High School Physics

Modern and Nuclear Physics近代物理与核物理

Modern physics rewrote two millennia of intuition in a single generation. This guide covers the quantum revolution (the photoelectric effect and the photon, wave-particle duality), the story of atomic structure from Thomson through Bohr to quantum models, the physics of the nucleus (nucleons, binding energy and mass defect), the three radioactive decay modes ($\alpha$, $\beta$, $\gamma$), the half-life exponential $N = N_0(\tfrac{1}{2})^{t/T}$, and finally fission, fusion, and Einstein's mass-energy equation $E = mc^2$. Every concept is grounded in the curriculum documents of US NGSS, Ontario SPH3U/SPH4U, BC Physics 12, and Alberta Physics 30 Unit D.近代物理在短短一代人的时间里改写了两千年的直觉。本指南涵盖量子革命（光电效应与光子，波粒二象性）、从汤姆孙到玻尔再到量子模型的原子结构历程、原子核物理（核子、结合能与质量亏损）、三种放射性衰变模式（$\alpha$、$\beta$、$\gamma$）、半衰期指数规律 $N = N_0(\tfrac{1}{2})^{t/T}$，以及裂变、聚变与爱因斯坦质能方程 $E = mc^2$。每个概念都以美国 NGSS、安大略 SPH3U/SPH4U、BC Physics 12、阿尔伯塔 Physics 30 D 单元的课纲文件为依据。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB $E = hf$ · $E = mc^2$ · half-life$E = hf$ · $E = mc^2$ · 半衰期
Read me first先读这里

How to use this guide如何使用本指南

Modern and nuclear physics is the capstone unit of the high school physics sequence. The four curricula we map to agree on the quantum and nuclear core — photoelectric effect, atomic spectra, radioactive decay, half-life, and the mass-energy equation — but differ in emphasis. US NGSS (HS-PS1-8) keeps nuclear content qualitative, with no quantitative energy calculations. Ontario SPH3U introduces nuclear energy and half-life in Strand D; SPH4U's Strand F adds quantum mechanics and special relativity at a more conceptual level. Alberta Physics 30 Unit D goes deepest quantitatively: nuclear equations, half-life arithmetic, mass defect calculations, and the photoelectric effect as a worked quantum problem. BC Physics 12 approaches modern physics through special relativity; nuclear content is lighter than Alberta's. The table below tells you which sections are core for your course right now.近代物理与核物理是高中物理序列的顶点单元。我们对照的四套大纲在量子与核心核物理方面——光电效应、原子光谱、放射性衰变、半衰期、质能方程——基本一致，但侧重点各异。美国 NGSS（HS-PS1-8）把核物理内容保持在定性层面，不涉及能量的定量计算。安大略 SPH3U 在 D 单元引入核能与半衰期；SPH4U 的 F 单元在更概念性的层面补充量子力学与狭义相对论。阿尔伯塔 Physics 30 D 单元定量要求最高：核方程、半衰期计算、质量亏损计算，以及作为量子例题的光电效应。BC Physics 12 以狭义相对论为切入口进入近代物理；核物理内容比阿尔伯塔浅。下表告诉你当前哪些节属于你的核心内容。

If you are in…如果你在… Focus on these sections重点学习 Depth note深度说明 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1 (photoelectric effect), §2 (wave-particle duality), §5 (radioactivity), §6 (half-life), §7 (fission, fusion, $E=mc^2$) — covered by HS-PS1-8 and HS-PS4-3§1（光电效应）、§2（波粒二象性）、§5（放射性）、§6（半衰期）、§7（裂变、聚变、$E=mc^2$）——由 HS-PS1-8 与 HS-PS4-3 覆盖 NGSS keeps nuclear physics qualitative — Assessment Boundary for HS-PS1-8 explicitly excludes quantitative energy calculations; emphasis is on models and scaleNGSS 将核物理保持在定性层面——HS-PS1-8 评估边界明确排除能量的定量计算；重点是模型与规模 ngss_hs_ps_extract.md — HS-PS1-8 (nuclear processes) + HS-PS4-3 (wave-particle / photoelectric)— HS-PS1-8（核过程）+ HS-PS4-3（波粒 / 光电）
🇨🇦 ON Grade 11/12 — SPH3U / SPH4U安大略 11/12 年级 — SPH3U / SPH4U SPH3U: §5 (radioactivity), §6 (half-life, D3.11), §7 (fission/fusion via nuclear power plant, D3.12). SPH4U adds: §1 (photoelectric effect, F3), §2 (wave-particle duality, F3), §3 (atomic models/spectra, F3)SPH3U：§5（放射性）、§6（半衰期，D3.11）、§7（裂变/聚变通过核电站，D3.12）。SPH4U 增加：§1（光电效应，F3）、§2（波粒二象性，F3）、§3（原子模型/光谱，F3） SPH3U folds nuclear energy into the "Energy and Society" strand; the full quantum picture (wave-particle duality, Bohr model) is Grade 12 SPH4U Strand FSPH3U 将核能纳入"能量与社会"单元；完整的量子图像（波粒二象性、玻尔模型）属于 12 年级 SPH4U F 单元 science_11-12_physics_extract.md — SPH3U Strand D (D3.11, D3.12); SPH4U Strand F (F1–F3)— SPH3U D 单元（D3.11、D3.12）；SPH4U F 单元（F1–F3）
🇨🇦 BC Grade 12 — Physics 12BC 12 年级 — Physics 12 §1 and §2 (wave-particle framing connects to postulates of special relativity); §7 ($E = mc^2$ as the conceptual capstone of the relativistic framework). Sections §3–§6 are useful background but lighter in BC Physics 12§1 和 §2（波粒框架与狭义相对论假设相连）；§7（$E = mc^2$ 作为相对论框架的概念顶点）。§3–§6 是有用背景知识，但在 BC Physics 12 中较浅 BC treats modern physics primarily through special relativity (frames of reference, relativistic effects); nuclear content is not a dedicated strand — focus on conceptual understanding rather than nuclear equation balancingBC 主要通过狭义相对论（参考系、相对论效应）处理近代物理；核物理不是专门单元——侧重概念理解而非核方程配平 physics_11-12_extract.md — Physics 12 Content: postulates of special relativity, relativistic effects; Big Idea: measurement of motion depends on reference frame— Physics 12 内容：狭义相对论假设，相对论效应；大概念：运动的测量取决于参考系
🇨🇦 AB Grade 12 — Physics 30阿尔伯塔 12 年级 — Physics 30 All seven sections in full. Physics 30 Unit D covers the complete arc: atomic models (§3), photoelectric effect (§1), wave-particle duality (§2), spectra (§3), nuclear structure (§4), radioactive decay (§5), half-life (§6), fission/fusion/$E=mc^2$ (§7). Quantitative depth required throughout全部七节完整学习。Physics 30 D 单元覆盖完整弧线：原子模型（§3）、光电效应（§1）、波粒二象性（§2）、光谱（§3）、核结构（§4）、放射性衰变（§5）、半衰期（§6）、裂变/聚变/$E=mc^2$（§7）。全程要求定量深度 AB is the most quantitative: nuclear equations with isotope notation, half-life arithmetic, mass-defect $\to$ energy calculations (30–D3.6k), photon energy and photoelectric emission (30–C2.1k–30–C2.4k)AB 定量要求最高：带同位素符号的核方程、半衰期计算、质量亏损 $\to$ 能量计算（30–D3.6k）、光子能量与光电发射（30–C2.1k–30–C2.4k） physics_20-30_extract.md — Physics 30 Unit C GO2 (photoelectric/Compton); Unit D GO1–GO4 (atomic physics, spectra, nuclear fission/fusion)— Physics 30 C 单元 GO2（光电/康普顿）；D 单元 GO1–GO4（原子物理、光谱、核裂变/聚变）

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise five anchors: $E = hf$ (photon energy); light behaves as both wave and particle; the atom has a tiny dense nucleus surrounded by electrons in quantised orbits; radioactive decay comes in three types ($\alpha$, $\beta$, $\gamma$); half-life uses $N = N_0(\tfrac{1}{2})^{t/T}$; and $E = mc^2$ converts mass defect to energy. Read every cram-cheat box. The going-deeper derivations are for top-mark candidates and AB students.背熟五个锚点：$E = hf$（光子能量）；光既是波又是粒子；原子有一个微小致密的原子核，外有处于量子化轨道的电子；放射性衰变有三种（$\alpha$、$\beta$、$\gamma$）；半衰期用 $N = N_0(\tfrac{1}{2})^{t/T}$；$E = mc^2$ 把质量亏损转化为能量。读每个速记框。深入推导留给目标顶分及阿尔伯塔考生。

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If you are going for the top mark如果你目标顶分

For the photoelectric effect, know both the qualitative story (intensity increases current, not stopping voltage; frequency must exceed threshold) and the quantitative equation $E_k = hf - \phi$. For nuclear equations, practise balancing mass number and atomic number independently. For half-life, be comfortable with both the exponential formula and successive halving. For $E = mc^2$, know that 1 u of mass-defect releases about 931 MeV. Alberta 30 D requires all four of these at full calculation depth.对于光电效应，既要掌握定性内容（强度增加电流而非截止电压；频率必须超过阈值），也要掌握定量方程 $E_k = hf - \phi$。对于核方程，练习独立地配平质量数与原子序数。对于半衰期，对指数公式和逐次减半都要得心应手。对于 $E = mc^2$，知道 1 u 的质量亏损释放约 931 MeV。阿尔伯塔 30 D 要求以上四点均达到完整计算深度。

Curriculum depth note.课纲深度说明。 US NGSS explicitly limits nuclear-energy content to qualitative models (Assessment Boundary of HS-PS1-8: "does not include quantitative calculation of energy released"). Ontario SPH3U covers nuclear energy in the context of power plants; SPH4U adds the quantum mechanics framing. BC Physics 12 approaches modern physics through special relativity rather than nuclear equations. Alberta Physics 30 Unit D requires the full quantitative treatment. If your curriculum sits closer to the NGSS or BC end, focus on understanding the concepts in each section; the worked examples with numbers are designed for Ontario/Alberta depth but help everyone build intuition.美国 NGSS 明确把核能内容限定在定性模型（HS-PS1-8 评估边界："不包括能量释放的定量计算"）。安大略 SPH3U 在核电站背景下涵盖核能；SPH4U 补充量子力学框架。BC Physics 12 通过狭义相对论而非核方程进入近代物理。阿尔伯塔 Physics 30 D 单元要求完整定量处理。若你的课纲更接近 NGSS 或 BC 一端，侧重理解每节的概念；带数字的例题为安大略/阿尔伯塔深度设计，但有助于所有人建立直觉。
Section 1 · The Photoelectric Effect & the Photon第 1 节 · 光电效应与光子

