Unit E.1: Structure of the Atom单元 E.1：原子结构
The opening unit of Theme E "Nuclear and quantum physics". The nuclear model and the Geiger-Marsden alpha-scattering evidence for it, nuclear notation and isotopes, the unified atomic mass unit, discrete atomic energy levels read off emission and absorption line spectra, the photon and the relation $E = hf$, nuclear energy levels seen through discrete gamma emission, and the strong nuclear force that holds the nucleus together. The HL extension adds the Bohr model of hydrogen, which lets you compute the wavelengths of the hydrogen spectral lines from first principles.主题 E"核物理与量子物理"的开篇。原子核模型及其卢瑟福（盖革-马斯登）α 散射证据、核素符号与同位素、统一原子质量单位、由发射光谱与吸收光谱读出的分立原子能级、光子及关系式 $E = hf$、由分立 γ 辐射体现的核能级，以及把原子核束缚在一起的强核力。HL 扩展加入氢原子玻尔模型，使你能够从基本原理计算氢光谱线的波长。
How to use this guide本指南使用说明
E.1 is the conceptual gateway to nuclear and quantum physics. The arithmetic is light; the marks come from precise reasoning about evidence (what does alpha-scattering actually show?) and from the single quantitative idea that runs through the whole theme: energy is exchanged in discrete photons, $\Delta E = hf$. Learn the evidence chain and the photon relation cold, and most of E.1 falls out.E.1 是核物理与量子物理的概念入口。计算量很轻；分数来自对证据的精确推理（α 散射究竟说明了什么？），以及贯穿整个主题的一个定量核心：能量以分立光子交换，$\Delta E = hf$。把证据链与光子关系式背熟，E.1 大部分内容自然展开。
Memorise $E = hf = \frac{hc}{\lambda}$ and that a transition between levels emits or absorbs one photon of energy $\Delta E = hf$. Know that alpha-scattering shows the atom is mostly empty with a tiny, dense, positive nucleus. Know nuclear notation $^{A}_{Z}X$: $A$ nucleons, $Z$ protons, $A - Z$ neutrons. Line spectra are discrete because energy levels are discrete.
背熟 $E = hf = \frac{hc}{\lambda}$，以及能级间跃迁发射或吸收一个能量为 $\Delta E = hf$ 的光子。知道 α 散射表明原子大部分是空的、含有一个微小、致密、带正电的原子核。知道核素符号 $^{A}_{Z}X$：$A$ 个核子、$Z$ 个质子、$A - Z$ 个中子。线状光谱分立是因为能级分立。
Be able to argue from the alpha-scattering data to each conclusion (most pass straight through ⇒ mostly empty; rare large-angle deflections ⇒ concentrated charge and mass). Convert fluently between eV and joules. Use $E_n = -\frac{13.6}{n^2}\ \mathrm{eV}$ HL to compute spectral wavelengths. Explain why both atomic and nuclear spectra are discrete, and what the discreteness of gamma spectra tells us about the nucleus.
能够从 α 散射数据论证每一条结论（多数直穿 ⇒ 大部分是空的；偶尔大角度偏转 ⇒ 电荷与质量集中）。熟练在 eV 与焦耳间换算。用 $E_n = -\frac{13.6}{n^2}\ \mathrm{eV}$ HL 计算光谱波长。解释为何原子光谱与核光谱都是分立的，以及 γ 谱的分立性对原子核意味着什么。
The Nuclear Model and Geiger-Marsden Scattering原子核模型与盖革-马斯登散射 E.1 SL+HL
- The atom is mostly empty space with a tiny, dense, positively charged nucleus (原子核) at the centre; electrons occupy the surrounding volume.
- Nearly all the mass and all the positive charge sit in the nucleus.
- Most pass straight through ⇒ the atom is mostly empty.
- A few deflect by large angles, a very few bounce back ⇒ the positive charge and mass are concentrated in a tiny region.
