IB Physics HL · 鼎睿学苑

Unit E.2: Quantum physics HL only单元 E.2：量子物理 HL only

A wholly HL-only super-topic of Theme E "Nuclear and quantum physics". Light, which Theme C treated as a wave, here also behaves as a stream of particles (photons), and electrons, treated as particles in Theme A, here also behave as waves. The photoelectric effect forces the photon idea; the de Broglie relation forces the matter-wave idea; the Heisenberg uncertainty principle ties the two together. Every result rests on one constant: Planck's constant $h$.主题 E"核物理与量子物理"中一个完全属于 HL 的超级专题。主题 C 中作为波处理的光，在这里同时表现为粒子流（光子）；主题 A 中作为粒子处理的电子，在这里同时表现为波。光电效应迫使我们接受光子概念；德布罗意关系迫使我们接受物质波概念；海森堡不确定性原理把二者联系起来。所有结论都依赖一个常数：普朗克常数 $h$。

IB Physics · Theme E.2 · First Assessment 2025 Papers 1 · 2 6 Topics · HL only6 个核心专题 · 仅 HL
Read me first先读这里

How to use this guide本指南使用说明

E.2 is a "tell the story right" unit. The algebra is light — almost everything is $E = hf$, $E_{max} = hf - \Phi$, $p = h/\lambda$, and $\lambda = h/p$ — but the marks come from explaining why the classical wave model fails and what the photon and matter-wave pictures fix. Pair each equation with the experimental observation it explains, and keep your units in joules or electronvolts straight.E.2 是一个"把故事讲对"的单元。代数很轻——几乎全是 $E = hf$、$E_{max} = hf - \Phi$、$p = h/\lambda$ 与 $\lambda = h/p$——但分数来自解释经典波动模型为何失败，以及光子图像与物质波图像修补了什么。把每个公式与它所解释的实验观察配对，并把单位（焦耳还是电子伏特）理清。

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If you are cramming如果你在临阵磨枪

Memorise $E = hf = hc/\lambda$, the photoelectric equation $E_{max} = hf - \Phi$, the stopping-voltage link $eV_s = E_{max}$, photon momentum $p = h/\lambda$, and de Broglie $\lambda = h/p$. Know the three classical failures: a threshold frequency exists, emission is instant, and maximum KE does not depend on intensity.

背熟 $E = hf = hc/\lambda$、光电方程 $E_{max} = hf - \Phi$、遏止电压关系 $eV_s = E_{max}$、光子动量 $p = h/\lambda$ 与德布罗意 $\lambda = h/p$。记住三大经典失败：存在截止频率、瞬时发射、最大动能与强度无关。

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If you are going for a 7如果你目标是 7 分

Be able to argue each classical failure quantitatively and link it to a one-photon-one-electron exchange. Read off $h$ and $\Phi$ from an $E_{max}$-vs-$f$ graph (gradient and intercepts). Explain electron diffraction (Davisson–Germer) as direct evidence of matter waves, and state the uncertainty principle as a fundamental limit, not a measurement error.

能定量论证每个经典失败，并把它与"一个光子—一个电子"的交换联系起来。从 $E_{max}$-$f$ 图读出 $h$ 与 $\Phi$（斜率与截距）。把电子衍射（戴维孙—革末实验）解释为物质波的直接证据，并把不确定性原理表述为一个基本极限，而非测量误差。

HL flagHL 标记说明 Super-topic E.2 in its entirety is HL only — SL students do not study quantum physics. Every section below is examinable only at HL on Papers 1 and 2.超级专题 E.2 整体均为 HL only——SL 学生不学习量子物理。下文每一节仅在 HL 的 Paper 1 与 Paper 2 中考核。
Topic E2.1 · The Photoelectric EffectTopic E2.1 · 光电效应

The Photoelectric Effect: What Classical Waves Cannot Explain光电效应：经典波动无法解释的现象 E.2 HL

The experiment. Light shines on a clean metal surface; electrons (photoelectrons, 光电子) are emitted. A classical wave model predicts certain behaviour — and every prediction is wrong.
  • Threshold frequency $f_0$ (截止频率/阈频率): below it, no electrons are emitted, however bright the light. Classical waves predict any frequency should work if intense enough.
  • Instant emission: electrons appear with no measurable delay even at very low intensity. Classical waves predict a build-up time to accumulate enough energy.
  • Maximum KE independent of intensity: $E_{max}$ depends only on frequency, not brightness. Intensity changes the number of electrons, not their maximum KE.
The photoelectric effect (光电效应) is the emission of electrons from a metal surface when electromagnetic radiation above the threshold frequency is incident on it.
实验现象。光照射在洁净金属表面上，发射出电子（光电子，photoelectrons）。经典波动模型会做出若干预言——而每一条预言都错了。
  • 截止频率（threshold frequency）$f_0$：低于它时，无论光多亮都不发射电子。经典波动预言只要足够强，任何频率都应可行。
  • 瞬时发射：即使强度极低，电子也几乎无可测延迟地出现。经典波动预言需要累积时间来积攒能量。
  • 最大动能与强度无关：$E_{max}$ 只依赖频率，与亮度无关。强度改变电子数目，而非其最大动能。
光电效应（photoelectric effect）指当频率高于截止频率的电磁辐射入射到金属表面时，电子从金属表面发射出来的现象。
Worked Example E2.1 (the classical contradiction)E2.1 例题（经典矛盾）

Red light ($f = 4.3 \times 10^{14}\ \mathrm{Hz}$) is shone on a sodium surface at high intensity for a long time, but no electrons are emitted. Blue light ($f = 6.7 \times 10^{14}\ \mathrm{Hz}$) at very low intensity emits electrons immediately. Explain why a classical wave model cannot account for this.高强度红光（$f = 4.3 \times 10^{14}\ \mathrm{Hz}$）长时间照射钠表面，但不发射电子；极低强度的蓝光（$f = 6.7 \times 10^{14}\ \mathrm{Hz}$）却立即发射电子。解释经典波动模型为何无法说明这一现象。

Identify. The deciding variable is frequency, not intensity or exposure time.

识别。决定因素是频率，而非强度或照射时长。

Classical prediction. A wave delivers energy continuously and at a rate set by intensity. Bright red light, given time, should pump enough energy into an electron to free it. It never does.

经典预言。波连续传递能量，传递速率由强度决定。明亮的红光只要有足够时间，应能向电子注入足够能量使其逸出——但它从不发生。

Observation. Below a threshold frequency, no intensity and no exposure time produce emission; above it, emission is instant even at low intensity.

观察。低于截止频率时，任何强度、任何照射时长都不产生发射；高于它时，即使低强度也立即发射。

Evaluate. A continuous-wave picture cannot encode a sharp frequency cut-off, nor instant emission at vanishing intensity. The energy must arrive in frequency-sized lumps absorbed one at a time — a particle (photon) picture.