The Photoelectric Effect and the Photon光电效应与光子

Light comes in packets — the photoelectric effect proves it.光以包的形式出现 — 光电效应证明了这一点。
  • The observation.实验现象。 Shining ultraviolet light on a metal surface ejects electrons. Increasing the frequency (not brightness) is what releases higher-energy electrons; below a threshold frequency $f_0$, no electrons are emitted at all, no matter how bright the light.用紫外线照射金属表面会打出电子。增加频率（而非亮度）才能释放更高能量的电子；低于阈频 $f_0$ 时，无论光多亮都没有电子射出。
  • Einstein's explanation (1905).爱因斯坦的解释（1905年）。 Light consists of discrete packets called photons (光子). Each photon carries energy $E = hf$, where $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$ is Planck's constant.光由称为光子的离散粒子组成。每个光子携带能量 $E = hf$，其中 $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$ 为普朗克常数。
  • Work function $\phi$.逸出功 $\phi$。 The minimum energy needed to free an electron from the metal surface. The kinetic energy of the ejected electron is $E_k = hf - \phi$ (zero if $hf < \phi$).从金属表面释放电子所需的最小能量。被射出电子的动能为 $E_k = hf - \phi$（若 $hf < \phi$ 则为零）。
$$ E = hf \qquad E_k = hf - \phi \qquad (\text{photoelectric equation}) $$ NGSS HS-PS4-3 names the photoelectric effect as an example of a phenomenon where "one model [particle] is more useful than the other." Alberta 30–C2.3k requires describing the photoelectric effect "in terms of the intensity and wavelength or frequency of the incident light."NGSS HS-PS4-3 把光电效应列为"一种模型（粒子）比另一种更有用"的现象例子。阿尔伯塔 30–C2.3k 要求"根据入射光的强度和波长或频率描述光电效应"。
Worked Example 1 · Photon energy and photoelectric threshold例题 1 · 光子能量与光电阈值

The work function of sodium is $\phi = 2.28\ \mathrm{eV}$. (a) What is the minimum frequency of light needed to eject an electron? (b) If light of frequency $8.00 \times 10^{14}\ \mathrm{Hz}$ hits the surface, what is the maximum kinetic energy of the ejected electron? Use $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$ and $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$.钠的逸出功 $\phi = 2.28\ \mathrm{eV}$。(a) 打出电子所需的最低频率是多少？(b) 若频率 $8.00 \times 10^{14}\ \mathrm{Hz}$ 的光照射表面，被射出电子的最大动能是多少？取 $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$，$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$。

(a) Threshold frequency: set $E_k = 0$.(a) 阈频：令 $E_k = 0$。

$$ hf_0 = \phi \;\Longrightarrow\; f_0 = \frac{\phi}{h} = \frac{2.28 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = \frac{3.648 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.50 \times 10^{14}\ \mathrm{Hz}. $$

(b) Kinetic energy at $f = 8.00 \times 10^{14}\ \mathrm{Hz}$.(b) $f = 8.00 \times 10^{14}\ \mathrm{Hz}$ 时的动能。

$$ E_k = hf - \phi = (6.63 \times 10^{-34})(8.00 \times 10^{14}) - 3.648 \times 10^{-19} $$ $$ = 5.304 \times 10^{-19} - 3.648 \times 10^{-19} = 1.656 \times 10^{-19}\ \mathrm{J} \approx 1.04\ \mathrm{eV}. $$

Since $f > f_0$, electrons are ejected and carry away $1.04$ eV of kinetic energy. Doubling the intensity (not frequency) would double the number of ejected electrons but would not change their individual kinetic energies.由于 $f > f_0$，电子被射出并带走 $1.04$ eV 的动能。将强度（而非频率）加倍会使射出电子数加倍，但不会改变每个电子的动能。

Light of a fixed frequency shines on a metal and ejects electrons. If the intensity of the light is doubled (frequency unchanged), what happens?固定频率的光照射金属并射出电子。若光的强度加倍（频率不变），会发生什么？
§1 · Q1
The kinetic energy of each ejected electron doubles每个被射出电子的动能加倍
The number of electrons ejected per second doubles; individual kinetic energy is unchanged每秒射出电子数加倍；每个电子的动能不变
No electrons are ejected because the frequency is below threshold没有电子射出，因为频率低于阈值
The kinetic energy of each ejected electron halves每个被射出电子的动能减半
In the photon model, intensity means more photons per second. Each photon still has the same energy $hf$ and each electron still gets the same $E_k = hf - \phi$. More photons $\to$ more electrons ejected, but individual kinetic energies are unchanged.在光子模型中，强度意味着每秒更多光子。每个光子的能量仍为 $hf$，每个电子的动能仍为 $E_k = hf - \phi$。更多光子 $\to$ 更多电子射出，但单个电子的动能不变。
Kinetic energy depends on frequency ($E_k = hf - \phi$), not intensity. Intensity controls the number of photons, hence the number of ejected electrons, not their individual energies.动能取决于频率（$E_k = hf - \phi$），而非强度。强度控制光子数量，因而控制射出电子数量，而非单个电子的能量。
A photon has frequency $5.0 \times 10^{14}\ \mathrm{Hz}$. What is its energy? ($h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$)一个光子频率为 $5.0 \times 10^{14}\ \mathrm{Hz}$。其能量是多少？（$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$）
§1 · Q2
$5.0 \times 10^{14}\ \mathrm{J}$
$6.63 \times 10^{-34}\ \mathrm{J}$
$1.3 \times 10^{-48}\ \mathrm{J}$
$3.3 \times 10^{-19}\ \mathrm{J}$
$E = hf = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.315 \times 10^{-19} \approx 3.3 \times 10^{-19}\ \mathrm{J}$.$E = hf = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.315 \times 10^{-19} \approx 3.3 \times 10^{-19}\ \mathrm{J}$。
Photon energy is $E = hf$. Multiply Planck's constant by the frequency.光子能量为 $E = hf$。将普朗克常数乘以频率。
Going deeper — why the wave model fails and the photon wins (NGSS HS-PS4-3, AB 30–C2.4k–30–C2.5k)深入 — 波动模型为何失败而光子模型获胜（NGSS HS-PS4-3，AB 30–C2.4k–30–C2.5k）

The classical wave model of light predicts that any frequency, given sufficient intensity, should eventually eject electrons — the energy just builds up over time. Experiments completely contradict this: below the threshold frequency $f_0$, no electrons are ejected even after hours of illumination, no matter how bright. And above $f_0$, electrons appear instantaneously, even in very dim light.

Einstein's photon model resolves both puzzles. Because each photon delivers its entire energy $hf$ to a single electron in one hit, either the energy is enough (if $hf \ge \phi$) or it is not — there is no gradual accumulation. Instantaneous emission follows naturally because only one photon is needed to free one electron. The photoelectric equation $E_k = hf - \phi$ was confirmed quantitatively by Millikan in 1916, winning Einstein the 1921 Nobel Prize for precisely this explanation, not for relativity.经典波动模型预测：在足够强度下，任何频率的光最终都应能射出电子——能量只是随时间积累。实验完全与此矛盾：低于阈频 $f_0$，无论多亮、照多长时间都没有电子射出；而在 $f_0$ 以上，即使极暗的光也会即刻产生电子。

爱因斯坦的光子模型同时解决了两个疑惑。由于每个光子将全部能量 $hf$ 一次性传给单个电子，能量要么足够（若 $hf \ge \phi$），要么不够——没有逐渐积累。即时发射自然而然，因为释放一个电子只需一个光子。光电方程 $E_k = hf - \phi$ 于 1916 年经密立根定量验证，爱因斯坦因此而非相对论赢得 1921 年诺贝尔物理学奖。

Section 2 · Wave-Particle Duality第 2 节 · 波粒二象性

Wave-Particle Duality波粒二象性

Light — and matter — behave as both wave and particle depending on the experiment.光——以及物质——在不同实验中既表现为波也表现为粒子。
  • Wave evidence for light.光的波动证据。 Double-slit interference, diffraction, and polarization are all wave phenomena that light displays. These cannot be explained by particles alone.双缝干涉、衍射与偏振都是光所展现的波动现象，单纯用粒子无法解释。
  • Particle evidence for light.光的粒子证据。 The photoelectric effect and the Compton effect (photon-electron scattering) show light behaving as discrete energy packets (photons). Wave models predict neither correctly.光电效应和康普顿效应（光子-电子散射）表明光以离散能量包（光子）运动。波动模型无法正确预测两者。
  • de Broglie hypothesis (1924).德布罗意假设（1924年）。 If light (a wave) can act as a particle, perhaps matter (particles) can act as waves. The de Broglie wavelength of any particle is $\lambda = h/p = h/(mv)$. Confirmed by electron diffraction (Davisson-Germer 1927).若光（波）能表现为粒子，也许物质（粒子）也能表现为波。任何粒子的德布罗意波长为 $\lambda = h/p = h/(mv)$。由电子衍射（戴维孙-革末，1927年）证实。
$$ \lambda = \frac{h}{p} = \frac{h}{mv} \qquad \text{(de Broglie wavelength)} $$ NGSS HS-PS4-3 requires evaluating "claims, evidence, and reasoning behind the idea that electromagnetic radiation can be described either by a wave model or a particle model." Alberta 30–D2.7k: "the two-slit electron interference experiment shows that quantum systems, like photons and electrons, may be modelled as particles or waves."NGSS HS-PS4-3 要求评估"电磁辐射可用波动或粒子模型描述这一论断背后的主张、证据与推理"。阿尔伯塔 30–D2.7k："双缝电子干涉实验表明，光子和电子等量子系统可以被建模为粒子或波。"
Worked Example 2 · de Broglie wavelength of an electron例题 2 · 电子的德布罗意波长

An electron ($m = 9.11 \times 10^{-31}\ \mathrm{kg}$) moves at $v = 1.00 \times 10^6\ \mathrm{m/s}$. Find its de Broglie wavelength. ($h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$)一个电子（$m = 9.11 \times 10^{-31}\ \mathrm{kg}$）以 $v = 1.00 \times 10^6\ \mathrm{m/s}$ 运动。求其德布罗意波长。（$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$）