- 原子绝大部分是空的，中心有一个微小、致密、带正电的原子核（nucleus）；电子占据周围空间。
- 几乎全部质量与全部正电荷集中于原子核。
- 多数直穿而过 ⇒ 原子大部分是空的。
- 少数大角度偏转，极少数被弹回 ⇒ 正电荷与质量集中在极小区域。
In the Geiger-Marsden experiment, roughly 1 in 8000 alpha particles was deflected by more than $90^{\circ}$. State what each of the following observations shows about the atom: (a) the vast majority pass nearly undeflected; (b) a tiny fraction bounce back through more than $90^{\circ}$.在盖革-马斯登实验中，约每 8000 个 α 粒子中有 1 个偏转超过 $90^{\circ}$。指出下列各观测说明了原子的什么：(a) 绝大多数几乎不偏转；(b) 极少数被弹回、偏转超过 $90^{\circ}$。
Identify. Each observation is evidence for one feature of the nuclear model. Link cause (a property of the atom) to effect (the trajectory of the alpha).
识别。每条观测都是原子核模型某一特征的证据。把原因（原子的某一属性）与结果（α 的轨迹）联系起来。
Set Up. An alpha particle is positive; it is deflected by the Coulomb repulsion of positive charge. Large deflection needs a strong, close-range force, hence a concentrated charge.
列式。α 粒子带正电；它被正电荷的库仑斥力偏转。大角度偏转需要近距离的强作用力，因而需要集中的电荷。
Execute. (a) Most alphas pass nearly straight through ⇒ most of the atom is empty space; the atom is not a uniform "plum pudding". (b) A tiny fraction reverse direction ⇒ they met something both very massive (so it barely recoils) and of concentrated positive charge (so the repulsion is huge at close range): the nucleus.
求解。(a) 多数 α 近乎直穿 ⇒ 原子大部分是空的；原子不是均匀的"葡萄干布丁"。(b) 极少数反向 ⇒ 它们遇到了既很重（几乎不反冲）又正电荷集中（近距离斥力极大）的东西：原子核。
Evaluate. The 1-in-8000 frequency is itself quantitative evidence that the nucleus is extremely small compared with the atom: most alphas never come close enough to feel a strong force.
评估。1/8000 的频率本身就是定量证据，表明原子核相对原子极小：多数 α 从未靠近到感受强作用力。
Going deeper: why "plum pudding" failed and a closest-approach estimate深入：为何"葡萄干布丁"被否定与最近距离估算
In Thomson's "plum pudding" model, positive charge is spread uniformly over the whole atom. The electric field inside such a smear is weak everywhere, so it could deflect a fast alpha by at most a fraction of a degree. Large-angle scattering is impossible in that model; observing it forces the charge into a tiny central region.
在汤姆孙"葡萄干布丁"模型中，正电荷均匀弥散于整个原子。这种弥散内部电场处处很弱，至多使快速 α 偏转零点几度。该模型无法产生大角度散射；观测到它就迫使电荷集中到极小的中心区域。
For a head-on collision, the alpha stops when all its kinetic energy has become electric potential energy. Setting $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d}$ and solving for $d$ gives the closest approach — an upper bound on the nuclear radius. For a $5\ \mathrm{MeV}$ alpha on gold ($Z = 79$), $d \approx 4.5 \times 10^{-14}\ \mathrm{m}$, consistent with a femtometre-scale nucleus.
正碰时，α 在全部动能转化为电势能处停下。令 $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d}$ 并解出 $d$，即得最近距离——核半径的上界。对金（$Z = 79$）上的 $5\ \mathrm{MeV}$ α，$d \approx 4.5 \times 10^{-14}\ \mathrm{m}$，与飞米尺度的原子核一致。
Nuclear Notation, Isotopes and the Unified Atomic Mass Unit核素符号、同位素与统一原子质量单位 E.1 SL+HL
${}^{A}_{Z}X$:
$$ {}^{A}_{Z}X, \qquad A = \text{nucleon (mass) number}, \quad Z = \text{proton number}, \quad N = A - Z = \text{neutrons}. $$
- Nucleons (核子) = protons + neutrons. $Z$ fixes the element; $A$ counts the nucleons.
- Isotopes (同位素): same $Z$, different $N$ (hence different $A$). Same chemistry, different mass.
