Shenzhen Hong Kong Pui Kiu College Longhua Xinyi School

2026-2027 Mid-term Assessment · Physics2026-2027 期中考试 · 物理详解

Grade 12 Physics  |  21 Oct 2026  |  Paper A (24 MCQs) + Paper B (3 structured questions)

Candidate 張珩 ZHANG Shuheng Lucas Class 12H2 · (30) Total 37 / 60 (61.7%) Topics Photoelectric effect · Atomic models · Hydrogen spectrum · Matter waves · Resolving power · Nanomaterials
Paper total
61.7 %
总分
Section A · MC
20 / 30
选择题部分
Section B · Long
17 / 30
结构题部分
MCQs correct
16 / 24
答对题数
Wrong items
3·4·5·6·7
12·14·16
错题编号

Jump to · Section A （red = answered wrongly）

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Section B · Structured questions

B1 · Photoelectric effect B2 · Hydrogen-like ion & Rydberg formula B3 · Scanning tunnelling microscope Formulae & constants Top ten pitfalls

0
Score & Knowledge-Gap Analysis得分与知识点结构分析

Understand where the marks were lost first, then read the item-by-item solutions — that is the fastest route.先看清分数丢在哪里，再逐题看解析，效率最高。

Mark breakdown成绩结构

PartQuestionsMaxScoreRate
Section A — Multiple choice1 – 24302066.7%
Section B — Q1 (Photoelectric effect)1(a)–(c)10770%
Section B — Q2 (Hydrogen-like ion)2(a)–(b)10550%
Section B — Q3 (STM)3(a)–(e)10550%
Total合计—603761.7%

Where the marks went丢分地图

8 MCQs wrong — 10 marks lost选择题错 8 题

  • Q3 / Q5 — graph questions: the relationship was not written down before reading intercepts and slopes.图像题：未先写出关系式再判断截距与斜率代表什么。
  • Q6 / Q7 — concept confusion between "must be true", the stopping potential and the threshold frequency."必然成立"、遏止电压与截止频率三者被混淆。
  • Q12 — excited-state counting: the fourth excited state is n = 5, not n = 4.激发态计数：第四激发态为 n=5。
  • Q14 — which Bohr assumption is newly added and which is compatible with Rutherford.玻尔假设中哪条新增、哪条与卢瑟福相容分不清。
  • Q16 — the Lyman series ends on n = 1, so absorption arrows must start at n = 1.Lyman 系终点为 n=1，吸收箭头须由 n=1 向上。

Main losses in Section B结构题主要失分点

  • 1(a) 0/2 — never stated that wave energy is accumulated continuously; this is the award point.未点出"波模型下能量可连续累积"这一采分核心。
  • 1(c) partial — for blue light at equal intensity the maximum photocurrent must decrease (fewer photons), not increase.蓝光同强度时最大光电流应减小。
  • 2(b)(ii) 1/3 — wrote only "No"; the modified formula (replace R by Z²R = 4R) was missing.只写 "No"，未给出修正公式。
  • 3(b) 0/3 — the imaging mechanism (scan → exponential dependence of current on gap → map) was missing.缺"扫描 → 电流随间距指数变化 → 成像"机制。
  • 3(d) 0/1 — "modification" is not an accepted answer; atomic manipulation was required.应写作 atomic manipulation。

Knowledge map知识点地图

TopicItemsCore requirement
Photoelectric effect & photonsA1–A8, B1hf = φ + Kmax; eVs = Kmax; I–V and Vs–f graphs; the three failures of the wave model光电方程、遏止电压、两类图像、波动说失效的三点理由
Evolution of atomic modelsA9–A11, A14Thomson plum-pudding → Rutherford nuclear → Bohr quantisation: what each can and cannot explain汤姆孙 → 卢瑟福 → 玻尔，各自能解释什么
Hydrogen spectrumA12, A13, A15, A16, B2En = −13.6/n² eV; series limits and excited-state numbering; emission vs absorption arrows; derivation of R能级公式、谱系编号、发射/吸收箭头方向、R 的推导
Matter waves & electron diffractionA17, A18, A21λ = h/p; higher kinetic energy → shorter λ → narrower fringes德布罗意波长与条纹间距
Resolving power & microscopesA20, A22, A23, B3Rayleigh criterion θ = 1.22λ/D; TEM vs STM principles and limits瑞利判据；TEM 与 STM 的原理与限制
Nanomaterials & allotropesA19, A24Same atoms, different arrangement; surface and quantum effects同素异形体；表面效应与量子效应