Apply the de Broglie formula.应用德布罗意公式。

$$ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(1.00 \times 10^6)} = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-25}} \approx 7.28 \times 10^{-10}\ \mathrm{m}. $$

This is $0.728\ \mathrm{nm}$ — in the X-ray range — which is why electrons can diffract off crystal lattices (spacing $\sim 0.1$–$0.3\ \mathrm{nm}$). The wavelength of a tennis ball at $50\ \mathrm{m/s}$ is $\sim 10^{-34}\ \mathrm{m}$, far too small to detect, which is why we don't notice matter waves in everyday life.这是 $0.728\ \mathrm{nm}$——处于 X 射线范围——这就是为什么电子能在晶格（间距约 $0.1$–$0.3\ \mathrm{nm}$）上衍射。网球以 $50\ \mathrm{m/s}$ 运动时波长约 $10^{-34}\ \mathrm{m}$，太小无法探测，这也是为什么日常生活中察觉不到物质波。

Which of the following is evidence that light behaves as a wave?以下哪项是光表现为波的证据？
§2 · Q1
The photoelectric effect光电效应
The Compton effect康普顿效应
Double-slit interference双缝干涉
The emission of photons from an LEDLED 发射光子
Double-slit interference produces a pattern of alternating bright and dark fringes that only waves can create through constructive and destructive interference. The photoelectric and Compton effects are particle evidence; LED emission involves discrete photons.双缝干涉产生明暗相间的条纹，只有波通过相长和相消干涉才能形成。光电效应和康普顿效应是粒子证据；LED 发光涉及离散光子。
Interference and diffraction are wave phenomena. The photoelectric and Compton effects require a particle (photon) picture to explain correctly.干涉和衍射是波动现象。光电效应和康普顿效应需要粒子（光子）图像才能正确解释。
According to the de Broglie hypothesis, if an electron's speed is doubled, its wavelength:根据德布罗意假设，若电子速率加倍，其波长：
§2 · Q2
Halves减半
Doubles加倍
Stays the same不变
Quadruples变为四倍
$\lambda = h/(mv)$. If $v$ doubles and $m$ is unchanged, the denominator doubles, so $\lambda$ halves. A faster particle has a shorter de Broglie wavelength.$\lambda = h/(mv)$。若 $v$ 加倍而 $m$ 不变，分母加倍，故 $\lambda$ 减半。粒子越快，德布罗意波长越短。
$\lambda = h/(mv)$. Wavelength is inversely proportional to speed (and momentum). Double the speed $\to$ halve the wavelength.$\lambda = h/(mv)$。波长与速度（及动量）成反比。速率加倍 $\to$ 波长减半。
Going deeper — the two-slit experiment with single electrons (AB 30–D2.7k)深入 — 单电子双缝实验（AB 30–D2.7k）

Fire electrons one at a time at a double slit. Classical intuition says each electron must go through one slit or the other and land at a predictable spot. Instead, over thousands of electrons, an interference pattern builds up — identical to the one produced by a continuous wave. Each individual electron seems to "interfere with itself," passing through both slits simultaneously as a wave, yet landing as a dot (a particle). If you try to detect which slit the electron used, the interference pattern disappears: the act of measurement collapses the wave behaviour. This is the central paradox of quantum mechanics, and it applies to photons just as well as to electrons. Alberta 30–D2.7k captures this exactly: "quantum systems, like photons and electrons, may be modelled as particles or waves, contrary to intuition."逐一向双缝发射电子。经典直觉认为每个电子必然通过其中一条缝并落在可预测的位置。然而，经过数千个电子后，却形成了与连续波完全相同的干涉图样。每个电子似乎"自我干涉"，同时以波的形式穿过两条缝，却以一个点（粒子）落下。如果试图探测电子通过哪条缝，干涉图样随即消失：测量行为使波动行为坍缩。这是量子力学的核心悖论，对光子和电子同样适用。阿尔伯塔 30–D2.7k 准确地描述了这一点："量子系统（如光子和电子）可以被建模为粒子或波，这与直觉相悖。"

Section 3 · Atomic Models & Spectra第 3 节 · 原子模型与光谱

Atomic Models and Spectra原子模型与光谱

The atom's structure was rebuilt four times in forty years.四十年间，原子结构被重建了四次。
  • Thomson (1897).汤姆孙（1897年）。 "Plum pudding" — electrons embedded in a diffuse positive charge. Predicted uniform scattering, which did not match Rutherford's results."葡萄干布丁"——电子嵌在弥散的正电荷中。预测均匀散射，与卢瑟福实验结果不符。
  • Rutherford (1911).卢瑟福（1911年）。 Gold-foil experiment: most $\alpha$-particles pass straight through; a few bounce back. Conclusion: nearly all mass and positive charge concentrated in a tiny nucleus. Electrons orbit the nucleus in mostly empty space.金箔实验：大多数 $\alpha$ 粒子直接穿过；少数大角度反弹。结论：几乎所有质量和正电荷集中在微小的原子核中，电子在大量空的空间中绕核运动。
  • Bohr (1913).玻尔（1913年）。 Electrons occupy discrete, stable energy levels. When an electron drops from a higher level $E_2$ to a lower level $E_1$, a photon is emitted with energy $E = hf = E_2 - E_1$. This explains the unique line spectrum (光谱) of each element.电子占据离散稳定的能级。电子从较高能级 $E_2$ 跃迁到较低能级 $E_1$ 时，发射能量为 $E = hf = E_2 - E_1$ 的光子。这解释了每种元素特有的线光谱。
$$ E_{\text{photon}} = hf = E_2 - E_1 \qquad \text{(emission transition)} $$ Alberta 30–D2.2k: "describe that each element has a unique line spectrum." 30–D2.5k: "calculate the energy difference between states, using the law of conservation of energy and the observed characteristics of an emitted photon." Ontario SPH4U Strand F evidence for quantum mechanics includes spectra and the photoelectric effect.阿尔伯塔 30–D2.2k："描述每种元素具有唯一线光谱。" 30–D2.5k："利用能量守恒定律和发射光子的观测特征计算能级差。" 安大略 SPH4U F 单元量子力学证据包括光谱与光电效应。
Worked Example 3 · Photon emitted in an atomic transition例题 3 · 原子跃迁发射的光子

A hydrogen atom transitions from an energy level of $-1.51\ \mathrm{eV}$ to $-3.40\ \mathrm{eV}$. Find the frequency and wavelength of the photon emitted. ($h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$, $c = 3.00 \times 10^8\ \mathrm{m/s}$, $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$)氢原子从能级 $-1.51\ \mathrm{eV}$ 跃迁到 $-3.40\ \mathrm{eV}$。求发射光子的频率和波长。（$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$，$c = 3.00 \times 10^8\ \mathrm{m/s}$，$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$）

Energy of photon.光子能量。

$$ \Delta E = E_2 - E_1 = (-1.51) - (-3.40) = 1.89\ \mathrm{eV} = 1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19}\ \mathrm{J}. $$

Frequency.频率。

$$ f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.56 \times 10^{14}\ \mathrm{Hz}. $$

Wavelength.波长。

$$ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{4.56 \times 10^{14}} \approx 6.58 \times 10^{-7}\ \mathrm{m} = 658\ \mathrm{nm}. $$

This is the red $H_\alpha$ line in the hydrogen spectrum — visible to the naked eye. Every element has its own unique set of such transitions, producing its unique line spectrum that serves as a "fingerprint" in astronomy.这正是氢光谱中的红色 $H_\alpha$ 线——肉眼可见。每种元素都有自己独特的跃迁集合，产生独特的线光谱，在天文学中用作"指纹"。

Rutherford's gold-foil experiment showed that most $\alpha$-particles passed straight through the foil. What did this prove?卢瑟福的金箔实验显示大多数 $\alpha$ 粒子直接穿过金箔。这证明了什么？
§3 · Q1
Atoms are solid spheres of positive charge原子是正电荷的实心球
Electrons are uniformly distributed throughout the atom电子均匀分布在原子中
Alpha particles are unchargedα 粒子不带电
Atoms are mostly empty space with a tiny, dense, positive nucleus原子大部分是空的，有一个微小致密的正电荷核
Most $\alpha$-particles passing through undeflected means most of the atom is empty space. The few that bounced back at large angles proved the positive charge (and mass) is concentrated in a tiny nucleus — the Rutherford nuclear model.大多数 $\alpha$ 粒子无偏转地穿过，说明原子大部分是空的。少数大角度反弹的粒子证明正电荷（和质量）集中在微小的原子核中——卢瑟福核模型。
The Thomson "plum-pudding" model predicted uniform small deflections (solid positive sphere). Rutherford's result — rare large deflections and mostly straight-through — required a concentrated nucleus in mostly empty space.汤姆孙"葡萄干布丁"模型预测均匀的小偏转（实心正球）。卢瑟福的结果——偶有大偏转、大多直穿——需要在大量空间中的集中核。
In the Bohr model, an electron drops from energy level $-0.85\ \mathrm{eV}$ to $-3.40\ \mathrm{eV}$. What is the energy of the emitted photon?在玻尔模型中，电子从能级 $-0.85\ \mathrm{eV}$ 跃迁到 $-3.40\ \mathrm{eV}$。发射光子的能量是多少？
§3 · Q2
$-4.25\ \mathrm{eV}$
$2.55\ \mathrm{eV}$
$0.85\ \mathrm{eV}$
$3.40\ \mathrm{eV}$
Photon energy $= E_{\text{high}} - E_{\text{low}} = (-0.85) - (-3.40) = 2.55\ \mathrm{eV}$. The photon carries away the energy difference between the two levels.光子能量 $= E_{\text{高}} - E_{\text{低}} = (-0.85) - (-3.40) = 2.55\ \mathrm{eV}$。光子带走两能级之间的能量差。
Subtract the lower (more negative) level from the upper (less negative) level: $(-0.85) - (-3.40) = +2.55\ \mathrm{eV}$. Photon energy is always positive.用上能级（较小负值）减下能级（较大负值）：$(-0.85) - (-3.40) = +2.55\ \mathrm{eV}$。光子能量始终为正。
Section 4 · Nuclear Structure & Binding Energy第 4 节 · 原子核与结合能