${}^{A}_{Z}X$：
$$ {}^{A}_{Z}X, \qquad A = \text{核子数（质量数）}, \quad Z = \text{质子数}, \quad N = A - Z = \text{中子数}. $$
- 核子（nucleons）= 质子 + 中子。$Z$ 决定元素；$A$ 计核子数。
- 同位素（isotopes）：$Z$ 相同、$N$（因而 $A$）不同。化学性质相同，质量不同。
For the nuclide $^{40}_{19}\mathrm{K}$: (a) how many protons, neutrons and nucleons does it contain? (b) Name one isotope of it. (c) Estimate its mass in kg.对核素 $^{40}_{19}\mathrm{K}$：(a) 含多少质子、中子与核子？(b) 写出它的一个同位素。(c) 估算其质量（kg）。
Identify. $A = 40$, $Z = 19$, element K (potassium).
识别。$A = 40$、$Z = 19$，元素 K（钾）。
Set Up. Protons $= Z$; neutrons $= A - Z$; nucleons $= A$. Isotope = same $Z$, different $A$. Mass $\approx A \times 1\ \mathrm{u}$.
列式。质子 $= Z$；中子 $= A - Z$；核子 $= A$。同位素 = $Z$ 相同、$A$ 不同。质量 $\approx A \times 1\ \mathrm{u}$。
Execute. (a) $19$ protons, $40 - 19 = 21$ neutrons, $40$ nucleons. (b) E.g. $^{39}_{19}\mathrm{K}$ (19 protons, 20 neutrons) — same $Z$, different $N$. (c) $m \approx 40 \times 1.661 \times 10^{-27} \approx 6.6 \times 10^{-26}\ \mathrm{kg}$.
求解。(a) $19$ 个质子，$40 - 19 = 21$ 个中子，$40$ 个核子。(b) 例如 $^{39}_{19}\mathrm{K}$（19 质子、20 中子）——$Z$ 相同、$N$ 不同。(c) $m \approx 40 \times 1.661 \times 10^{-27} \approx 6.6 \times 10^{-26}\ \mathrm{kg}$。
Evaluate. Using $1\ \mathrm{u}$ per nucleon ignores the small mass defect (binding energy) and the electron masses, but it is the right order of magnitude — exactly what an exam estimate asks for.
评估。每个核子取 $1\ \mathrm{u}$ 忽略了微小的质量亏损（结合能）与电子质量，但数量级正确——正是考试估算所要求的。
Going deeper: why $1\ \mathrm{u}$ is defined via carbon-12深入：为何用碳-12 定义 $1\ \mathrm{u}$
Masses on the nuclear scale are too small to weigh directly, so they are compared with a standard. Carbon-12 is chosen because it is abundant, stable, and easy to handle in a mass spectrometer; defining $1\ \mathrm{u} = \frac{1}{12}m(^{12}\mathrm{C})$ makes the mass number $A$ a good first estimate of the atomic mass in u. The proton ($1.0073\ \mathrm{u}$) and neutron ($1.0087\ \mathrm{u}$) are each slightly more than $1\ \mathrm{u}$; a bound nucleus is slightly less than $A\ \mathrm{u}$ because of the mass lost as binding energy — the topic of E.4 fission and E.5 fusion.
核尺度的质量太小无法直接称量，故与一个标准比较。选碳-12 是因为它丰度高、稳定、易于在质谱仪中处理；定义 $1\ \mathrm{u} = \frac{1}{12}m(^{12}\mathrm{C})$ 使质量数 $A$ 成为原子质量（以 u 计）的良好初估。质子（$1.0073\ \mathrm{u}$）与中子（$1.0087\ \mathrm{u}$）各略大于 $1\ \mathrm{u}$；束缚的原子核则略小于 $A\ \mathrm{u}$，因为结合能对应的质量损失了——这是 E.4 裂变与 E.5 聚变的主题。
Discrete Energy Levels and Line Spectra分立能级与线状光谱 E.1 SL+HL
A cool hydrogen gas is placed between a white-light source and a spectrometer; separately, the same gas is excited in a discharge tube and viewed directly. Describe each spectrum and explain why the dark lines of the first coincide in wavelength with the bright lines of the second.将低温氢气置于白光源与光谱仪之间；另外，将同种气体在放电管中激发后直接观测。描述两种光谱，并解释为何前者的暗线与后者的亮线波长一致。
Identify. Configuration 1 = absorption (white light through cool gas). Configuration 2 = emission (excited gas viewed directly).