评估。连续波图像既无法编码尖锐的频率阈值，也无法解释极低强度下的瞬时发射。能量必须以与频率成正比的"小块"到达、被逐一吸收——这是一种粒子（光子）图像。

Going deeper: why intensity controls current, not KE深入：为何强度控制电流而非动能

In the photon picture, intensity is the number of photons arriving per second. Each photon, if energetic enough, ejects at most one electron. Doubling the intensity doubles the photon flux, hence doubles the photoelectron count and the photocurrent — but each photon still carries the same energy $hf$, so each freed electron has the same maximum KE.

在光子图像中，强度即每秒到达的光子数。若每个光子足够"有劲"，至多击出一个电子。强度加倍则光子通量加倍，故光电子数与光电流加倍——但每个光子仍携带相同能量 $hf$，因而每个被释放的电子具有相同的最大动能。

This cleanly separates two experimental knobs: frequency sets the energy per photon (and thus $E_{max}$), while intensity sets the photon count (and thus the saturation current). No classical wave model can split these two.

这就把两个实验旋钮干净地分开：频率决定每个光子的能量（从而决定 $E_{max}$），强度决定光子数（从而决定饱和电流）。任何经典波动模型都无法把这两者分开。

Which observation of the photoelectric effect is impossible to explain with a classical wave model of light?光电效应的哪个观察用经典波动模型无法解释？
E2.1 · Q1
The photocurrent increases with light intensity.光电流随光强增大。
The photocurrent can be reduced to zero by a reverse voltage.反向电压可使光电流减小到零。
Below a threshold frequency, no electrons are emitted at any intensity.低于截止频率时，任何强度都不发射电子。
Electrons gain kinetic energy when emitted.电子发射时获得动能。
A classical wave delivers energy continuously, so a bright enough beam of any frequency should eventually free electrons. The sharp threshold frequency, with no emission below it however intense the light, is the killer observation.经典波连续传递能量，故任何频率只要足够亮最终都应释放电子。存在尖锐的截止频率、低于它时再亮也不发射，是致命的观察。
Intensity-current dependence, stopping voltage, and KE gain are all compatible with a wave picture. The frequency threshold is what classical theory cannot produce.强度—电流关系、遏止电压与获得动能都与波动图像相容。经典理论无法产生的是频率阈值。
Doubling the intensity of monochromatic light above threshold, at fixed frequency, changes:在频率固定且高于阈值时，把单色光强度加倍，会改变：
E2.1 · Q2
The maximum KE of the emitted electrons only.仅发射电子的最大动能。
The number of electrons emitted per second only.仅每秒发射的电子数。
Both the number and the maximum KE.数目与最大动能都改变。
Neither; nothing changes.都不改变。
Intensity sets the photon flux, hence the rate of electron emission (the saturation photocurrent). Maximum KE is fixed by photon energy $hf$, which depends on frequency, not intensity.强度决定光子通量，因而决定电子发射率（饱和光电流）。最大动能由光子能量 $hf$ 决定，只依赖频率而非强度。
Keep the two knobs apart: frequency controls $E_{max}$; intensity controls the count. At fixed frequency, only the number changes.把两个旋钮分开：频率控制 $E_{max}$；强度控制数目。频率固定时，只有数目改变。
Topic E2.2 · Einstein's Photon ExplanationTopic E2.2 · 爱因斯坦的光子解释

Einstein's Photon Equation, Work Function, Threshold Frequency爱因斯坦光子方程、逸出功、截止频率 E.2 HL

The photon (光子). Light of frequency $f$ comes in quanta each of energy E = hf, where $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$ (data booklet). Equivalently $E = hc/\lambda$. $$ E = h f = \frac{h c}{\lambda}. $$ Work function $\Phi$ (逸出功). The minimum energy needed to free the least tightly bound electron from the metal surface. Einstein's photoelectric equation. One photon gives all its energy to one electron; what is left after paying the work function is kinetic energy: $$ E_{max} = h f - \Phi. $$ Threshold frequency (截止频率) $f_0$. Emission just begins when $E_{max} = 0$: $$ f_0 = \frac{\Phi}{h}, \qquad \Phi = h f_0. $$
光子（photon）。频率为 $f$ 的光以量子形式出现，每个量子能量为 E = hf，其中 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$（数据手册）。等价地 $E = hc/\lambda$。 $$ E = h f = \frac{h c}{\lambda}. $$ 逸出功 $\Phi$（work function）。把束缚最松的电子从金属表面释放所需的最小能量。 爱因斯坦光电方程。一个光子把全部能量给一个电子；付清逸出功后剩下的就是动能： $$ E_{max} = h f - \Phi. $$ 截止频率 $f_0$（threshold frequency）。当 $E_{max} = 0$ 时发射恰好开始： $$ f_0 = \frac{\Phi}{h}, \qquad \Phi = h f_0. $$
Worked Example E2.2 (work function and threshold)E2.2 例题（逸出功与阈频率）

A metal has work function $\Phi = 2.3\ \mathrm{eV}$. (a) Find its threshold frequency. (b) Light of wavelength $\lambda = 400\ \mathrm{nm}$ strikes it. Find the maximum KE of the photoelectrons, in eV and in joules. Use $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$, $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$, $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$.某金属逸出功 $\Phi = 2.3\ \mathrm{eV}$。(a) 求其截止频率。(b) 波长 $\lambda = 400\ \mathrm{nm}$ 的光照射其上，求光电子的最大动能（以 eV 与焦耳表示）。取 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$、$c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$、$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$。

Identify. Convert $\Phi$ to joules: $\Phi = 2.3 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}\ \mathrm{J}$.

识别。把 $\Phi$ 换算为焦耳：$\Phi = 2.3 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}\ \mathrm{J}$。

(a) Threshold frequency. $f_0 = \Phi / h$:

(a) 截止频率。$f_0 = \Phi / h$：

$$ f_0 = \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.55 \times 10^{14}\ \mathrm{Hz}. $$

(b) Photon energy at 400 nm. $E = hc/\lambda$:

(b) 400 nm 光子能量。$E = hc/\lambda$：

$$ E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{400 \times 10^{-9}} \approx 4.97 \times 10^{-19}\ \mathrm{J} = 3.11\ \mathrm{eV}. $$

Maximum KE. $E_{max} = hf - \Phi = 3.11 - 2.3 = 0.81\ \mathrm{eV} = 1.30 \times 10^{-19}\ \mathrm{J}$.

最大动能。$E_{max} = hf - \Phi = 3.11 - 2.3 = 0.81\ \mathrm{eV} = 1.30 \times 10^{-19}\ \mathrm{J}$。

Evaluate. The 400 nm photon (3.11 eV) exceeds the 2.3 eV work function, so emission occurs and the surplus 0.81 eV is the maximum KE. Working in eV is faster here; convert at the end only if joules are required.

评估。400 nm 光子（3.11 eV）超过 2.3 eV 的逸出功，故发生发射，多出的 0.81 eV 即最大动能。这里用 eV 计算更快；只在需要焦耳时最后换算。

Going deeper: "maximum" KE and the spread below it深入："最大"动能与其下方的分布

Why maximum? Electrons sit at a range of depths in the metal. The work function $\Phi$ is the minimum escape energy, paid only by the least tightly bound electrons at the very surface. Electrons deeper in must surrender more energy to escape, so they emerge with less KE.