Nuclear Structure and Binding Energy原子核结构与结合能

The nucleus is held together by the strong nuclear force against electrostatic repulsion.原子核由强核力对抗静电斥力而结合在一起。
  • Notation ${}^A_Z X$.符号 ${}^A_Z X$。 $A$ = mass number (protons + neutrons), $Z$ = atomic number (protons), $A - Z$ = neutrons. Example: ${}^{12}_6\mathrm{C}$ has 6 protons and 6 neutrons.$A$ = 质量数（质子 + 中子），$Z$ = 原子序数（质子数），$A - Z$ = 中子数。例如：${}^{12}_6\mathrm{C}$ 有 6 个质子和 6 个中子。
  • Mass defect $\Delta m$.质量亏损 $\Delta m$。 The nucleus is less massive than the sum of its free nucleons. The missing mass has been converted to the binding energy that holds the nucleus together: $E_B = \Delta m \cdot c^2$.原子核的质量小于其自由核子的质量之和。缺失的质量转化为结合原子核的结合能：$E_B = \Delta m \cdot c^2$。
  • Binding energy per nucleon.每个核子的结合能。 The most stable nuclei (near iron, $A \approx 56$) have the highest binding energy per nucleon. This is why both fission of heavy nuclei and fusion of light nuclei release energy — both move toward the iron peak.最稳定的原子核（接近铁，$A \approx 56$）具有最高的每核子结合能。这就是为什么重核裂变和轻核聚变都释放能量——两者都向铁峰靠近。
$$ E_B = \Delta m \cdot c^2 \qquad \text{(binding energy)} \qquad 1\ \mathrm{u} \approx 931\ \mathrm{MeV}/c^2 $$ NGSS HS-PS1-8 requires models illustrating "changes in the composition of the nucleus" during nuclear processes. Alberta 30–D3.6k: "relate, qualitatively and quantitatively, the mass defect of the nucleus to the energy released in nuclear reactions, using Einstein's concept of mass-energy equivalence."NGSS HS-PS1-8 要求建立模型说明核过程中"原子核组成的变化"。阿尔伯塔 30–D3.6k："定性与定量地用爱因斯坦质能等价概念关联原子核质量亏损与核反应释放能量。"
Worked Example 4 · Binding energy of helium-4例题 4 · 氦-4 的结合能

The mass of ${}^4_2\mathrm{He}$ is $4.00260\ \mathrm{u}$. Proton mass $m_p = 1.00728\ \mathrm{u}$, neutron mass $m_n = 1.00867\ \mathrm{u}$. Find (a) the mass defect and (b) the binding energy in MeV. ($1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$)${}^4_2\mathrm{He}$ 的质量为 $4.00260\ \mathrm{u}$。质子质量 $m_p = 1.00728\ \mathrm{u}$，中子质量 $m_n = 1.00867\ \mathrm{u}$。求 (a) 质量亏损和 (b) 以 MeV 表示的结合能。（$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$）

(a) Mass defect.(a) 质量亏损。 He-4 has 2 protons and 2 neutrons. Sum of free nucleon masses:He-4 有 2 个质子和 2 个中子。自由核子质量之和：

$$ m_{\text{free}} = 2(1.00728) + 2(1.00867) = 2.01456 + 2.01734 = 4.03190\ \mathrm{u}. $$ $$ \Delta m = m_{\text{free}} - m_{\text{nucleus}} = 4.03190 - 4.00260 = 0.02930\ \mathrm{u}. $$

(b) Binding energy.(b) 结合能。

$$ E_B = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.02930 \times 931.5 \approx 27.3\ \mathrm{MeV}. $$

Binding energy per nucleon: $27.3 / 4 \approx 6.83\ \mathrm{MeV/nucleon}$ — comfortably on the stable side of the binding-energy curve.每核子结合能：$27.3 / 4 \approx 6.83\ \mathrm{MeV/nucleon}$——处于结合能曲线稳定一侧。

The nucleus of ${}^{16}_8\mathrm{O}$ contains how many neutrons?${}^{16}_8\mathrm{O}$ 的原子核中有多少个中子？
§4 · Q1
1616
2424
88
44
Neutrons $= A - Z = 16 - 8 = 8$. The superscript is the mass number (total nucleons); the subscript is the atomic number (protons only).中子数 $= A - Z = 16 - 8 = 8$。上标是质量数（核子总数）；下标是原子序数（仅质子数）。
Neutrons $= $ mass number $-$ atomic number $= A - Z$. For ${}^{16}_8\mathrm{O}$: $16 - 8 = 8$ neutrons.中子数 $= $ 质量数 $-$ 原子序数 $= A - Z$。对 ${}^{16}_8\mathrm{O}$：$16 - 8 = 8$ 个中子。
Why does the mass of a nucleus equal less than the sum of the masses of its individual protons and neutrons?为什么原子核的质量小于其各质子和中子质量之和？
§4 · Q2
Mass has been converted to binding energy ($E = mc^2$) that holds the nucleus together质量已通过 $E = mc^2$ 转化为将原子核结合在一起的结合能
Some neutrons escape from the nucleus during formation原子核形成时一些中子逃逸
Electrons inside the nucleus reduce its total mass原子核内的电子减少了其总质量
Protons are lighter inside the nucleus than when free质子在原子核内比自由时更轻
The mass defect $\Delta m$ represents energy ($E_B = \Delta m \cdot c^2$) that was released when the nucleus formed and is now stored as binding energy. By conservation of mass-energy, the bound nucleus is less massive than its free parts.质量亏损 $\Delta m$ 代表原子核形成时释放并储存为结合能的能量（$E_B = \Delta m \cdot c^2$）。根据质能守恒，束缚态核比其自由部分质量小。
The "missing" mass became binding energy via $E = mc^2$ when the nucleons came together. This is the mass defect — a direct consequence of mass-energy equivalence."缺失"的质量在核子聚合时通过 $E = mc^2$ 转化为结合能。这就是质量亏损——质能等价的直接结果。
Section 5 · Radioactivity & Decay第 5 节 · 放射性与衰变

Radioactivity and Decay放射性与衰变

Three decay types — know their charges, masses, and penetrations.三种衰变类型——掌握其电荷、质量与穿透力。
  • Alpha ($\alpha$) decay.Alpha（$\alpha$）衰变。 Nucleus emits a helium-4 nucleus (${}^4_2\mathrm{He}$): $A$ decreases by 4, $Z$ by 2. Least penetrating (stopped by paper or a few cm of air); most ionising.原子核发射氦-4 核（${}^4_2\mathrm{He}$）：$A$ 减少 4，$Z$ 减少 2。穿透力最弱（被纸或几厘米空气阻挡）；电离能力最强。
  • Beta ($\beta^-$) decay.Beta（$\beta^-$）衰变。 A neutron converts to a proton, emitting an electron ($e^-$) and antineutrino: $Z$ increases by 1, $A$ unchanged. Penetrating enough to pass through a few mm of aluminium.中子转变为质子，发射电子（$e^-$）和反中微子：$Z$ 增加 1，$A$ 不变。穿透力足以穿过几毫米的铝。
  • Gamma ($\gamma$) decay.Gamma（$\gamma$）衰变。 Nucleus releases excess energy as a high-energy photon. No change in $A$ or $Z$. Most penetrating (requires centimetres of lead or metres of concrete to attenuate).原子核以高能光子的形式释放多余能量。$A$ 和 $Z$ 均不变。穿透力最强（需要厘米级铅或米级混凝土来衰减）。
Nuclear equations must balance: total $A$ (top) and total $Z$ (bottom) are conserved on both sides. Alberta 30–D3.2k: "write nuclear equations, using isotope notation, for alpha, beta-negative and beta-positive decays, including the appropriate neutrino and antineutrino." NGSS HS-PS1-8 Assessment Boundary: "limited to alpha, beta, and gamma radioactive decays."核方程必须配平：两边的总 $A$（上标）和总 $Z$（下标）均守恒。阿尔伯塔 30–D3.2k："用同位素符号书写 alpha、beta-negative 和 beta-positive 衰变的核方程，包括相应的中微子和反中微子。" NGSS HS-PS1-8 评估边界："限于 alpha、beta 和 gamma 放射性衰变。"
Worked Example 5 · Writing and balancing decay equations例题 5 · 书写与配平衰变方程

(a) Write the alpha-decay equation for ${}^{226}_{88}\mathrm{Ra}$ (radium-226). (b) Write the beta-minus decay equation for ${}^{14}_6\mathrm{C}$ (carbon-14).(a) 写出 ${}^{226}_{88}\mathrm{Ra}$（镭-226）的 alpha 衰变方程。(b) 写出 ${}^{14}_6\mathrm{C}$（碳-14）的 beta 负衰变方程。

(a) Alpha decay of Ra-226.(a) Ra-226 的 alpha 衰变。 Emit ${}^4_2\mathrm{He}$: daughter has $A = 226 - 4 = 222$, $Z = 88 - 2 = 86$ (Radon, Rn).发射 ${}^4_2\mathrm{He}$：子核 $A = 226 - 4 = 222$，$Z = 88 - 2 = 86$（氡，Rn）。

$$ {}^{226}_{88}\mathrm{Ra} \;\longrightarrow\; {}^{222}_{86}\mathrm{Rn} \;+\; {}^{4}_{2}\mathrm{He} $$

Check: top $226 = 222 + 4$ ✓; bottom $88 = 86 + 2$ ✓.核验：上标 $226 = 222 + 4$ ✓；下标 $88 = 86 + 2$ ✓。

(b) Beta-minus decay of C-14.(b) C-14 的 beta 负衰变。 Emit ${}^0_{-1}e$ (electron) and $\bar\nu_e$ (antineutrino): daughter has $A = 14$, $Z = 7$ (Nitrogen, N).发射 ${}^0_{-1}e$（电子）和 $\bar\nu_e$（反中微子）：子核 $A = 14$，$Z = 7$（氮，N）。

$$ {}^{14}_{6}\mathrm{C} \;\longrightarrow\; {}^{14}_{7}\mathrm{N} \;+\; {}^{0}_{-1}e \;+\; \bar\nu_e $$

This is the basis of radiocarbon dating: living organisms maintain a steady ${}^{14}\mathrm{C}$ ratio; after death the ${}^{14}\mathrm{C}$ decays with a half-life of $\approx 5730$ years, allowing age determination.这是放射性碳定年法的基础：活体生物维持稳定的 ${}^{14}\mathrm{C}$ 比例；死后 ${}^{14}\mathrm{C}$ 以约 $5730$ 年的半衰期衰变，从而确定年代。