识别。配置 1 = 吸收（白光穿过低温气体）。配置 2 = 发射（激发气体直接观测）。
Set Up. Both processes involve the same set of energy levels, so the same set of $\Delta E$ values, so the same set of photon energies $E = hf$.
列式。两过程涉及同一套能级，故同一套 $\Delta E$，故同一套光子能量 $E = hf$。
Execute. Absorption: a continuous (rainbow) spectrum crossed by dark lines, where photons matching an upward transition were absorbed. Emission: a dark background with bright coloured lines, where electrons falling between the same levels emitted photons. Because the level differences are identical, the dark absorption lines fall at exactly the wavelengths of the bright emission lines.
求解。吸收：连续（彩虹）谱上叠加暗线，对应被吸收的、匹配向上跃迁的光子。发射：暗背景上的亮彩色线，对应电子在同一能级间下落所发射的光子。由于能级差相同，吸收暗线恰好落在发射亮线的波长处。
Evaluate. The coincidence is the key piece of evidence: the same discrete level set governs both emission and absorption, which is exactly what a quantised-energy-level model predicts.
评估。这种重合是关键证据：同一套分立能级同时支配发射与吸收，这正是能级量子化模型所预言的。
Going deeper: why the Sun's spectrum has dark Fraunhofer lines深入：为何太阳光谱有暗夫琅禾费线
The Sun's hot dense interior emits a near-continuous spectrum. As that light passes outward through the cooler gas of the Sun's atmosphere, atoms there absorb photons at their characteristic transition energies, punching dark lines into the continuum. Each set of lines fingerprints a particular element, so the absorption spectrum tells us what the Sun (and distant stars) are made of without ever sampling them — astrophysics built entirely on E.1's discrete energy levels.
太阳炽热致密的内部发出近连续谱。该光向外穿过太阳大气较冷的气体时，那里的原子在其特征跃迁能量处吸收光子，在连续谱上打出暗线。每组谱线是某种元素的"指纹"，故吸收光谱无需取样即可告诉我们太阳（及遥远恒星）的成分——完全建立在 E.1 分立能级之上的天体物理学。
The Photon, Photon Energy and Transitions光子、光子能量与跃迁 E.1 SL+HL
$E = hf$ and $c = f\lambda$. Constants: $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$, $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.
Transitions. A jump between two levels emits/absorbs one photon of energy
$$ \Delta E = hf = E_{\text{high}} - E_{\text{low}}. $$
The electronvolt. $1\ \mathrm{eV}$ (电子伏特) is the energy gained by an electron accelerated through $1\ \mathrm{V}$: $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$. Multiply eV by $1.60\times10^{-19}$ to get joules.
$E = hf$ 与 $c = f\lambda$。常量：$h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$、$c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。
跃迁。两能级间跃迁发射/吸收一个能量为下式的光子
$$ \Delta E = hf = E_{\text{high}} - E_{\text{low}}. $$
电子伏特。$1\ \mathrm{eV}$（电子伏特）是电子经 $1\ \mathrm{V}$ 加速所获得的能量：$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$。eV 乘以 $1.60\times10^{-19}$ 得焦耳。
An electron in an atom drops from a level at $-3.40\ \mathrm{eV}$ to a level at $-13.6\ \mathrm{eV}$. Find the energy, frequency and wavelength of the emitted photon. ($h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$.)原子中电子从 $-3.40\ \mathrm{eV}$ 能级跃迁到 $-13.6\ \mathrm{eV}$ 能级。求所发射光子的能量、频率与波长。（$h = 6.63\times10^{-34}\ \mathrm{J\,s}$，$c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$。）
Identify. Downward transition ⇒ emission. Photon energy $\Delta E = E_{\text{high}} - E_{\text{low}}$.