为何是最大动能？金属中电子处于不同深度。逸出功 $\Phi$ 是最小逸出能量，只有最表面、束缚最松的电子才付这么少。更深处的电子逸出时要交出更多能量，故它们以更小的动能逸出。

So $E_{max} = hf - \Phi$ is the upper edge of a continuous distribution of photoelectron energies running from $0$ up to $E_{max}$. The stopping voltage (next section) measures exactly this upper edge.

因此 $E_{max} = hf - \Phi$ 是光电子能量连续分布的上边界，分布从 $0$ 延伸到 $E_{max}$。遏止电压（下一节）测量的正是这个上边界。

A metal's work function is $\Phi$. The threshold frequency $f_0$ is given by:某金属逸出功为 $\Phi$。截止频率 $f_0$ 为：
E2.2 · Q1
$f_0 = \Phi / h$
$f_0 = h \Phi$
$f_0 = h / \Phi$
$f_0 = \Phi / (h c)$
Threshold is where $E_{max} = hf_0 - \Phi = 0$, so $hf_0 = \Phi$ and $f_0 = \Phi / h$. Check units: $\mathrm{J} / (\mathrm{J\,s}) = \mathrm{s^{-1}} = \mathrm{Hz}$. ✓阈值处 $E_{max} = hf_0 - \Phi = 0$，故 $hf_0 = \Phi$，$f_0 = \Phi / h$。验单位：$\mathrm{J} / (\mathrm{J\,s}) = \mathrm{Hz}$。✓
Set $E_{max} = hf_0 - \Phi = 0$ and solve. Only $f_0 = \Phi / h$ has units of hertz.令 $E_{max} = hf_0 - \Phi = 0$ 求解。只有 $f_0 = \Phi / h$ 的单位是赫兹。
Photons of energy $3.0\ \mathrm{eV}$ strike a metal of work function $2.0\ \mathrm{eV}$. The maximum KE of the photoelectrons is:能量 $3.0\ \mathrm{eV}$ 的光子打在逸出功 $2.0\ \mathrm{eV}$ 的金属上。光电子最大动能为：
E2.2 · Q2
$5.0\ \mathrm{eV}$
$2.0\ \mathrm{eV}$
$1.5\ \mathrm{eV}$
$1.0\ \mathrm{eV}$
$E_{max} = hf - \Phi = 3.0 - 2.0 = 1.0\ \mathrm{eV}$. The work function is subtracted, not added.$E_{max} = hf - \Phi = 3.0 - 2.0 = 1.0\ \mathrm{eV}$。逸出功是减去，不是加上。
Einstein's equation subtracts the work function from the photon energy: $E_{max} = hf - \Phi$.爱因斯坦方程从光子能量中减去逸出功：$E_{max} = hf - \Phi$。
Topic E2.3 · Stopping Voltage and the $E_{max}$-$f$ GraphTopic E2.3 · 遏止电压与 $E_{max}$-$f$ 图

Stopping Voltage and Reading $h$ and $\Phi$ off a Graph遏止电压与从图像读出 $h$ 与 $\Phi$ E.2 HL

Stopping voltage $V_s$ (遏止电压). Apply a reverse potential across the photocell. As it rises, fewer electrons reach the collector; at $V = V_s$ even the fastest electron is just turned back, so the photocurrent falls to zero. The work done against the field equals the electron's maximum KE: $$ e V_s = E_{max}. $$ Combine with Einstein. $e V_s = hf - \Phi$, so $$ V_s = \frac{h}{e} f - \frac{\Phi}{e}. $$ The graph. Plotting $E_{max}$ (i.e. $e V_s$) against $f$ gives a straight line:
  • Gradient $= h$ (Planck's constant) — the same for every metal.
  • $f$-intercept $= f_0$ (the threshold frequency).
  • $E_{max}$-intercept $= -\Phi$ (extend the line back to $f = 0$).
遏止电压 $V_s$（stopping voltage）。在光电管两端加反向电势。电势升高时到达收集极的电子减少；当 $V = V_s$ 时连最快的电子也恰被挡回，光电流降为零。逆电场所做的功等于电子的最大动能： $$ e V_s = E_{max}. $$ 与爱因斯坦方程结合。$e V_s = hf - \Phi$，故 $$ V_s = \frac{h}{e} f - \frac{\Phi}{e}. $$ 图像。把 $E_{max}$（即 $e V_s$）对 $f$ 作图得到一条直线：
  • 斜率$= h$（普朗克常数）——对每种金属都相同。
  • $f$ 轴截距$= f_0$（截止频率）。
  • $E_{max}$ 轴截距$= -\Phi$（把直线延伸回 $f = 0$）。
Worked Example E2.3 (gradient gives $h$, intercept gives $\Phi$)E2.3 例题（斜率给出 $h$，截距给出 $\Phi$）

In a photoelectric experiment, $E_{max}$ is measured at two frequencies: $E_{max} = 0.50\ \mathrm{eV}$ at $f = 6.0 \times 10^{14}\ \mathrm{Hz}$ and $E_{max} = 2.16\ \mathrm{eV}$ at $f = 10.0 \times 10^{14}\ \mathrm{Hz}$. The data fall on a straight line. Find (a) Planck's constant from the gradient and (b) the work function.在光电实验中，测得两个频率下的 $E_{max}$：$f = 6.0 \times 10^{14}\ \mathrm{Hz}$ 时 $E_{max} = 0.50\ \mathrm{eV}$，$f = 10.0 \times 10^{14}\ \mathrm{Hz}$ 时 $E_{max} = 2.16\ \mathrm{eV}$。数据落在一条直线上。求 (a) 由斜率得普朗克常数，(b) 逸出功。

Identify. Convert energies to joules: $0.50\ \mathrm{eV} = 8.0 \times 10^{-20}\ \mathrm{J}$, $2.16\ \mathrm{eV} = 3.456 \times 10^{-19}\ \mathrm{J}$.

识别。把能量换算为焦耳：$0.50\ \mathrm{eV} = 8.0 \times 10^{-20}\ \mathrm{J}$，$2.16\ \mathrm{eV} = 3.456 \times 10^{-19}\ \mathrm{J}$。

(a) Gradient = $h$. Since $E_{max} = hf - \Phi$, the slope of $E_{max}$ vs $f$ is $h$:

(a) 斜率 = $h$。由 $E_{max} = hf - \Phi$，$E_{max}$ 对 $f$ 的斜率即 $h$：

$$ h = \frac{\Delta E_{max}}{\Delta f} = \frac{3.456 \times 10^{-19} - 8.0 \times 10^{-20}}{(10.0 - 6.0) \times 10^{14}} = \frac{2.656 \times 10^{-19}}{4.0 \times 10^{14}} \approx 6.64 \times 10^{-34}\ \mathrm{J\,s}. $$

(b) Work function. Use one data point: $\Phi = hf - E_{max}$ at $f = 6.0 \times 10^{14}\ \mathrm{Hz}$:

(b) 逸出功。代入一个数据点：在 $f = 6.0 \times 10^{14}\ \mathrm{Hz}$ 处 $\Phi = hf - E_{max}$：

$$ \Phi = (6.64 \times 10^{-34})(6.0 \times 10^{14}) - 8.0 \times 10^{-20} \approx 3.18 \times 10^{-19}\ \mathrm{J} \approx 1.99\ \mathrm{eV}. $$

Evaluate. The gradient recovers $h$ to within $0.2\%$ of the booklet value $6.63 \times 10^{-34}\ \mathrm{J\,s}$, and $\Phi \approx 2.0\ \mathrm{eV}$ is a sensible metal work function. The graph method needs no absolute calibration — only the slope and intercept matter.