Which type of radiation is the most penetrating?哪种辐射穿透力最强？
§5 · Q1
Alpha ($\alpha$)Alpha（$\alpha$）
Gamma ($\gamma$)Gamma（$\gamma$）
Beta ($\beta$)Beta（$\beta$）
All three are equally penetrating三种穿透力相同
Gamma rays are high-energy photons with no charge and no mass — they require centimetres of lead or metres of concrete to attenuate significantly. Alpha particles (stopped by paper) and beta particles (stopped by a few mm of aluminium) are far less penetrating.Gamma 射线是无电荷无质量的高能光子——需要厘米级铅或米级混凝土才能显著衰减。Alpha 粒子（被纸阻挡）和 beta 粒子（被几毫米铝阻挡）穿透力差得多。
Penetrating power: $\alpha$ (paper) < $\beta$ (mm of Al) < $\gamma$ (cm of Pb). Gamma rays are uncharged photons and penetrate the most deeply.穿透力：$\alpha$（纸）< $\beta$（毫米铝）< $\gamma$（厘米铅）。Gamma 射线是不带电光子，穿透最深。
In the alpha decay of ${}^{238}_{92}\mathrm{U}$, what are the mass number and atomic number of the daughter nucleus?${}^{238}_{92}\mathrm{U}$ 的 alpha 衰变中，子核的质量数和原子序数分别是多少？
§5 · Q2
$A = 238,\ Z = 90$$A = 238,\ Z = 90$
$A = 242,\ Z = 94$$A = 242,\ Z = 94$
$A = 234,\ Z = 92$$A = 234,\ Z = 92$
$A = 234,\ Z = 90$$A = 234,\ Z = 90$
Alpha emission takes away 4 mass units and 2 protons: daughter $A = 238 - 4 = 234$, $Z = 92 - 2 = 90$ (Thorium-234).Alpha 发射带走 4 个质量单位和 2 个质子：子核 $A = 238 - 4 = 234$，$Z = 92 - 2 = 90$（钍-234）。
An alpha particle is ${}^4_2\mathrm{He}$. Subtract from parent: $A$ decreases by 4 (to 234), $Z$ decreases by 2 (to 90).Alpha 粒子是 ${}^4_2\mathrm{He}$。从母核减去：$A$ 减少 4（变为 234），$Z$ 减少 2（变为 90）。
Section 6 · Half-Life第 6 节 · 半衰期

Half-Life and Radioactive Decay Rates半衰期与放射性衰变速率

Half-life: the time for half the nuclei in a sample to decay.半衰期：样品中一半原子核衰变所需的时间。
  • The half-life formula.半衰期公式。 $$ N = N_0 \!\left(\tfrac{1}{2}\right)^{t/T} $$ where $N_0$ = initial number of nuclei, $N$ = number remaining at time $t$, $T$ = half-life. Each period $T$, the sample halves: $N_0 \to N_0/2 \to N_0/4 \to \cdots$其中 $N_0$ 为初始核数，$N$ 为时刻 $t$ 剩余核数，$T$ 为半衰期。每经过一个 $T$，样品减半：$N_0 \to N_0/2 \to N_0/4 \to \cdots$
  • Activity $A$.放射性活度 $A$。 The number of decays per second (unit: becquerel, Bq, $= 1\ \mathrm{s^{-1}}$). Activity halves with the same half-life as the number of nuclei.每秒衰变数（单位：贝可勒尔，Bq，$= 1\ \mathrm{s^{-1}}$）。活度以与核数相同的半衰期减半。
  • Half-life is constant.半衰期是常数。 It does not depend on temperature, pressure, or chemical state — radioactive decay is a purely nuclear process. Different isotopes have wildly different half-lives (milliseconds to billions of years).它不依赖温度、压力或化学状态——放射性衰变是纯核过程。不同同位素的半衰期差异巨大（从毫秒到数十亿年）。
Ontario SPH3U D3.11: "explain radioactive half-life for a given radioisotope, and describe its applications and their consequences." Alberta 30–D3.3k: "perform simple, nonlogarithmic half-life calculations."安大略 SPH3U D3.11："解释给定放射性同位素的放射性半衰期，并描述其应用及后果。" 阿尔伯塔 30–D3.3k："进行简单的非对数半衰期计算。"
Worked Example 6 · Half-life calculation for iodine-131例题 6 · 碘-131 的半衰期计算

Iodine-131 (${}^{131}_{53}\mathrm{I}$) is used in thyroid cancer treatment. Its half-life is $T = 8.02\ \mathrm{days}$. A patient receives a dose containing $N_0 = 1.60 \times 10^{12}$ atoms. (a) How many atoms remain after $24.06\ \mathrm{days}$? (b) What fraction of the original activity remains after $32.08\ \mathrm{days}$?碘-131（${}^{131}_{53}\mathrm{I}$）用于甲状腺癌治疗，半衰期 $T = 8.02$ 天。患者接受含 $N_0 = 1.60 \times 10^{12}$ 个原子的剂量。(a) $24.06$ 天后剩余多少原子？(b) $32.08$ 天后原始活度的多少分数仍存在？

(a) Number of half-lives elapsed: $t/T = 24.06/8.02 = 3$. Apply the formula.(a) 经历的半衰期数：$t/T = 24.06/8.02 = 3$。应用公式。

$$ N = N_0 \!\left(\tfrac{1}{2}\right)^3 = 1.60 \times 10^{12} \times \tfrac{1}{8} = 2.00 \times 10^{11}\ \text{atoms}. $$

(b) Fraction after 4 half-lives ($32.08/8.02 = 4$).(b) 4 个半衰期（$32.08/8.02 = 4$）后的分数。

$$ \frac{N}{N_0} = \left(\tfrac{1}{2}\right)^4 = \frac{1}{16} \approx 6.25\%. $$

Since activity is proportional to $N$, the activity also drops to $1/16$ of the original. Medical physicists use this to plan dosing schedules — after $\approx 5$ half-lives ($\approx 40$ days) the activity is $\approx 3\%$ of the original, considered negligible for clinical purposes.由于活度正比于 $N$，活度也降至原来的 $1/16$。医学物理学家据此制定剂量方案——经过约 5 个半衰期（约 40 天）后，活度约为原来的 3%，在临床上视为可忽略。

A sample starts with $800$ radioactive nuclei. After 3 half-lives, how many nuclei remain?一个样品开始时有 $800$ 个放射性核。经过 3 个半衰期后，剩余多少个核？
§6 · Q1
400400
200200
100100
00
Each half-life halves the count: $800 \to 400 \to 200 \to 100$. After 3 half-lives: $N = 800 \times (1/2)^3 = 800/8 = 100$.每个半衰期数量减半：$800 \to 400 \to 200 \to 100$。3 个半衰期后：$N = 800 \times (1/2)^3 = 800/8 = 100$。
Apply $(1/2)^3 = 1/8$ to the initial count: $800 \times (1/8) = 100$, not zero — the sample never fully decays in a finite number of half-lives.将 $(1/2)^3 = 1/8$ 应用于初始数量：$800 \times (1/8) = 100$，不是零——样品在有限个半衰期内永远不会完全衰变。
The half-life of carbon-14 is approximately $5730$ years. A wood sample is found to have $25\%$ of the ${}^{14}\mathrm{C}$ activity of a living tree today. Approximately how old is the sample?碳-14 的半衰期约为 $5730$ 年。一个木样品的 ${}^{14}\mathrm{C}$ 活度为今天活树的 $25\%$。该样品约多少年历史？
§6 · Q2
$5\,730$ years$5\,730$ 年
$11\,460$ years$11\,460$ 年
$17\,190$ years$17\,190$ 年
$22\,920$ years$22\,920$ 年
$25\% = (1/2)^2$, so 2 half-lives have passed: $2 \times 5730 = 11\,460$ years. Each halving: $100\% \to 50\% \to 25\%$.$25\% = (1/2)^2$，故经历了 2 个半衰期：$2 \times 5730 = 11\,460$ 年。逐次减半：$100\% \to 50\% \to 25\%$。
Find how many halvings get from $100\%$ to $25\%$: $(1/2)^n = 0.25 \Rightarrow n = 2$. Age $= 2 \times 5730 = 11\,460$ years.找出从 $100\%$ 到 $25\%$ 需要几次减半：$(1/2)^n = 0.25 \Rightarrow n = 2$。年龄 $= 2 \times 5730 = 11\,460$ 年。
Section 7 · Fission, Fusion & Mass-Energy第 7 节 · 裂变、聚变与质能方程

Fission, Fusion and Mass-Energy ($E = mc^2$)裂变、聚变与质能方程（$E = mc^2$）

$E = mc^2$: mass and energy are interchangeable; the conversion factor is $c^2 = (3 \times 10^8)^2\ \mathrm{m^2/s^2}$.$E = mc^2$：质量与能量可以互换；转换因子为 $c^2 = (3 \times 10^8)^2\ \mathrm{m^2/s^2}$。
  • Fission (裂变).裂变。 A heavy nucleus (e.g. ${}^{235}\mathrm{U}$) absorbs a neutron and splits into two mid-sized nuclei plus more neutrons, releasing $\sim 200\ \mathrm{MeV}$ per event. The product nuclei sit higher on the binding-energy-per-nucleon curve than uranium, so energy is released by the mass defect.重核（如 ${}^{235}\mathrm{U}$）吸收中子后裂变为两个中等质量核加上更多中子，每次释放约 $200\ \mathrm{MeV}$。产物核在每核子结合能曲线上高于铀，故通过质量亏损释放能量。
  • Fusion (聚变).聚变。 Two light nuclei (e.g. ${}^2\mathrm{H} + {}^3\mathrm{H} \to {}^4\mathrm{He} + n$) combine into a heavier nucleus, releasing $\sim 17.6\ \mathrm{MeV}$. The product also sits higher on the curve than the light inputs. Powers stars and the Sun.两个轻核（如 ${}^2\mathrm{H} + {}^3\mathrm{H} \to {}^4\mathrm{He} + n$）结合成较重核，释放约 $17.6\ \mathrm{MeV}$。产物在曲线上也高于轻核原料。这是星体和太阳的能源。
  • Chain reaction.链式反应。 Fission of ${}^{235}\mathrm{U}$ releases $2$–$3$ neutrons per event, each capable of triggering further fissions. A controlled chain reaction powers a reactor; an uncontrolled one produces an explosion.${}^{235}\mathrm{U}$ 裂变每次释放 $2$–$3$ 个中子，每个都能引发进一步裂变。受控链式反应驱动反应堆；不受控的产生爆炸。
$$ E = mc^2 \qquad \text{(mass-energy equivalence)} \qquad 1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2 $$ NGSS HS-PS1-8: "models to illustrate ... energy released during the processes of fission, fusion, and radioactive decay." Alberta 30–D3.5k: "compare and contrast the characteristics of fission and fusion reactions." 30–D3.6k: "relate ... the mass defect ... to the energy released ... using Einstein's concept of mass-energy equivalence." Ontario SPH3U D3.12: "nuclear fission results in the liberation of energy, which is converted into thermal energy."NGSS HS-PS1-8："建立模型说明裂变、聚变和放射性衰变过程中释放的能量。" 阿尔伯塔 30–D3.5k："比较裂变与聚变反应的特征。" 30–D3.6k："用爱因斯坦质能等价概念关联质量亏损与核反应释放能量。" 安大略 SPH3U D3.12："核裂变释放能量，转化为热能。"
Worked Example 7 · Energy released in a D-T fusion reaction例题 7 · D-T 聚变反应释放的能量