识别。向下跃迁 ⇒ 发射。光子能量 $\Delta E = E_{\text{high}} - E_{\text{low}}$。
Set Up. $\Delta E = (-3.40) - (-13.6) = 10.2\ \mathrm{eV}$. Convert to joules, then use $\Delta E = hf$ and $\lambda = hc/\Delta E$.
列式。$\Delta E = (-3.40) - (-13.6) = 10.2\ \mathrm{eV}$。换算为焦耳，再用 $\Delta E = hf$ 与 $\lambda = hc/\Delta E$。
Execute.
求解。
$$ \Delta E = 10.2 \times 1.60\times10^{-19} = 1.63\times10^{-18}\ \mathrm{J}. $$ $$ f = \frac{\Delta E}{h} = \frac{1.63\times10^{-18}}{6.63\times10^{-34}} \approx 2.46\times10^{15}\ \mathrm{Hz}. $$ $$ \lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{1.63\times10^{-18}} \approx 1.22\times10^{-7}\ \mathrm{m} = 122\ \mathrm{nm}. $$Evaluate. $122\ \mathrm{nm}$ is ultraviolet — this is the Lyman-$\alpha$ line of hydrogen (transition to the $n = 1$ ground state). A real, checkable answer.
评估。$122\ \mathrm{nm}$ 属紫外——即氢的莱曼-$\alpha$ 线（跃迁到 $n = 1$ 基态）。是可核验的真实结果。
A green photon has wavelength $\lambda = 520\ \mathrm{nm}$. Find its energy in joules and in electronvolts.一个绿光光子波长 $\lambda = 520\ \mathrm{nm}$。求其能量（焦耳与电子伏特）。
Identify. Use $E = hc/\lambda$, then divide by $1.60\times10^{-19}$ to convert J to eV.
识别。用 $E = hc/\lambda$，再除以 $1.60\times10^{-19}$ 把 J 换为 eV。
Set Up & Execute.
列式与求解。
$$ E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{520\times10^{-9}} \approx 3.82\times10^{-19}\ \mathrm{J}. $$ $$ E = \frac{3.82\times10^{-19}}{1.60\times10^{-19}} \approx 2.39\ \mathrm{eV}. $$Evaluate. A few eV is the right scale for a visible photon — visible-light transitions in atoms typically span $\sim 1.5$–$3\ \mathrm{eV}$, which is exactly why eV is the natural unit here.
评估。几个 eV 正是可见光光子的合理量级——原子中可见光跃迁一般在 $\sim 1.5$–$3\ \mathrm{eV}$，这正是此处以 eV 为自然单位的原因。
Going deeper: why eV is convenient and the $\lambda(\mathrm{nm}) = 1240/E(\mathrm{eV})$ shortcut深入：为何 eV 方便，以及 $\lambda(\mathrm{nm}) = 1240/E(\mathrm{eV})$ 速算
Combining $hc = (6.63\times10^{-34})(3.00\times10^{8}) = 1.99\times10^{-25}\ \mathrm{J\,m}$ and dividing through by $1.60\times10^{-19}\ \mathrm{J/eV}$ and $10^{-9}\ \mathrm{m/nm}$ gives the handy shortcut
把 $hc = (6.63\times10^{-34})(3.00\times10^{8}) = 1.99\times10^{-25}\ \mathrm{J\,m}$ 除以 $1.60\times10^{-19}\ \mathrm{J/eV}$ 与 $10^{-9}\ \mathrm{m/nm}$，得到便捷速算式
$$ E(\mathrm{eV}) \approx \frac{1240}{\lambda(\mathrm{nm})}. $$So a $620\ \mathrm{nm}$ photon is about $2.0\ \mathrm{eV}$; a $1240\ \mathrm{nm}$ infrared photon is about $1.0\ \mathrm{eV}$. Use this only as a fast check — show the full $E = hc/\lambda$ working for marks.
于是 $620\ \mathrm{nm}$ 光子约 $2.0\ \mathrm{eV}$；$1240\ \mathrm{nm}$ 红外光子约 $1.0\ \mathrm{eV}$。仅作快速核验——为拿分仍要写出完整的 $E = hc/\lambda$ 过程。
The Bohr Model of Hydrogen氢原子玻尔模型 HL E.1 HL
$E = -\dfrac{13.6}{n^2}\ \mathrm{eV}$.