评估。斜率给出的 $h$ 与手册值 $6.63 \times 10^{-34}\ \mathrm{J\,s}$ 相差约 $0.2\%$，$\Phi \approx 2.0\ \mathrm{eV}$ 是合理的金属逸出功。图像法不需要绝对标定——只看斜率与截距。

Going deeper: why all metals share one gradient深入：为何所有金属共用同一斜率

Rewrite the line as $E_{max} = h f - \Phi$. The slope $h$ multiplies $f$ and carries no reference to the metal at all — it is a universal constant of nature. The metal enters only through the intercept $\Phi$, which shifts the line up or down (and moves $f_0 = \Phi/h$ left or right) but never tilts it.

把直线写成 $E_{max} = h f - \Phi$。斜率 $h$ 乘以 $f$，完全不涉及金属——它是自然界的普适常数。金属只通过截距 $\Phi$ 进入，截距上下平移直线（并使 $f_0 = \Phi/h$ 左右移动），但绝不改变倾斜度。

So a plot for several different metals is a family of parallel lines, all of slope $h$, each with its own threshold frequency. Historically this is exactly how Millikan (1916) measured $h$ to confirm Einstein's equation.

因此几种不同金属的图是一族平行直线，斜率都是 $h$，各有自己的截止频率。历史上密立根（1916）正是用此法测得 $h$，验证了爱因斯坦方程。

On a graph of maximum photoelectron KE against frequency, the gradient of the line represents:在光电子最大动能对频率的图像中，直线斜率表示：
E2.3 · Q1
The work function $\Phi$.逸出功 $\Phi$。
Planck's constant $h$.普朗克常数 $h$。
The threshold frequency $f_0$.截止频率 $f_0$。
The elementary charge $e$.基本电荷 $e$。
$E_{max} = hf - \Phi$ has the form $y = mx + c$ with $m = h$. The gradient is Planck's constant; the $f$-intercept is $f_0$ and the vertical intercept is $-\Phi$.$E_{max} = hf - \Phi$ 形如 $y = mx + c$，$m = h$。斜率是普朗克常数；$f$ 轴截距是 $f_0$，纵轴截距是 $-\Phi$。
Compare with $y = mx + c$: $E_{max} = hf - \Phi$. The coefficient of $f$ — the gradient — is $h$.与 $y = mx + c$ 比较：$E_{max} = hf - \Phi$。$f$ 的系数（斜率）是 $h$。
The stopping voltage for a photocell is $1.2\ \mathrm{V}$. The maximum KE of the photoelectrons is:某光电管的遏止电压为 $1.2\ \mathrm{V}$。光电子最大动能为：
E2.3 · Q2
$1.2\ \mathrm{J}$
$0\ \mathrm{eV}$
$1.2\ \mathrm{eV}$
$1.92 \times 10^{-19}\ \mathrm{eV}$
$E_{max} = eV_s$. With $V_s = 1.2\ \mathrm{V}$, $E_{max} = e \times 1.2\ \mathrm{V} = 1.2\ \mathrm{eV}$ (or $1.92 \times 10^{-19}\ \mathrm{J}$). The electronvolt is defined so $E_{max}$ in eV equals $V_s$ in volts.$E_{max} = eV_s$。$V_s = 1.2\ \mathrm{V}$ 时，$E_{max} = e \times 1.2\ \mathrm{V} = 1.2\ \mathrm{eV}$（即 $1.92 \times 10^{-19}\ \mathrm{J}$）。电子伏特的定义使得以 eV 为单位的 $E_{max}$ 数值等于以伏特为单位的 $V_s$。
$E_{max} = eV_s$. An electron through $1.2\ \mathrm{V}$ gains $1.2\ \mathrm{eV}$ by definition of the electronvolt.$E_{max} = eV_s$。电子经过 $1.2\ \mathrm{V}$ 按电子伏特定义获得 $1.2\ \mathrm{eV}$。
Topic E2.4 · The Photon as a Particle: MomentumTopic E2.4 · 光子作为粒子：动量

Photon Momentum $p = h/\lambda$光子动量 $p = h/\lambda$ E.2 HL

A photon carries momentum. Although it has zero rest mass, a photon has momentum p = h / λ: $$ p = \frac{h}{\lambda} = \frac{h f}{c} = \frac{E}{c}. $$ Three equivalent forms. Use $\lambda = c/f$ and $E = hf$ to move between them. The last form $p = E/c$ is the cleanest when you already know the photon energy. Direction. Momentum points along the direction of travel of the light. This is what lets light exert radiation pressure and lets a photon recoil an electron (Compton scattering, beyond IB detail but the same $p = h/\lambda$).
光子携带动量。尽管静止质量为零，光子仍具有动量 p = h / λ： $$ p = \frac{h}{\lambda} = \frac{h f}{c} = \frac{E}{c}. $$ 三种等价形式。用 $\lambda = c/f$ 与 $E = hf$ 在它们之间转换。已知光子能量时，最后一式 $p = E/c$ 最简洁。 方向。动量沿光的传播方向。正是它让光能施加辐射压，并让光子能把电子反冲（康普顿散射，超出 IB 细节，但仍是同一 $p = h/\lambda$）。
Worked Example E2.4 (photon momentum)E2.4 例题（光子动量）

A green laser emits light of wavelength $\lambda = 532\ \mathrm{nm}$. (a) Find the momentum of one photon. (b) Find its energy in joules. Use $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$, $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.绿色激光发出波长 $\lambda = 532\ \mathrm{nm}$ 的光。(a) 求一个光子的动量。(b) 求其能量（焦耳）。取 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$、$c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。

Identify. $\lambda = 532 \times 10^{-9}\ \mathrm{m}$. Use $p = h/\lambda$ directly.

识别。$\lambda = 532 \times 10^{-9}\ \mathrm{m}$。直接用 $p = h/\lambda$。

(a) Momentum.