In the deuterium-tritium fusion reaction ${}^2_1\mathrm{H} + {}^3_1\mathrm{H} \to {}^4_2\mathrm{He} + {}^1_0\mathrm{n}$, the masses are: $m({}^2\mathrm{H}) = 2.01410\ \mathrm{u}$, $m({}^3\mathrm{H}) = 3.01605\ \mathrm{u}$, $m({}^4\mathrm{He}) = 4.00260\ \mathrm{u}$, $m_n = 1.00867\ \mathrm{u}$. Find the energy released. ($1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$)在氘-氚聚变反应 ${}^2_1\mathrm{H} + {}^3_1\mathrm{H} \to {}^4_2\mathrm{He} + {}^1_0\mathrm{n}$ 中，质量如下：$m({}^2\mathrm{H}) = 2.01410\ \mathrm{u}$，$m({}^3\mathrm{H}) = 3.01605\ \mathrm{u}$，$m({}^4\mathrm{He}) = 4.00260\ \mathrm{u}$，$m_n = 1.00867\ \mathrm{u}$。求释放的能量。（$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$）

Mass before and after.反应前后质量。

$$ m_{\text{before}} = 2.01410 + 3.01605 = 5.03015\ \mathrm{u}. $$ $$ m_{\text{after}} = 4.00260 + 1.00867 = 5.01127\ \mathrm{u}. $$

Mass defect and energy.质量亏损与能量。

$$ \Delta m = 5.03015 - 5.01127 = 0.01888\ \mathrm{u}. $$ $$ E = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.01888 \times 931.5 \approx 17.6\ \mathrm{MeV}. $$

To put this in perspective: burning one carbon atom releases about $4\ \mathrm{eV}$ (chemical energy). One D-T fusion event releases $17.6\ \mathrm{MeV}$ — about $4\,400\,000$ times more energy per reaction. This enormous ratio is why nuclear fusion is pursued as a potential power source.作为对比：燃烧一个碳原子释放约 $4\ \mathrm{eV}$（化学能）。一次 D-T 聚变释放 $17.6\ \mathrm{MeV}$——每次反应约多出 $4\,400\,000$ 倍能量。这一巨大比值正是核聚变被作为潜在能源追求的原因。

Both fission of uranium-235 and fusion of hydrogen isotopes release energy. Why?铀-235 裂变和氢同位素聚变都释放能量。为什么？
§7 · Q1
In both cases the products have higher binding energy per nucleon than the reactants, so the mass of the products is less (mass defect converted to energy via $E = mc^2$)在两种情况下，产物的每核子结合能都高于反应物，因此产物质量更小（质量亏损通过 $E = mc^2$ 转化为能量）
Neutrons are released, and neutrons carry kinetic energy释放了中子，中子携带动能
The reactions produce heat by burning the nuclear fuel chemically反应通过化学燃烧核燃料产生热量
Fission releases energy but fusion absorbs energy裂变释放能量，但聚变吸收能量
Both processes move nuclei toward the peak of the binding-energy-per-nucleon curve (near iron). The products are more tightly bound, so the total mass decreases (mass defect). This "missing" mass becomes energy via $E = mc^2$. The neutrons released in fission carry some kinetic energy too, but the fundamental source is the mass defect.两个过程都使核向每核子结合能曲线顶峰（铁附近）移动。产物结合更紧，总质量减少（质量亏损）。这些"缺失"质量通过 $E = mc^2$ 转化为能量。裂变释放的中子也携带一些动能，但根本来源是质量亏损。
Nuclear energy comes from mass-energy equivalence, not chemical burning. The key is that products have more binding energy per nucleon (sit higher on the curve) than reactants — so the total mass decreases and energy is released.核能来自质能等价，而非化学燃烧。关键是产物的每核子结合能高于反应物（在曲线上位置更高）——总质量减少，能量释放。
A nuclear reaction has a mass defect of $0.00200\ \mathrm{u}$. How much energy is released? ($1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$)一个核反应的质量亏损为 $0.00200\ \mathrm{u}$。释放多少能量？（$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$）
§7 · Q2
$0.00200\ \mathrm{MeV}$
$465.8\ \mathrm{MeV}$
$1.863\ \mathrm{MeV}$
$18.63\ \mathrm{MeV}$
$E = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.00200 \times 931.5 = 1.863\ \mathrm{MeV}$.$E = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.00200 \times 931.5 = 1.863\ \mathrm{MeV}$。
Multiply the mass defect in atomic mass units by $931.5\ \mathrm{MeV/u}$: $0.00200 \times 931.5 = 1.863\ \mathrm{MeV}$.将原子质量单位的质量亏损乘以 $931.5\ \mathrm{MeV/u}$：$0.00200 \times 931.5 = 1.863\ \mathrm{MeV}$。
Going deeper — the chain reaction and critical mass (NGSS HS-PS1-8, AB 30–D3.5k)深入 — 链式反应与临界质量（NGSS HS-PS1-8，AB 30–D3.5k）

When ${}^{235}\mathrm{U}$ absorbs a slow (thermal) neutron, it fissions into two fragments and releases on average $2.5$ new neutrons. If each new neutron triggers another fission event, the number of reactions doubles each generation: $1 \to 2.5 \to 6.25 \to \cdots$ — exponential growth. This is a chain reaction. For a chain reaction to sustain itself, the sample must be large enough that the neutrons are likely to hit another ${}^{235}\mathrm{U}$ nucleus before escaping — this minimum mass is the critical mass ($\approx 52\ \mathrm{kg}$ for pure ${}^{235}\mathrm{U}$, less with a neutron reflector). A reactor controls the reaction by inserting neutron-absorbing control rods (boron or cadmium) to keep the multiplication factor exactly at $1$. A weapon removes the control and allows exponential growth. Ontario SPH3U D3.12 traces the energy path: fission $\to$ kinetic energy of fragments $\to$ thermal energy $\to$ steam $\to$ turbine $\to$ electrical energy.当 ${}^{235}\mathrm{U}$ 吸收一个慢（热）中子时，裂变为两个碎片，平均释放 $2.5$ 个新中子。若每个新中子又引发一次裂变，反应数每代翻倍：$1 \to 2.5 \to 6.25 \to \cdots$——指数增长。这就是链式反应。要使链式反应自持，样品必须足够大，使中子在逃逸前很可能击中另一个 ${}^{235}\mathrm{U}$ 核——这个最小质量称为临界质量（纯 ${}^{235}\mathrm{U}$ 约 $52\ \mathrm{kg}$，加中子反射体可减少）。反应堆通过插入吸收中子的控制棒（硼或镉）使增殖系数精确保持为 $1$。武器则撤去控制，允许指数增长。安大略 SPH3U D3.12 追踪能量路径：裂变 $\to$ 碎片动能 $\to$ 热能 $\to$ 蒸汽 $\to$ 涡轮机 $\to$ 电能。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline for nuclear problems核物理题的解题纪律
  • Balance $A$ and $Z$ separately.分别配平 $A$ 和 $Z$。 In every nuclear equation, check that the sum of mass numbers (top superscripts) and the sum of atomic numbers (bottom subscripts) both equal on left and right. Do this in two independent steps.在每个核方程中，检查质量数之和（上标）和原子序数之和（下标）在左右两边都相等。分两个独立步骤进行。
  • Mass defect in atomic mass units, energy in MeV.质量亏损用原子质量单位，能量用 MeV。 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$. Convert mass defect to energy by multiplying by $931.5$. Never mix SI and atomic units in the same calculation without converting.$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$。将质量亏损乘以 $931.5$ 转换为能量。切勿在同一计算中混用国际单位和原子单位而不转换。
  • Half-life: count halvings first.半衰期：先数减半次数。 Divide total time by half-life to get $n$. Then apply $(1/2)^n$. If $n$ is not an integer, use the exponential formula. Never subtract nuclei linearly.用总时间除以半衰期得 $n$。然后应用 $(1/2)^n$。若 $n$ 不是整数，使用指数公式。切勿线性地减去核数。
Quantum and atomic pitfalls (§1–§3)量子与原子的常见陷阱（§1–§3）
  • Intensity $\ne$ photon energy.强度 $\ne$ 光子能量。 Doubling intensity doubles the number of photons, not their individual energies. Kinetic energy of ejected electrons depends on frequency, not brightness.强度加倍使光子数加倍，而非每个光子的能量。被射出电子的动能取决于频率，而非亮度。
  • Threshold is about frequency, not intensity.阈值关乎频率，不是强度。 Below the threshold frequency $f_0$, no electrons are ejected no matter how bright the light. Above $f_0$, electrons are emitted instantly even in very dim light.低于阈频 $f_0$，无论光多亮都没有电子射出。超过 $f_0$ 后，即使极暗的光也立即射出电子。
  • Photon energy: $E = hf$, not $hv$.光子能量：$E = hf$，不是 $hv$。 The symbol for frequency is $f$ (or $\nu$ in some texts). Do not confuse it with velocity. Use $c = f\lambda$ to convert between frequency and wavelength.频率的符号是 $f$（某些教材用 $\nu$）。不要与速度混淆。用 $c = f\lambda$ 在频率和波长之间转换。
Nuclear equation and decay pitfalls (§4–§5)核方程与衰变陷阱（§4–§5）
  • Alpha: $A$ drops by 4, $Z$ drops by 2.Alpha：$A$ 减少 4，$Z$ 减少 2。 The emitted ${}^4_2\mathrm{He}$ carries 4 mass units and 2 protons. Verify the daughter element using a periodic table.发射的 ${}^4_2\mathrm{He}$ 带走 4 个质量单位和 2 个质子。用元素周期表核验子核元素。
  • Beta-minus: $A$ unchanged, $Z$ increases by 1.Beta 负：$A$ 不变，$Z$ 增加 1。 A neutron becomes a proton — no mass-number change, but the element changes. Do not forget the antineutrino ($\bar\nu_e$) if your curriculum (AB Physics 30) requires it.中子变为质子——质量数不变，但元素改变。若你的课纲（阿尔伯塔 Physics 30）要求，不要忘记反中微子（$\bar\nu_e$）。
  • Gamma: no change in $A$ or $Z$.Gamma：$A$ 和 $Z$ 均不变。 Gamma decay is an energy release — the nucleus just drops to a lower energy state and emits a photon. The isotope does not change.Gamma 衰变是能量释放——原子核只是跃迁到较低能量状态并发射光子。同位素不变。
Answer hygiene for modern physics近代物理的作答规范
  • State your model.说明你用的模型。 When a question asks "which model explains this," name the model (wave or particle) and cite the specific experimental evidence (interference, photoelectric effect, etc.).当题目要求"哪种模型解释这个"时，说明模型（波或粒子）并引用具体实验证据（干涉、光电效应等）。
  • Use correct isotope notation.使用正确的同位素符号。 Write ${}^A_Z X$ with superscript for mass number and subscript for atomic number. In Alberta exams, omitting the subscript is an error.用 ${}^A_Z X$ 书写，上标为质量数，下标为原子序数。在阿尔伯塔考试中，省略下标是错误。
  • Scale check for nuclear energy.核能的量级核验。 Nuclear energies are typically in the range of MeV (millions of eV), not eV or keV. If your answer for a fission or fusion energy comes out in the eV range, recheck your units.核能通常在 MeV（百万电子伏特）范围内，而非 eV 或 keV。若裂变或聚变能量答案在 eV 范围，请核查单位。
Active Recall主动复习