- $n = 1$: ground state, $E_1 = -13.6\ \mathrm{eV}$ (most bound).
- Levels get closer together as $n$ rises; $E_n \to 0$ as $n \to \infty$ (ionisation).
- Ionisation energy from the ground state $= +13.6\ \mathrm{eV}$.
$E = -\dfrac{13.6}{n^2}\ \mathrm{eV}$。
- $n = 1$：基态，$E_1 = -13.6\ \mathrm{eV}$（束缚最强）。
- $n$ 越大能级越密；$n \to \infty$ 时 $E_n \to 0$（电离）。
- 从基态的电离能$= +13.6\ \mathrm{eV}$。
Using the Bohr model $E_n = -13.6/n^2\ \mathrm{eV}$, find the wavelength of the photon emitted in the hydrogen transition $n = 3 \to n = 2$ (the red H-$\alpha$ line). ($h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$.)用玻尔模型 $E_n = -13.6/n^2\ \mathrm{eV}$，求氢原子跃迁 $n = 3 \to n = 2$（红色 H-$\alpha$ 线）所发射光子的波长。（$h = 6.63\times10^{-34}\ \mathrm{J\,s}$，$c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$。）
Identify. HL transition with $n_i = 3$, $n_f = 2$. Compute $\Delta E$ in eV, convert to J, then $\lambda = hc/\Delta E$.
识别。HL 跃迁，$n_i = 3$、$n_f = 2$。先算 $\Delta E$（eV），换算为 J，再 $\lambda = hc/\Delta E$。
Set Up. $E_3 = -13.6/9 = -1.51\ \mathrm{eV}$; $E_2 = -13.6/4 = -3.40\ \mathrm{eV}$.
列式。$E_3 = -13.6/9 = -1.51\ \mathrm{eV}$；$E_2 = -13.6/4 = -3.40\ \mathrm{eV}$。
Execute.
求解。
$$ \Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \mathrm{eV} = 1.89 \times 1.60\times10^{-19} = 3.02\times10^{-19}\ \mathrm{J}. $$ $$ \lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{3.02\times10^{-19}} \approx 6.58\times10^{-7}\ \mathrm{m} = 658\ \mathrm{nm}. $$Evaluate. $658\ \mathrm{nm}$ is red light, matching the observed H-$\alpha$ Balmer line at $656\ \mathrm{nm}$. The tiny discrepancy comes from rounding $h$ and the $13.6\ \mathrm{eV}$ constant.
评估。$658\ \mathrm{nm}$ 为红光，与观测到的 $656\ \mathrm{nm}$ H-$\alpha$ 巴尔末线吻合。微小偏差源于对 $h$ 与 $13.6\ \mathrm{eV}$ 常量的取整。
Going deeper: spectral series and where Bohr breaks down深入：光谱系与玻尔模型的失效
Transitions ending on $n_f = 1$ form the Lyman series (ultraviolet); on $n_f = 2$ the Balmer series (visible); on $n_f = 3$ the Paschen series (infrared). Each series crowds toward a short-wavelength limit as $n_i \to \infty$. The Bohr model gets hydrogen's wavelengths almost perfectly, but it is a stop-gap: it cannot explain multi-electron atoms, the relative brightness of lines, or fine structure. The full account waits for the quantum (Schrödinger) treatment in HL E.2 — Bohr's quantised levels survive as a special case, but the orbiting-electron picture does not.
终态为 $n_f = 1$ 的跃迁组成莱曼系（紫外）；$n_f = 2$ 为巴尔末系（可见）；$n_f = 3$ 为帕邢系（红外）。每个系在 $n_i \to \infty$ 时向短波极限聚集。玻尔模型几乎完美给出氢的波长，但只是权宜之计：无法解释多电子原子、谱线相对亮度或精细结构。完整解释要等 HL E.2 的量子（薛定谔）处理——玻尔的量子化能级作为特例保留，但电子绕行轨道的图像并不成立。
Nuclear Energy Levels and the Strong Nuclear Force核能级与强核力 E.1 SL+HL
- Attractive between nucleons (proton-proton, proton-neutron, neutron-neutron) at short range.