(a) 动量。

$$ p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{532 \times 10^{-9}} \approx 1.25 \times 10^{-27}\ \mathrm{kg\,m\,s^{-1}}. $$

(b) Energy. Use $E = pc$ (or $E = hc/\lambda$):

(b) 能量。用 $E = pc$（或 $E = hc/\lambda$）：

$$ E = p c = (1.25 \times 10^{-27})(3.00 \times 10^{8}) \approx 3.74 \times 10^{-19}\ \mathrm{J} \approx 2.34\ \mathrm{eV}. $$

Evaluate. The momentum is tiny — a single photon barely nudges anything — but in a laser beam of $10^{18}$ photons per second the cumulative momentum transfer is measurable as radiation pressure. The energy $2.34\ \mathrm{eV}$ is in the visible range, as expected for green light.

评估。动量极小——单个光子几乎推不动任何东西——但在每秒 $10^{18}$ 个光子的激光束中，累积动量传递可作为辐射压被测得。能量 $2.34\ \mathrm{eV}$ 处于可见光范围，与绿光相符。

Going deeper: $p = E/c$ from relativistic energy深入：由相对论能量得 $p = E/c$

The relativistic energy-momentum relation is $E^{2} = (pc)^{2} + (m_0 c^{2})^{2}$. A photon has rest mass $m_0 = 0$, so the relation collapses to $E^{2} = (pc)^{2}$, i.e. $E = pc$, hence $p = E/c$.

相对论能量-动量关系为 $E^{2} = (pc)^{2} + (m_0 c^{2})^{2}$。光子静质量 $m_0 = 0$，关系塌缩为 $E^{2} = (pc)^{2}$，即 $E = pc$，故 $p = E/c$。

Substituting the photon energy $E = hf = hc/\lambda$ gives $p = hf/c = h/\lambda$. So photon momentum is not an extra postulate — it follows from $E = hf$ together with the massless-particle limit of special relativity (Theme A.5).

代入光子能量 $E = hf = hc/\lambda$ 得 $p = hf/c = h/\lambda$。可见光子动量并非额外假设——它由 $E = hf$ 与狭义相对论无质量粒子极限（主题 A.5）共同推出。

A photon's wavelength is halved. Its momentum:某光子波长减半。其动量：
E2.4 · Q1
Halves.减半。
Stays the same.不变。
Quarters.变为四分之一。
Doubles.加倍。
$p = h/\lambda$, so momentum is inversely proportional to wavelength. Halving $\lambda$ doubles $p$ (and also doubles the energy, since $E = pc$).$p = h/\lambda$，动量与波长成反比。$\lambda$ 减半则 $p$ 加倍（能量也加倍，因 $E = pc$）。
$p = h/\lambda$ is an inverse relationship: smaller wavelength means larger momentum. Halving $\lambda$ doubles $p$.$p = h/\lambda$ 为反比：波长越小动量越大。$\lambda$ 减半则 $p$ 加倍。
A photon has energy $E$. Its momentum is:某光子能量为 $E$。其动量为：
E2.4 · Q2
$E / c$
$E c$
$E c^{2}$
$E / c^{2}$
For a massless photon, $E = pc$, so $p = E/c$. Equivalently $p = hf/c = h/\lambda$.对无质量光子，$E = pc$，故 $p = E/c$。等价地 $p = hf/c = h/\lambda$。
A photon satisfies $E = pc$ (massless limit of $E^2 = (pc)^2 + (m_0 c^2)^2$), so $p = E/c$. The $E/c^2$ form is a mass, not a momentum.光子满足 $E = pc$（$E^2 = (pc)^2 + (m_0 c^2)^2$ 的无质量极限），故 $p = E/c$。$E/c^2$ 是质量，不是动量。
Topic E2.5 · Matter Waves and de BroglieTopic E2.5 · 物质波与德布罗意

The de Broglie Wavelength and Electron Diffraction德布罗意波长与电子衍射 E.2 HL

Matter waves (物质波). If light (a wave) carries particle-like momentum, de Broglie proposed that particles (matter) carry a wavelength. Every moving particle has a de Broglie wavelength (德布罗意波长) λ = h / p: $$ \lambda = \frac{h}{p} = \frac{h}{m v}. $$ For an electron accelerated through a voltage $V$: $eV = \tfrac{1}{2} m v^{2}$ gives $p = \sqrt{2 m e V}$, so $$ \lambda = \frac{h}{\sqrt{2 m e V}}. $$ Evidence: electron diffraction (电子衍射). Davisson and Germer (1927) fired electrons at a nickel crystal and saw a diffraction pattern — peaks and troughs at angles set by the de Broglie wavelength. Diffraction is a wave-only phenomenon; particles that diffract must have a wavelength.
物质波（matter waves）。若光（波）携带类粒子的动量，德布罗意提出粒子（物质）携带波长。每个运动粒子都有一个德布罗意波长（de Broglie wavelength） λ = h / p： $$ \lambda = \frac{h}{p} = \frac{h}{m v}. $$ 对经电压 $V$ 加速的电子：$eV = \tfrac{1}{2} m v^{2}$ 给出 $p = \sqrt{2 m e V}$，故 $$ \lambda = \frac{h}{\sqrt{2 m e V}}. $$ 证据：电子衍射（electron diffraction）。戴维孙与革末（1927）将电子射向镍晶体，观察到衍射图样——在由德布罗意波长决定的角度处出现极大与极小。衍射是仅波动才有的现象；能衍射的粒子必有波长。
Worked Example E2.5 (de Broglie wavelength of an electron)E2.5 例题（电子的德布罗意波长）

An electron is accelerated from rest through a potential difference of $54\ \mathrm{V}$ (the Davisson–Germer value). Find its de Broglie wavelength. Use $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$, $m_e = 9.11 \times 10^{-31}\ \mathrm{kg}$, $e = 1.60 \times 10^{-19}\ \mathrm{C}$.电子从静止经 $54\ \mathrm{V}$ 电势差加速（戴维孙—革末实验所用值）。求其德布罗意波长。取 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$、$m_e = 9.11 \times 10^{-31}\ \mathrm{kg}$、$e = 1.60 \times 10^{-19}\ \mathrm{C}$。

Identify. Energy gained $= eV$ becomes kinetic energy. Find momentum, then $\lambda = h/p$.

识别。获得的能量 $= eV$ 转化为动能。先求动量，再求 $\lambda = h/p$。

Momentum. $p = \sqrt{2 m_e e V}$:

动量。$p = \sqrt{2 m_e e V}$：

$$ p = \sqrt{2(9.11 \times 10^{-31})(1.60 \times 10^{-19})(54)} \approx 3.97 \times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$

Wavelength. $\lambda = h/p$:

波长。$\lambda = h/p$：

$$ \lambda = \frac{6.63 \times 10^{-34}}{3.97 \times 10^{-24}} \approx 1.67 \times 10^{-10}\ \mathrm{m} = 0.167\ \mathrm{nm}. $$

Evaluate. This wavelength ($\sim 0.17\ \mathrm{nm}$) is comparable to atomic spacing in a crystal ($\sim 0.2\ \mathrm{nm}$), which is exactly why electrons diffract off a crystal lattice. A baseball would have $\lambda \sim 10^{-34}\ \mathrm{m}$ — far too small to ever show wave effects.