Flashcards闪卡

0 / 14 flipped0 / 14 已翻
Photon energy formula?光子能量公式？
$$E = hf$$ $h = 6.63\times10^{-34}\ \mathrm{J\cdot s}$ (Planck's constant)$h = 6.63\times10^{-34}\ \mathrm{J\cdot s}$（普朗克常数）
Photoelectric equation?光电方程？
$$E_k = hf - \phi$$ $\phi$ = work function; $E_k = 0$ if $hf < \phi$$\phi$ = 逸出功；若 $hf < \phi$ 则 $E_k = 0$
de Broglie wavelength?德布罗意波长？
$$\lambda = \frac{h}{mv}$$ all moving particles have an associated wavelength所有运动粒子都有对应波长
Wave evidence vs particle evidence for light?光的波动证据与粒子证据？
Wave: interference, diffraction, polarization. Particle: photoelectric effect, Compton effect.波：干涉、衍射、偏振。粒子：光电效应、康普顿效应。
Bohr model: photon emission?玻尔模型：光子发射？
$$E_{\text{photon}} = hf = E_2 - E_1$$ electron drops from level $E_2$ to $E_1$ ($E_2 > E_1$)电子从 $E_2$ 跃迁到 $E_1$（$E_2 > E_1$）
Nuclear notation ${}^A_Z X$?核符号 ${}^A_Z X$？
$A$ = mass number (p+n); $Z$ = atomic number (p); neutrons $= A - Z$$A$ = 质量数（质子+中子）；$Z$ = 原子序数（质子）；中子 $= A - Z$
Alpha decay changes?Alpha 衰变的变化？
Emits ${}^4_2\mathrm{He}$; $A$ decreases by 4, $Z$ decreases by 2发射 ${}^4_2\mathrm{He}$；$A$ 减少 4，$Z$ 减少 2
Beta-minus decay changes?Beta 负衰变的变化？
Emits $e^-$ + $\bar\nu_e$; $A$ unchanged, $Z$ increases by 1发射 $e^-$ + $\bar\nu_e$；$A$ 不变，$Z$ 增加 1
Penetrating power: $\alpha$, $\beta$, $\gamma$?穿透力：$\alpha$、$\beta$、$\gamma$？
$\alpha$ (paper) < $\beta$ (mm Al) < $\gamma$ (cm Pb). Ionising power is the reverse.$\alpha$（纸）< $\beta$（毫米铝）< $\gamma$（厘米铅）。电离能力相反。
Half-life formula?半衰期公式？
$$N = N_0\!\left(\tfrac{1}{2}\right)^{t/T}$$ $T$ = half-life; after $n$ half-lives: $N = N_0/2^n$$T$ = 半衰期；经 $n$ 个半衰期：$N = N_0/2^n$
Mass-energy equivalence?质能等价？
$$E = mc^2$$ $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$
Binding energy from mass defect?由质量亏损求结合能？
$$E_B = \Delta m \cdot c^2$$ $\Delta m = $ (sum of free nucleon masses) $-$ (nucleus mass)$\Delta m = $（自由核子质量之和）$-$（核质量）
Fission vs fusion?裂变与聚变？
Fission: heavy nucleus splits (e.g. ${}^{235}\mathrm{U}$). Fusion: light nuclei combine (e.g. D+T). Both release energy by moving toward the iron peak on the binding-energy curve.裂变：重核分裂（如 ${}^{235}\mathrm{U}$）。聚变：轻核结合（如 D+T）。两者都通过向结合能曲线铁峰靠近而释放能量。
Why does increasing light intensity not increase ejected electron energy?为何增加光强不能增加射出电子的能量？
Each photon interacts with one electron. Energy per photon $= hf$ depends only on frequency, not how many photons there are.每个光子与一个电子相互作用。每个光子的能量 $= hf$ 只取决于频率，而非光子数量。
Mixed Practice综合练习