- Very short range ($\sim 10^{-15}\ \mathrm{m}$, about one nucleon diameter); essentially zero beyond a few fm.
- Becomes repulsive at extremely small separations, which stops the nucleus collapsing.
- 核子之间（质子-质子、质子-中子、中子-中子）在短程内吸引。
- 程极短（$\sim 10^{-15}\ \mathrm{m}$，约一个核子直径）；超过几 fm 基本为零。
- 在极小间距处变为排斥，从而阻止原子核坍缩。
An excited nucleus de-excites between two nuclear energy levels separated by $1.33\ \mathrm{MeV}$, emitting a gamma photon. Find the photon's frequency. ($h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$.)某激发态原子核在相距 $1.33\ \mathrm{MeV}$ 的两核能级间退激，发射一个 γ 光子。求该光子频率。（$h = 6.63\times10^{-34}\ \mathrm{J\,s}$，$1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$。）
Identify. Same $\Delta E = hf$ rule as atomic transitions, but at MeV scale. Convert MeV → J, then $f = \Delta E / h$.
识别。与原子跃迁同样的 $\Delta E = hf$ 规则，但在 MeV 量级。先 MeV → J，再 $f = \Delta E / h$。
Set Up. $1.33\ \mathrm{MeV} = 1.33\times10^{6}\ \mathrm{eV}$.
列式。$1.33\ \mathrm{MeV} = 1.33\times10^{6}\ \mathrm{eV}$。
Execute.
求解。
$$ \Delta E = 1.33\times10^{6} \times 1.60\times10^{-19} = 2.13\times10^{-13}\ \mathrm{J}. $$ $$ f = \frac{\Delta E}{h} = \frac{2.13\times10^{-13}}{6.63\times10^{-34}} \approx 3.21\times10^{20}\ \mathrm{Hz}. $$Evaluate. $\sim 10^{20}\ \mathrm{Hz}$ is firmly in the gamma band, $\sim 10^{5}$ times the frequency of the visible atomic photon in E1.4 — exactly what the MeV-vs-eV energy ratio predicts.
评估。$\sim 10^{20}\ \mathrm{Hz}$ 稳处 γ 波段，约为 E1.4 中可见原子光子频率的 $10^{5}$ 倍——正是 MeV 与 eV 能量比所预言的。
Going deeper: why the strong force must be short range and charge-blind深入：为何强核力必须短程且与电荷无关
If the strong force reached as far as the Coulomb force, nuclei of all sizes would be infinitely stable and matter would clump without limit; its very short range ($\sim 1$–$3\ \mathrm{fm}$) means a nucleon only feels its nearest neighbours, which is why large nuclei become unstable once Coulomb repulsion (long range) outpaces the strong attraction (short range). The force is also charge-independent: it acts equally between p-p, p-n and n-n pairs, which is why both protons and neutrons are needed to bind heavy nuclei. The short-range repulsive core at very small separation is what gives nuclei a roughly constant density rather than letting them collapse to a point.
若强核力像库仑力一样作用很远，各种大小的原子核都将无限稳定、物质会无限聚团；其极短程（$\sim 1$–$3\ \mathrm{fm}$）意味着一个核子只感受到最近邻，这正是大原子核在库仑斥力（长程）超过强吸引（短程）后变得不稳定的原因。该力还与电荷无关：对 p-p、p-n、n-n 同样作用，这正是束缚重核需要质子与中子并存的原因。极小间距处的短程排斥核使原子核密度大致恒定，而非坍缩为一点。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- "State what the observation shows" wants a cause-and-effect sentence, not a description. Pair each scattering observation with the atomic feature it implies.
- "说明该观测表明什么"要的是因果句，不是描述。把每条散射观测与它所暗示的原子特征配对。
- Keep "mostly empty" and "concentrated nucleus" as two separate conclusions. They come from two different observations and markschemes credit them separately.