评估。此波长（约 $0.17\ \mathrm{nm}$）与晶体原子间距（约 $0.2\ \mathrm{nm}$）相当，这正是电子能在晶格上衍射的原因。一个棒球的 $\lambda \sim 10^{-34}\ \mathrm{m}$——太小，永远不会显示波动效应。

Going deeper: why Davisson–Germer settles the argument深入：为何戴维孙—革末实验一锤定音

Electrons are unambiguously particles: they carry definite charge and mass, and you can count them one at a time. Yet a beam of them, scattered off a crystal, produces interference maxima at angles obeying the Bragg condition $n\lambda = 2 d \sin\theta$ — with $\lambda$ equal to the de Broglie value $h/p$. Plugging in the measured angle reproduces $\lambda$ within experimental error.

电子无疑是粒子：带确定电荷与质量，可逐个计数。然而一束电子被晶体散射后，在满足布拉格条件 $n\lambda = 2 d \sin\theta$ 的角度处产生干涉极大——其中 $\lambda$ 等于德布罗意值 $h/p$。代入实测角度，可在实验误差内重现 $\lambda$。

Crucially, the pattern persists even when electrons are sent through one at a time: each lands as a single dot, but the accumulated dots build the diffraction pattern. The wave describes the probability of where each particle lands — the seed of the quantum interpretation.

关键在于，即使电子逐个发射，图样仍然出现：每个电子落为一个点，但累积的点构成衍射图样。波描述的是每个粒子落点的概率——这正是量子诠释的种子。

The de Broglie wavelength of a particle of momentum $p$ is:动量为 $p$ 的粒子的德布罗意波长为：
E2.5 · Q1
$\lambda = h / p$
$\lambda = p / h$
$\lambda = h p$
$\lambda = h / (p c)$
$\lambda = h/p$ — identical in form to the photon relation $p = h/\lambda$ rearranged. Larger momentum means shorter wavelength.$\lambda = h/p$——与光子关系 $p = h/\lambda$ 移项后形式相同。动量越大，波长越短。
de Broglie proposed $\lambda = h/p$ for matter, mirroring the photon's $p = h/\lambda$. Wavelength is inversely proportional to momentum.德布罗意对物质提出 $\lambda = h/p$，与光子的 $p = h/\lambda$ 对应。波长与动量成反比。
The Davisson–Germer experiment is taken as direct evidence that:戴维孙—革末实验被视为以下哪项的直接证据：
E2.5 · Q2
Light is made of photons.光由光子构成。
Electrons have a work function.电子具有逸出功。
Electrons exhibit wave behaviour (diffraction).电子表现出波动行为（衍射）。
Electrons have no momentum.电子没有动量。
Electrons scattered off a nickel crystal produced a diffraction pattern matching the de Broglie wavelength $\lambda = h/p$. Diffraction is a wave phenomenon, so this confirms matter waves.电子被镍晶体散射产生与德布罗意波长 $\lambda = h/p$ 相符的衍射图样。衍射是波动现象，故证实了物质波。
The photoelectric effect is the photon evidence. Davisson–Germer is the matter-wave evidence: electrons (particles) diffracting like waves.光电效应是光子的证据。戴维孙—革末是物质波的证据：电子（粒子）像波一样衍射。
Topic E2.6 · Heisenberg Uncertainty and DualityTopic E2.6 · 不确定性原理与二象性

The Heisenberg Uncertainty Principle and Wave–Particle Duality海森堡不确定性原理与波粒二象性 E.2 HL

Heisenberg uncertainty principle (不确定性原理), qualitative. You cannot know both the position and the momentum of a particle to arbitrary precision at the same time. The product of the uncertainties has a floor set by $h$: $$ \Delta x \, \Delta p \gtrsim \frac{h}{4\pi}. $$ Read it carefully. Pinning down position ($\Delta x$ small) forces momentum to become uncertain ($\Delta p$ large), and vice versa. This is a fundamental limit of nature, not a flaw in your apparatus or technique. Wave–particle duality (波粒二象性). Light and matter each show wave behaviour (interference, diffraction) and particle behaviour (photons, localized electrons) depending on the experiment. Neither picture alone is complete; the uncertainty principle is the price of trying to use both at once.
海森堡不确定性原理（uncertainty principle），定性。你无法同时以任意精度确定粒子的位置与动量。两个不确定度之积有一个由 $h$ 决定的下限： $$ \Delta x \, \Delta p \gtrsim \frac{h}{4\pi}. $$ 仔细理解。把位置确定得很准（$\Delta x$ 小）会迫使动量变得不确定（$\Delta p$ 大），反之亦然。这是自然界的基本极限，并非仪器或技术的缺陷。 波粒二象性（wave–particle duality）。光与物质都依实验既显示波动行为（干涉、衍射），又显示粒子行为（光子、定域电子）。任一图像单独都不完整；不确定性原理正是试图同时使用两者所付的代价。
Worked Example E2.6 (estimating a momentum uncertainty)E2.6 例题（估算动量不确定度）

An electron is confined to a region the size of an atom, $\Delta x \approx 1.0 \times 10^{-10}\ \mathrm{m}$. Estimate the minimum uncertainty in its momentum, and comment on the implication. Use $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$.某电子被约束在原子尺度的区域内，$\Delta x \approx 1.0 \times 10^{-10}\ \mathrm{m}$。估算其动量的最小不确定度，并讨论其含义。取 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$。

Identify. Use the equality limit $\Delta x \, \Delta p \approx h/(4\pi)$ to estimate the minimum $\Delta p$.

识别。取等号极限 $\Delta x \, \Delta p \approx h/(4\pi)$ 估算最小 $\Delta p$。

Momentum uncertainty.

动量不确定度。

$$ \Delta p \approx \frac{h}{4\pi \, \Delta x} = \frac{6.63 \times 10^{-34}}{4\pi (1.0 \times 10^{-10})} \approx 5.3 \times 10^{-25}\ \mathrm{kg\,m\,s^{-1}}. $$

Implication. Dividing by the electron mass gives a velocity uncertainty $\Delta v = \Delta p / m_e \approx 5.8 \times 10^{5}\ \mathrm{m\,s^{-1}}$ — comparable to actual electron speeds in atoms.

含义。除以电子质量得速度不确定度 $\Delta v = \Delta p / m_e \approx 5.8 \times 10^{5}\ \mathrm{m\,s^{-1}}$——与原子中电子的实际速率相当。

Evaluate. Confining the electron to atomic dimensions makes its momentum hugely uncertain. This is why electrons in atoms cannot "sit still" in tiny orbits as classical physics would allow — uncertainty forbids it, and it underpins the very stability and size of atoms.

评估。把电子约束到原子尺度使其动量极其不确定。这正是原子中电子不能像经典物理允许的那样在微小轨道上"静止"的原因——不确定性原理禁止它，并支撑了原子的稳定性与尺度。

Going deeper: synthesis — when to use which picture深入：综合——何时用哪种图像

The two halves of E.2 are mirror images. The photoelectric effect makes a wave (light) behave like a particle: discrete photons of energy $E = hf$ and momentum $p = h/\lambda$. Electron diffraction makes a particle (the electron) behave like a wave: a de Broglie wavelength $\lambda = h/p$. The very same constant $h$, and the very same equation $p = h/\lambda$, links the pair.