Practice Quiz综合测验

Light hits a metal surface and no electrons are ejected. Increasing the intensity of the light (same frequency) still produces no electrons. What is the best explanation?光照射金属表面，没有电子射出。增加光强（频率不变）仍没有电子产生。最佳解释是什么？
Q1
The metal surface is too thick金属表面太厚
There are not enough photons光子数量不够
The light wavelength is too long but could be shortened by increasing intensity光波长太长，但可以通过增加强度来缩短
The photon frequency is below the threshold; individual photon energy ($hf$) is too low to overcome the work function, regardless of intensity光子频率低于阈值；单个光子能量（$hf$）太低，无法克服逸出功，与强度无关
In the photon model, each electron can only absorb one photon at a time. If $hf < \phi$, no single photon has enough energy to eject an electron, no matter how many photons there are (intensity).在光子模型中，每个电子一次只能吸收一个光子。若 $hf < \phi$，无论有多少光子（强度），单个光子都没有足够能量射出电子。
Intensity controls the count of photons, not their individual energies. Only increasing the frequency (and thus $hf$) can overcome the work function threshold.强度控制光子数量，而非单个光子的能量。只有增加频率（从而增加 $hf$）才能克服逸出功阈值。
A hydrogen atom emits a photon of wavelength $486\ \mathrm{nm}$. What type of electromagnetic radiation is this? ($c = 3.0\times10^8\ \mathrm{m/s}$; recall the visible range is $\approx 400$–$700\ \mathrm{nm}$)氢原子发射波长 $486\ \mathrm{nm}$ 的光子。这是什么类型的电磁辐射？（$c = 3.0\times10^8\ \mathrm{m/s}$；可见光范围约 $400$–$700\ \mathrm{nm}$）
Q2
X-rayX 射线
Visible light (blue-green)可见光（蓝绿色）
Ultraviolet紫外线
Infrared红外线
$486\ \mathrm{nm}$ falls in the visible range ($400$–$700\ \mathrm{nm}$), specifically in the blue-green region. This is the $H_\beta$ line of the hydrogen Balmer series.$486\ \mathrm{nm}$ 在可见光范围（$400$–$700\ \mathrm{nm}$）内，具体在蓝绿色区域。这是氢巴尔末系的 $H_\beta$ 线。
Compare $486\ \mathrm{nm}$ with the visible range ($400$–$700\ \mathrm{nm}$). It falls inside, so it is visible light.将 $486\ \mathrm{nm}$ 与可见光范围（$400$–$700\ \mathrm{nm}$）比较。它在范围内，故为可见光。
How many neutrons are in a nucleus of ${}^{235}_{92}\mathrm{U}$?${}^{235}_{92}\mathrm{U}$ 的原子核中有多少个中子？
Q3
9292
235235
143143
327327
Neutrons $= A - Z = 235 - 92 = 143$.中子数 $= A - Z = 235 - 92 = 143$。
Neutrons $= $ mass number $-$ atomic number $= 235 - 92 = 143$.中子数 $= $ 质量数 $-$ 原子序数 $= 235 - 92 = 143$。
A radioactive isotope has a half-life of $6.0\ \mathrm{hours}$. A $1600\ \mathrm{Bq}$ sample is measured. What will the activity be after $18\ \mathrm{hours}$? 🇨🇦 ON D3.11 / AB 30–D3.3k一种放射性同位素的半衰期为 $6.0\ \mathrm{小时}$。测量样品活度为 $1600\ \mathrm{Bq}$。$18\ \mathrm{小时}$ 后活度是多少？🇨🇦 ON D3.11 / AB 30–D3.3k
Q4
$200\ \mathrm{Bq}$
$400\ \mathrm{Bq}$
$533\ \mathrm{Bq}$
$800\ \mathrm{Bq}$
$t/T = 18/6 = 3$ half-lives. Activity $= 1600 \times (1/2)^3 = 1600/8 = 200\ \mathrm{Bq}$.$t/T = 18/6 = 3$ 个半衰期。活度 $= 1600 \times (1/2)^3 = 1600/8 = 200\ \mathrm{Bq}$。
Count half-lives: $18/6 = 3$. Apply $(1/2)^3 = 1/8$: $1600/8 = 200\ \mathrm{Bq}$.数半衰期数：$18/6 = 3$。应用 $(1/2)^3 = 1/8$：$1600/8 = 200\ \mathrm{Bq}$。
Complete the following alpha-decay equation: ${}^{210}_{84}\mathrm{Po} \to {}^{?}_{?}\mathrm{X} + {}^4_2\mathrm{He}$. What is the daughter nucleus? 🇨🇦 AB 30–D3.2k完成以下 alpha 衰变方程：${}^{210}_{84}\mathrm{Po} \to {}^{?}_{?}\mathrm{X} + {}^4_2\mathrm{He}$。子核是什么？🇨🇦 AB 30–D3.2k
Q5
${}^{206}_{82}\mathrm{Pb}$${}^{206}_{82}\mathrm{Pb}$
${}^{206}_{82}\mathrm{Pb}$ — Lead-206${}^{206}_{82}\mathrm{Pb}$ — 铅-206
${}^{214}_{86}\mathrm{Rn}$ — Radon-214${}^{214}_{86}\mathrm{Rn}$ — 氡-214
${}^{210}_{82}\mathrm{Pb}$ — Lead-210${}^{210}_{82}\mathrm{Pb}$ — 铅-210
Alpha emission: $A$ decreases by 4 ($210 - 4 = 206$), $Z$ decreases by 2 ($84 - 2 = 82$). Element with $Z = 82$ is Lead (Pb). Daughter is ${}^{206}_{82}\mathrm{Pb}$.Alpha 发射：$A$ 减少 4（$210 - 4 = 206$），$Z$ 减少 2（$84 - 2 = 82$）。$Z = 82$ 的元素是铅（Pb）。子核为 ${}^{206}_{82}\mathrm{Pb}$。
Alpha particle is ${}^4_2\mathrm{He}$. Subtract: $A = 210-4 = 206$, $Z = 84-2 = 82$ (Lead). Daughter: ${}^{206}_{82}\mathrm{Pb}$.Alpha 粒子是 ${}^4_2\mathrm{He}$。相减：$A = 210-4 = 206$，$Z = 84-2 = 82$（铅）。子核：${}^{206}_{82}\mathrm{Pb}$。
A nuclear reaction releases energy because the total mass of the products is less than the total mass of the reactants. What happened to the "missing" mass?核反应释放能量是因为产物总质量小于反应物总质量。"缺失"的质量去哪了？
Q6
It was lost as heat to the surroundings它以热量形式散失到周围环境
It escaped as radiation with no energy它以无能量的辐射形式逃逸
It was converted to energy via $E = mc^2$它通过 $E = mc^2$ 转化为能量
It remains in the products as potential energy它作为势能保留在产物中
Mass and energy are equivalent ($E = mc^2$). The mass defect is converted directly to kinetic energy of the products and radiation energy, consistent with conservation of mass-energy.质量和能量是等价的（$E = mc^2$）。质量亏损直接转化为产物的动能和辐射能，符合质能守恒。
Mass is not destroyed — it is converted to energy. Einstein's $E = mc^2$ quantifies this: the mass defect $\Delta m$ becomes energy $E = \Delta m \cdot c^2$.质量不是消失了——而是转化为能量。爱因斯坦的 $E = mc^2$ 量化了这一点：质量亏损 $\Delta m$ 转化为能量 $E = \Delta m \cdot c^2$。
A carbon-14 sample initially has $N_0$ atoms. After $11\,460$ years, approximately what fraction remains? (Half-life of ${}^{14}\mathrm{C} \approx 5730\ \mathrm{years}$) 🇨🇦 ON D3.11 / AB 30–D3.3k碳-14 样品初始有 $N_0$ 个原子。$11\,460$ 年后剩余约多少分数？（${}^{14}\mathrm{C}$ 半衰期约 $5730$ 年）🇨🇦 ON D3.11 / AB 30–D3.3k
Q7
$1/4$$1/4$
$1/2$$1/2$
$1/8$$1/8$
$1/16$$1/16$
$11\,460 / 5730 = 2$ half-lives. $(1/2)^2 = 1/4$ remains.$11\,460 / 5730 = 2$ 个半衰期。$(1/2)^2 = 1/4$ 剩余。
Number of half-lives $= 11\,460/5730 = 2$. Fraction remaining $= (1/2)^2 = 1/4$.半衰期数 $= 11\,460/5730 = 2$。剩余分数 $= (1/2)^2 = 1/4$。
Which statement best describes why nuclear fusion of hydrogen isotopes powers the Sun?以下哪句话最好地描述了为什么氢同位素核聚变为太阳提供能量？
Q8
Hydrogen nuclei break apart and release stored chemical energy氢核分裂并释放储存的化学能
Electrons fall to lower energy levels and emit photons电子跃迁到较低能级并发射光子
Heavy elements fission into lighter fragments, releasing neutrons重元素裂变为较轻碎片，释放中子
Light nuclei combine into a heavier nucleus; the mass defect is converted to energy via $E = mc^2$轻核结合成较重核；质量亏损通过 $E = mc^2$ 转化为能量
Fusion combines light nuclei (e.g. ${}^1\mathrm{H}$ and isotopes) into heavier helium. The products have less total mass than the reactants; this mass defect is converted to enormous amounts of energy via $E = mc^2$.聚变将轻核（如 ${}^1\mathrm{H}$ 及其同位素）结合成较重的氦。产物总质量小于反应物；此质量亏损通过 $E = mc^2$ 转化为巨大能量。
The Sun runs on nuclear fusion (light nuclei combining), not fission (heavy nuclei splitting) or chemistry. Energy comes from the mass defect via $E = mc^2$.太阳依靠核聚变（轻核结合）运行，而非裂变（重核分裂）或化学反应。能量来自质量亏损，通过 $E = mc^2$。
A nuclear reaction has a mass defect of $0.00500\ \mathrm{u}$. How much energy is released? ($1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$) 🇨🇦 AB 30–D3.6k一个核反应的质量亏损为 $0.00500\ \mathrm{u}$。释放多少能量？（$1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$）🇨🇦 AB 30–D3.6k
Q9
$4.66\ \mathrm{MeV}$
$4.66\ \mathrm{MeV}$ — $0.00500 \times 931.5$$0.00500 \times 931.5$
$931.5\ \mathrm{MeV}$
$0.00500\ \mathrm{MeV}$
$E = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.00500 \times 931.5 = 4.66\ \mathrm{MeV}$.$E = \Delta m \times 931.5\ \mathrm{MeV/u} = 0.00500 \times 931.5 = 4.66\ \mathrm{MeV}$。
Multiply the mass defect (in u) by $931.5\ \mathrm{MeV/u}$: $0.00500 \times 931.5 = 4.66\ \mathrm{MeV}$.将质量亏损（以 u 计）乘以 $931.5\ \mathrm{MeV/u}$：$0.00500 \times 931.5 = 4.66\ \mathrm{MeV}$。
The electron diffraction experiment (Davisson-Germer) demonstrated that electrons have wave properties. This is an example of:电子衍射实验（戴维孙-革末）证明了电子具有波动性质。这是以下的例子：
Q10
The photoelectric effect光电效应
Alpha decayAlpha 衰变
Wave-particle duality of matter物质的波粒二象性
The Compton effect康普顿效应
Electrons producing diffraction patterns when scattered off a crystal confirmed de Broglie's hypothesis that matter has an associated wavelength ($\lambda = h/mv$). This is wave-particle duality applied to matter, not just light.电子在晶体上散射产生衍射图样，证实了德布罗意关于物质具有对应波长（$\lambda = h/mv$）的假设。这是适用于物质（而非仅光）的波粒二象性。
Diffraction is a wave phenomenon. Electrons producing diffraction patterns proves they have wave properties — the wave-particle duality of matter (de Broglie's hypothesis, confirmed by Davisson and Germer).衍射是波动现象。电子产生衍射图样证明它们具有波动性质——物质的波粒二象性（德布罗意假设，由戴维孙和革末证实）。
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Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Modern and nuclear physics is the capstone of the high school physics sequence. The quantum ideas here (photon, wave-particle duality, quantised energy levels) are the foundation of solid-state physics, lasers, LEDs, medical imaging (PET scans, X-rays), and nuclear medicine. The mass-energy equation $E = mc^2$ underlies both nuclear power generation and the astrophysics of stars. There is no dedicated AP Physics course on modern/nuclear physics at the high school level in this repo; the natural next steps are IB Physics HL (which covers quantum and nuclear topics in Themes D and E) and AP Physics 2 (Algebra-Based), which includes modern physics as one of its seven units. The HS Physics Thermodynamics and Waves guides are prerequisite reading for the wave-particle framing in this unit.近代物理与核物理是高中物理序列的顶点。这里的量子思想（光子、波粒二象性、量子化能级）是固态物理、激光、LED、医学成像（PET 扫描、X 射线）和核医学的基础。质能方程 $E = mc^2$ 是核能发电和恒星天体物理学的基础。本仓库中没有针对高中阶段近代/核物理的专门 AP Physics 课程；自然的下一步是 IB Physics HL（在主题 D 和 E 中涵盖量子和核物理）以及 AP Physics 2（代数版），其中近代物理是七个单元之一。本仓库中的 HS Physics 热力学和波的指南是本单元波粒框架的先修内容。

Within High School Physics.在 HS Physics 内部。

Unit 6 (Waves and Sound) and Unit 7 (Light and Geometric Optics) introduce the wave model whose limitations the photoelectric effect exposes in §1. Unit 11 (Thermodynamics) provides the energy-conversion framework that contextualises nuclear power in §7. Unit 10 (Magnetism and EM Induction) builds the electromagnetic radiation picture that quantum physics then quantises into photons.第 6 单元（波与声音）和第 7 单元（光与几何光学）介绍波动模型，而光电效应在 §1 中揭示了其局限性。第 11 单元（热力学）提供能量转化框架，为 §7 的核能提供背景。第 10 单元（磁学与电磁感应）建立电磁辐射图像，量子物理随后将其量子化为光子。

Advanced courses in this repo — IB Physics HL and AP Physics 2.本仓库中的进阶课程——IB Physics HL 与 AP Physics 2。

No dedicated AP modern-physics unit exists in this repo. The next-level treatments of these topics live in IB Physics HL (Themes D: Waves, E: Nuclear and Quantum Physics) and AP Physics 2 (Unit 7: Modern Physics). Both go well beyond this guide: IB Physics HL covers the uncertainty principle, wave functions, and the Standard Model; AP Physics 2 adds the photoelectric effect quantitatively, mass-energy calculations, and nuclear reaction equations. If you are heading to either of those courses, this guide is your conceptual on-ramp.本仓库中没有专门的 AP 近代物理单元。这些主题的更高级处理位于 IB Physics HL（主题 D：波；主题 E：核物理与量子物理）和 AP Physics 2（第 7 单元：近代物理）。两者都远超本指南：IB Physics HL 涵盖不确定性原理、波函数和标准模型；AP Physics 2 增加了定量的光电效应、质能计算和核反应方程。若你正准备这两门课程，本指南是你的概念入门。

DINGRUI SCHOLARS

High School Physics · Modern and Nuclear Physics · v1高中物理 · 近代物理与核物理 · v1