- 把"大部分是空的"与"原子核集中"作为两条独立结论。它们来自两种不同观测，评分分别给分。
- Decide eV or joules before substituting. Use $\Delta E = hf$ in joules; if levels are in eV, multiply by $1.60\times10^{-19}$ first.
- 代入前先定 eV 还是焦耳。$\Delta E = hf$ 用焦耳；若能级以 eV 给出，先乘 $1.60\times10^{-19}$。
- $\Delta E = E_{\text{high}} - E_{\text{low}}$ is always positive for an emitted photon. A negative answer means you subtracted the wrong way round.
- 发射光子时 $\Delta E = E_{\text{high}} - E_{\text{low}}$ 恒为正。得负值说明减反了。
- Compute $E_{n_i}$ and $E_{n_f}$ separately, then subtract. Watch the double negatives; $E_n$ is always negative.
- 分别算 $E_{n_i}$ 与 $E_{n_f}$ 再相减。注意双重负号；$E_n$ 恒为负。
- For ionisation, the final state is $n \to \infty$ where $E = 0$. Ground-state ionisation energy is $+13.6\ \mathrm{eV}$.
- 电离时末态为 $n \to \infty$（$E = 0$）。基态电离能为 $+13.6\ \mathrm{eV}$。
- Emission = bright lines on dark; absorption = dark lines on a continuum. Both sit at the same wavelengths because the level set is the same.
- 发射 = 暗背景亮线；吸收 = 连续谱上暗线。两者波长相同，因为能级集相同。
- "Discreteness of the spectrum proves discreteness of the levels." Reach for that sentence whenever a spectrum question asks "what does this show".
- "谱的分立性证明能级的分立性。"凡光谱题问"说明了什么"，就用这句。
Flashcards闪卡
Unit E.1 Practice Quiz单元 E.1 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- State the nuclear model: a tiny dense positive nucleus in a mostly empty atom陈述原子核模型：大部分空旷的原子中有一个微小致密的带正电原子核
- Argue from each Geiger-Marsden observation to its conclusion about the atom从盖革-马斯登每条观测论证关于原子的结论
- Compare nucleus size ($\sim10^{-15}\,\mathrm{m}$) with atom size ($\sim10^{-10}\,\mathrm{m}$)比较原子核（$\sim10^{-15}\,\mathrm{m}$）与原子（$\sim10^{-10}\,\mathrm{m}$）尺度
- Read $Z$, $N$ and $A$ from nuclear notation $^{A}_{Z}X$ and identify isotopes从核素符号 $^{A}_{Z}X$ 读出 $Z, N, A$ 并辨认同位素
- Use the unified atomic mass unit, $1\,\mathrm{u} = 1.661\times10^{-27}\,\mathrm{kg}$, to estimate nuclear mass用统一原子质量单位 $1\,\mathrm{u} = 1.661\times10^{-27}\,\mathrm{kg}$ 估算核质量
- Explain how discrete line spectra are evidence for discrete energy levels解释分立线状光谱如何成为分立能级的证据
- Distinguish emission and absorption spectra and explain why their lines coincide区分发射与吸收光谱并解释其谱线为何重合
- Apply $E = hf = hc/\lambda$ and $\Delta E = hf$ to atomic transitions对原子跃迁应用 $E = hf = hc/\lambda$ 与 $\Delta E = hf$
- Convert between electronvolts and joules ($1\,\mathrm{eV} = 1.60\times10^{-19}\,\mathrm{J}$)在电子伏特与焦耳间换算（$1\,\mathrm{eV} = 1.60\times10^{-19}\,\mathrm{J}$）
- HL Use $E_n = -13.6/n^2\,\mathrm{eV}$ to find transition energies and spectral wavelengths用 $E_n = -13.6/n^2\,\mathrm{eV}$ 求跃迁能量与光谱波长
- Explain how discrete gamma spectra show the nucleus has discrete energy levels解释分立 γ 谱如何表明原子核具有分立能级
- Describe the strong nuclear force: attractive, very short range, charge-independent描述强核力：吸引、极短程、与电荷无关
IB Paper-Style PracticeIB 试卷风格练习
E.1 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_E1_*.html with the bilingual built-in pattern.
E.1 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_E1_*.html。