E.2 的两半是镜像。光电效应让波（光）表现得像粒子：能量 $E = hf$、动量 $p = h/\lambda$ 的离散光子。电子衍射让粒子（电子）表现得像波：德布罗意波长 $\lambda = h/p$。正是同一常数 $h$、同一方程 $p = h/\lambda$ 把二者联系起来。

Which face shows depends on the experiment. Detection events (a click, a dot, a freed electron) are particle-like; propagation and interference (diffraction fringes, build-up of a pattern) are wave-like. The uncertainty principle is the statement that you can never make both faces sharp at once — momentum (wave-like, $p = h/\lambda$) and position (particle-like, localized) are conjugate.

显示哪一面取决于实验。探测事件（咔哒声、亮点、被释放的电子）是类粒子的；传播与干涉（衍射条纹、图样的累积）是类波的。不确定性原理表明：你永远无法同时使两面都清晰——动量（类波，$p = h/\lambda$）与位置（类粒子，定域）是共轭量。

If the uncertainty in a particle's position $\Delta x$ is made smaller, then the minimum uncertainty in its momentum $\Delta p$:若把粒子位置的不确定度 $\Delta x$ 变小，则其动量最小不确定度 $\Delta p$：
E2.6 · Q1
Also becomes smaller.也变小。
Becomes larger.变大。
Stays the same.不变。
Becomes exactly zero.恰为零。
$\Delta x \, \Delta p \gtrsim h/(4\pi)$ is a fixed floor. If $\Delta x$ shrinks, $\Delta p$ must grow to keep the product above the limit. Position and momentum trade off against each other.$\Delta x \, \Delta p \gtrsim h/(4\pi)$ 是固定下限。$\Delta x$ 减小时，$\Delta p$ 必须增大以使乘积仍高于该限。位置与动量此消彼长。
The product $\Delta x \, \Delta p$ cannot drop below $h/(4\pi)$. Shrinking one uncertainty forces the other to grow.乘积 $\Delta x \, \Delta p$ 不能低于 $h/(4\pi)$。一个不确定度减小迫使另一个增大。
Which single relation most directly ties together the photon (light as particle) and de Broglie (particle as wave) pictures?哪一个关系最直接地把光子（光作粒子）与德布罗意（粒子作波）两种图像联系起来？
E2.6 · Q2
$E_{max} = hf - \Phi$
$eV_s = E_{max}$
$f_0 = \Phi / h$
$p = h / \lambda$
$p = h/\lambda$ assigns a momentum to a wave (photon) and, rearranged as $\lambda = h/p$, assigns a wavelength to a particle (de Broglie). The same equation runs both ways — the heart of duality.$p = h/\lambda$ 给波（光子）赋予动量，移项为 $\lambda = h/p$ 又给粒子（德布罗意）赋予波长。同一方程双向运行——二象性的核心。
The photoelectric and stopping-voltage relations are about light-as-particle only. The bridge to matter waves is $p = h/\lambda \Leftrightarrow \lambda = h/p$.光电与遏止电压关系只涉及光作粒子。通往物质波的桥梁是 $p = h/\lambda \Leftrightarrow \lambda = h/p$。
Exam Tactics考试技巧

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

Units: joules vs electronvolts (every paper)单位：焦耳与电子伏特（每张试卷）
  • Decide your unit before you start and convert work function and photon energy into the same one. Mixing eV and J inside $E_{max} = hf - \Phi$ is the most common slip.
  • 动笔前先定单位，把逸出功与光子能量都换算到同一单位。在 $E_{max} = hf - \Phi$ 中混用 eV 与 J 是最常见的失误。
  • Remember $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$, and that a stopping voltage of $V_s$ volts gives $E_{max} = V_s$ in eV directly.
  • 记住 $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$，且遏止电压为 $V_s$ 伏特时直接给出以 eV 为单位的 $E_{max} = V_s$。
Explaining classical failure (Paper 2 standard)解释经典失败（Paper 2 常考）
  • Name all three observations: threshold frequency, instant emission, KE independent of intensity. Then state, for each, the classical wave prediction it contradicts.
  • 列出三大观察：截止频率、瞬时发射、动能与强度无关。再对每一项说明它所违背的经典波动预言。
  • Tie the fix to a one-photon-one-electron exchange. "Each electron absorbs a single photon of energy $hf$" is the sentence markschemes reward.
  • 把修补归结为"一个光子—一个电子"的交换。"每个电子吸收一个能量为 $hf$ 的光子"是评分要点。
Graph questions (data-response Section B)图像题（数据题 B 部分）
  • $E_{max}$-vs-$f$: gradient is $h$, $f$-intercept is $f_0$, vertical intercept is $-\Phi$. State which feature you are reading and show the rise/run.
  • $E_{max}$-$f$ 图：斜率是 $h$，$f$ 轴截距是 $f_0$，纵轴截距是 $-\Phi$。说明你在读哪个特征，并给出"上升/水平"读数。
  • If the plot is $V_s$ vs $f$ instead, the gradient is $h/e$, so multiply by $e$ to recover $h$.
  • 若改画 $V_s$-$f$ 图，斜率为 $h/e$，故乘以 $e$ 才能得到 $h$。
Duality and uncertainty (explain-in-words)二象性与不确定性（文字解释）
  • State the uncertainty principle as a fundamental limit, not a measurement error. Examiners penalise "the apparatus disturbs the particle" as the sole explanation.
  • 把不确定性原理表述为基本极限，而非测量误差。仅以"仪器扰动粒子"作解释会被扣分。
  • For duality, name an experiment for each face: photoelectric effect (particle nature of light), electron diffraction (wave nature of matter).
  • 论二象性时，为每一面各举一个实验：光电效应（光的粒子性）、电子衍射（物质的波动性）。
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Photon energy?光子能量？
$$E = hf = \frac{hc}{\lambda}$$
Einstein's photoelectric equation?爱因斯坦光电方程？
$$E_{max} = hf - \Phi$$
Work function $\Phi$?逸出功 $\Phi$？
Minimum energy to free the least-bound surface electron.释放束缚最松的表面电子所需最小能量。
Threshold frequency?截止频率？
$$f_0 = \frac{\Phi}{h}$$
Stopping-voltage relation?遏止电压关系？
$$eV_s = E_{max}$$
Gradient of $E_{max}$-vs-$f$ graph?$E_{max}$-$f$ 图的斜率？
Planck's constant $h$.普朗克常数 $h$。
$E_{max}$-intercept of that graph?该图的 $E_{max}$ 轴截距？
$$-\Phi$$
Three classical failures of the photoelectric effect?光电效应的三大经典失败？
Threshold frequency; instant emission; $E_{max}$ independent of intensity.截止频率；瞬时发射；$E_{max}$ 与强度无关。
Photon momentum?光子动量？
$$p = \frac{h}{\lambda} = \frac{E}{c}$$
de Broglie wavelength?德布罗意波长？
$$\lambda = \frac{h}{p}$$
$\lambda$ of electron through voltage $V$?经电压 $V$ 加速电子的 $\lambda$？
$$\lambda = \frac{h}{\sqrt{2meV}}$$
Evidence for matter waves?物质波的证据？
Electron diffraction (Davisson–Germer).电子衍射（戴维孙—革末实验）。
Heisenberg uncertainty principle?海森堡不确定性原理？
$$\Delta x\,\Delta p \gtrsim \frac{h}{4\pi}$$
Wave–particle duality?波粒二象性？
Light and matter each show wave and particle behaviour depending on the experiment.光与物质依实验各自显示波动与粒子行为。
Mixed Practice综合练习

