.60 × 10^-19 就换算成焦耳。举个数：锌的功函数是 4.3 eV，也就是 6.9 × 10^-19 J，用 250 纳米的光照射，光子能量是 7.96 × 10^-19 J，剩下的 1.08 × 10^-19 J 就是最大动能，对应遏止电压约 0.67 V。锌的阈值波长约 289 纳米，落在紫外区，所以演示时要用紫外灯而不是白炽灯泡。
The photoelectric equation sets the photon energy equal to the work function plus the maximum kinetic energy of the emitted electron; on a graph of maximum kinetic energy against frequency, the gradient gives the Planck constant, the intercept on the frequency axis is the threshold frequency, and the vertical intercept is minus the work function. Intensity controls how many electrons leave the metal, never how much energy any single electron carries, and that distinction is where most marks are dropped. For zinc with a work function of 4.3 eV, light at 250 nm supplies photons of 7.96 × 10^-19 J, leaving 1.08 × 10^-19 J of kinetic energy and a stopping potential near 0.67 V. The threshold wavelength for zinc sits at about 289 nm in the ultraviolet, which is why a UV lamp rather than a filament bulb is used in the demonstration.
能级跃迁、电离与电子伏换算
电子只能停留在离散的能级上，跃迁时吸收或放出的光子能量恰好等于两能级之差，线性光谱就是这件事的证据。激发是把电子抬到更高的能级，电离则是把它彻底移出原子，氢原子基态的电离能是 13.6 eV。这两个词在简答题里经常被互换使用而丢分，写答案时先想清楚电子是留在原子内还是离开了原子。换算关系是 1 eV 等于 1.60 × 10^-19 J，两个方向都要能算，题目给的能级有时用 eV 有时用 J。荧光灯管的原理写在 3.2.2.2 里：电子与汞原子碰撞使汞原子激发，回到低能级时放出紫外光子，紫外光子被管壁的荧光粉吸收后放出可见光。氢原子从 -3.40 eV 跃到 -13.6 eV，能量差是 10.2 eV，换成焦耳再除以普朗克常数得到频率 2.46 × 10^15 Hz，波长 122 纳米，属于紫外区。
Electrons sit on discrete energy levels, and a transition absorbs or emits a photon whose energy matches the gap exactly; line spectra are the evidence for this; excitation lifts an electron to a higher level, while ionisation removes it from the atom altogether, and hydrogen needs 13.6 eV for that from the ground state. Since 1 eV equals 1.60 × 10^-19 J, AQA expects conversion in both directions, which matters when a question quotes levels in eV and asks for a wavelength. The fluorescent tube named in section 3.2.2.2 works by electron collisions exciting mercury atoms, whose ultraviolet emission is then converted to visible light by the coating. A jump from -3.40 eV to -13.6 eV releases 10.2 eV, a 122 nm photon in the ultraviolet.
德布罗意波长与电子衍射的判据
德布罗意波长等于普朗克常数除以动量，只要有了动量，宏观物体也能算出一个极小的波长。电子穿过石墨薄膜会出现同心衍射环，这个现象只能用量子力学解释，因为只有波才会衍射。加速电压越高，电子动量越大，波长越短，衍射环就越收紧，题目问衍射程度如何变化时答的正是这条逻辑。1927 年的戴维孙与革末实验用镍晶体测到了衍射峰，验证了德布罗意在 1924 年提出的假设。举个数：电子经 50 V 加速，动能是 8.0 × 10^-18 J，动量约 3.8 × 10^-24 kg m/s，波长约 1.7 × 10^-10 m，也就是 0.17 纳米，与原子间距同一数量级，所以衍射环才看得清楚。题目只要求解释动量和波长的关系，不要求记住具体实验装置的细节。
The de Broglie wavelength is the Planck constant divided by momentum, and because the constant is so small, everyday objects end up with wavelengths far too tiny to observe; fire electrons through graphite and you get concentric diffraction rings, which is evidence that particles behave as waves. Increase the accelerating voltage and the momentum rises, so the wavelength shortens and the rings tighten, and that relationship is normally what the question is testing; the Davisson and Germer experiment of 1927 measured diffraction peaks from a nickel crystal, confirming the hypothesis de Broglie published in 1924. An electron accelerated through 50 V carries 8.0 × 10^-18 J, giving a momentum near 3.8 × 10^-24 kg m/s and a wavelength of about 0.17 nm, comparable to atomic spacing.
第 1 卷里粒子题的检查顺序
按 AQA 7408 的官方安排，粒子与辐射在第 1 卷与第 1 到 5 章一起考，全卷 2 小时 85 分，其中 60 分短答与长答，25 分选择题。做计算之前先看单位：兆电子伏要换成焦耳，纳米和飞米要换成米，这三处是历年错误最集中的地方。有效数字按题目给的位数处理，数据表里的常数一般给三位，中间步骤多留一位可以减少累积误差。粒子类选择题有一个固定套路：先看电荷守恒，再看重子数和轻子数，最后看奇异数，四条里只要有一条不成立就可以排除。做完一题回读一遍题目问的是能量还是波长，阅卷时这两种答案互换的情况相当常见。考试当天把数据表放在手边，不要凭记忆写普朗克常数和电子电荷，AQA 的目的是考物理而不是考验背诵。
In the official AQA 7408 scheme this topic is examined in Paper 1 alongside sections 1 to 5, and that paper lasts two hours with 85 marks, 60 of them short and extended answers and 25 multiple choice; check units before substituting, converting MeV into joules and nanometres or femtometres into metres, because those conversions cost more marks than the physics itself. Round to the significant figures the question implies, keeping one extra digit through the intermediate steps. In multiple-choice particle questions, test charge first, then baryon number, lepton number and strangeness, and stop at the first law that fails. Reread whether the answer wanted energy or wavelength as well, since swapping the two is a very common slip.
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