Unit E.2 Practice Quiz单元 E.2 练习测验

A photon has frequency $5.0 \times 10^{14}\ \mathrm{Hz}$. Its energy is (use $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$):某光子频率 $5.0 \times 10^{14}\ \mathrm{Hz}$。其能量为（取 $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$）：
Q1
$1.3 \times 10^{-48}\ \mathrm{J}$
$7.5 \times 10^{-19}\ \mathrm{J}$
$3.3 \times 10^{-19}\ \mathrm{J}$
$3.3 \times 10^{-34}\ \mathrm{J}$
$E = hf = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.32 \times 10^{-19}\ \mathrm{J}$ (about $2.1\ \mathrm{eV}$).$E = hf = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.32 \times 10^{-19}\ \mathrm{J}$（约 $2.1\ \mathrm{eV}$）。
Multiply $h$ by $f$: $E = hf$. The exponents add: $-34 + 14 = -20$, and $6.63 \times 5.0 = 33$, giving $3.3 \times 10^{-19}\ \mathrm{J}$.$h$ 乘以 $f$：$E = hf$。指数相加 $-34 + 14 = -20$，$6.63 \times 5.0 = 33$，得 $3.3 \times 10^{-19}\ \mathrm{J}$。
A clean metal surface emits photoelectrons under blue light but not under red light of the same intensity. The best single explanation is:洁净金属表面在蓝光下发射光电子，但在同强度红光下不发射。最佳单一解释是：
Q2
Blue light is more intense than red light.蓝光比红光更强。
Each blue photon has energy above the work function; each red photon does not.每个蓝光子能量高于逸出功；每个红光子不够。
Blue light delivers energy faster, so electrons accumulate it quicker.蓝光传能更快，故电子积累得更快。
Red photons are absorbed before reaching the surface.红光子在到达表面前被吸收。
Blue has higher frequency, so $hf_{\text{blue}} > \Phi$ but $hf_{\text{red}} < \Phi$. One photon, one electron: if a single photon cannot supply $\Phi$, no emission occurs no matter how many arrive.蓝光频率更高，故 $hf_{蓝} > \Phi$ 而 $hf_{红} < \Phi$。一个光子对一个电子：若单个光子无法提供 $\Phi$，再多光子也不发射。
Emission depends on per-photon energy $hf$ versus $\Phi$, not on intensity or accumulation. Blue clears the threshold per photon; red does not.发射取决于单光子能量 $hf$ 与 $\Phi$ 的比较，而非强度或累积。蓝光每个光子越过阈值，红光不能。
On a graph of $E_{max}$ against frequency $f$ for a single metal, the line is extended back to $f = 0$. The vertical intercept equals:单一金属的 $E_{max}$-频率 $f$ 图中，直线延伸回 $f = 0$。纵轴截距等于：
Q3
$-\Phi$
$+\Phi$
$f_0$
$h$
$E_{max} = hf - \Phi$. At $f = 0$, $E_{max} = -\Phi$. The $f$-intercept (where $E_{max} = 0$) is $f_0 = \Phi/h$; the slope is $h$.$E_{max} = hf - \Phi$。$f = 0$ 时 $E_{max} = -\Phi$。$f$ 轴截距（$E_{max} = 0$ 处）为 $f_0 = \Phi/h$；斜率为 $h$。
Put $f = 0$ into $E_{max} = hf - \Phi$: the vertical intercept is $-\Phi$. Do not confuse it with the $f$-intercept $f_0$ or the slope $h$.把 $f = 0$ 代入 $E_{max} = hf - \Phi$：纵轴截距为 $-\Phi$。不要与 $f$ 轴截距 $f_0$ 或斜率 $h$ 混淆。
A proton and an electron move with the same speed. Compared with the electron, the proton's de Broglie wavelength is:质子与电子以相同速率运动。与电子相比，质子的德布罗意波长：
Q4
The same.相同。
Larger, because the proton is heavier.更大，因为质子更重。
Larger, because the proton has more charge.更大，因为质子电荷更多。
Much smaller, because its momentum is much larger.小得多，因为其动量大得多。
$\lambda = h/p = h/(mv)$. At equal speed, the heavier proton ($\approx 1836$ times the electron mass) has far greater momentum, so a far smaller wavelength.$\lambda = h/p = h/(mv)$。速率相同时，更重的质子（约为电子质量的 $1836$ 倍）动量大得多，故波长小得多。
$\lambda = h/(mv)$. Larger mass at equal speed means larger momentum, hence smaller wavelength. Charge is irrelevant here.$\lambda = h/(mv)$。速率相同时质量越大动量越大，故波长越小。电荷在此无关。
Which statement about the Heisenberg uncertainty principle is correct?关于海森堡不确定性原理，哪句正确？
Q5
It arises only because measuring apparatus is imperfect.它仅因测量仪器不完善而产生。
It lets us know position and momentum exactly if we are careful.只要小心，就能同时精确知道位置与动量。
It is a fundamental limit: $\Delta x\,\Delta p$ cannot fall below about $h/(4\pi)$.它是基本极限：$\Delta x\,\Delta p$ 不能低于约 $h/(4\pi)$。
It applies to photons but not to electrons.它只适用于光子而不适用于电子。
The principle is a fundamental feature of nature, applying to all quantum objects: the product $\Delta x\,\Delta p$ has a floor of order $h/(4\pi)$, independent of how good your instruments are.该原理是自然界的基本特征，适用于所有量子客体：乘积 $\Delta x\,\Delta p$ 有约 $h/(4\pi)$ 的下限，与仪器优劣无关。
It is not about apparatus quality and is not restricted to one kind of particle. It is a fundamental floor $\Delta x\,\Delta p \gtrsim h/(4\pi)$ for any quantum object.它与仪器优劣无关，也不限于某一种粒子。对任何量子客体它都是基本下限 $\Delta x\,\Delta p \gtrsim h/(4\pi)$。
Self-check自查

Readiness Checklist备考清单

Tick each item when you can do it cold, without notes, on your first attempt.

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IB Physics HL · Unit E.2 Quantum physics · HL only · v1 · First Assessment 2025IB 物理 HL · 单元 E.2 量子物理 · 仅 HL · v1 · 首次考核 2025