Answer key · Section A （red = your answer was wrong · 红色为你答错的题）

1B2D3B4B5C6D7A8D9A10C11D12C13A14A15C16D17B18D19B20C21A22A23B24C
1 Photoelectric circuit实验电路
answerB
youB✓
Which circuit correctly shows how the stopping voltage is measured with a photocell?下列哪一个电路正确地表示用光电池测量遏止电压？
A+VA
B+AV
C+AV
DAV
Four requirements: ① a reversing voltage (metal plate / cathode at the higher potential); ② ammeter in series to measure the photocurrent; ③ voltmeter in parallel across the photocell; ④ an adjustable supply (potentiometer) so that the current can be brought to zero. 测遏止电压的四个必要条件：① 加反向电压（金属板电位高于阳极）；② 电流表串联；③ 电压表并联；④ 电源可调，便于找到电流恰为零的临界点。
  • A The voltmeter is in series with the photocell (its huge resistance blocks the photocurrent) while the ammeter is placed in parallel — both meters are in the wrong position.电压表串入光电池支路、电流表却并联在两极之间，两个表都接错。
  • B Ammeter in series, voltmeter in parallel, and the cell polarity puts the metal plate (cathode) at the higher potential — a retarding voltage. Correct.电流表串联、电压表并联，且金属板接正极（加遏止电压），正确。
  • C The meters are placed correctly, but the battery is reversed, so an accelerating voltage is applied and the current can never be reduced to zero.电表位置正确，但电池极性反接，加的是加速电压，电流无法降为零。
  • D An a.c. supply is used, so the applied voltage keeps changing and the stopping voltage cannot be defined.用交流电源，电压不断变化，无法确定遏止电压。
You treated C as identical to B. Check meter connection first, then supply polarity — two independent tests. 你把 C 与 B 当成同一个电路。判断时分两步独立检查：先看电表串/并联，再看电源极性。
Only under a retarding voltage does the field do negative work on the photoelectrons, so the current falls to zero at eVs = Kmax. 只有在反向电压下电场力才做负功，电流才会随电压增大而减小到零。
2 Photons · concepts光子说
answerD
youD✓
Which statements about the concept of a photon are correct?
(1) All phenomena of light can be explained by the photon concept. (2) If the intensity of radiation of fixed wavelength increases, the number of photons increases. (3) For radiation of wavelength λ, each photon carries energy hc/λ.关于光子概念，下列陈述哪（些）正确？
  • (1) False — interference, diffraction and polarisation are wave properties and need the wave model. Light is both, so "all phenomena" is an absolute statement and therefore wrong.干涉、衍射、偏振必须用波动模型解释；光具有波粒二象性，"所有现象"这种绝对化表述错误。
  • (2) True — at fixed wavelength each photon carries the same energy hc/λ, so higher intensity means more photons per second.波长固定则单光子能量固定，强度增大即单位时间光子数增多。
  • (3) True — E = hf = hc/λ.E = hf = hc/λ。
(2) and (3) → D. 结论：(2)(3) 正确，选 D。
3 I–V graph光电流–电压图像
answerB
youA✗
Radiation of fixed frequency but different intensities falls on a photocell. Which graph of photoelectric current Ip against applied voltage V is correct? (Take V positive when the metal plate is at the lower potential.)固定频率、不同强度的辐射照射光电池，哪幅 Ip–V 图正确？
AVI
BVI
CVI
DVI
Write the relationship before reading the graph. ① eVs = hf − φ depends only on frequency, so all curves for the same frequency must meet the negative V-axis at the same point; ② the saturation current is proportional to the number of electrons per second, hence to intensity, so higher intensity gives a higher plateau. 先写出关系式：① 遏止电压只由频率决定，同频率曲线必须交于同一点；② 饱和光电流正比于强度，强度越大平台越高。
V I −Vs strong weak same f ⇒ same Vs
Correct graph: a common intercept at −Vs, different saturation levels正确图像：所有曲线交于 −Vs，饱和电流高度不同
  • A Different V-axis intercepts — implies the stopping potential changes with intensity. Violates ①.两条曲线截距不同，暗示遏止电压随强度变化，违反①。
  • B Same intercept, different plateaus — satisfies both tests.截距相同、平台不同，完全符合①②。
  • C Equal plateaus (current independent of intensity) and different intercepts — wrong twice.平台等高且截距不同，两处都错。
  • D Curves cross and intercepts differ — violates ①.曲线相交、截距不同，违反①。
You chose A. Graph questions are rarely about memorised shapes: decide the intercept from the frequency, then the plateau from the intensity. 你选了 A。图像题不要凭形状记忆：先用频率判断截距，再用强度判断平台。
4 Wave model vs photoelectric effect波动说的矛盾
answerB
youD✗
Which of the following are implications of the wave theory that are inconsistent with the photoelectric effect?
(1) The number of photons is proportional to the intensity. (2) The energy of an electromagnetic wave is transferred to the irradiated surface continuously. (3) The energy transfer rate of an electromagnetic wave is independent of its frequency.下列哪（些）是光的波动说的推论，且与光电效应矛盾？

Read the stem carefully: the statement must be both an implication of the wave theory and in conflict with the photoelectric effect.

  • (1) Does not qualify — this is a statement of the photon model, not of the wave model, and it is in fact consistent with the photoelectric effect (saturation current ∝ intensity). It is the distractor.这是光子模型的陈述，且与光电效应相符（饱和电流∝强度），不符合题干条件。
  • (2) Qualifies — a continuously distributed wave would let an electron accumulate energy, so weak light should show a measurable delay. Electrons actually appear almost instantaneously.连续传能应导致可测的延迟，而实验是瞬时发射，矛盾。
  • (3) Qualifies — if the transfer rate depended only on amplitude, any frequency would work at sufficient intensity, so no threshold frequency would exist. A threshold frequency is observed.若能流率与频率无关，则任何频率只要够强都能发射，就不存在截止频率，与实验矛盾。
(2) and (3) → B. 结论：(2)(3)，选 B。
You chose D, counting (1) as a contradiction. Statement (1) is a particle-model claim that agrees with the photoelectric effect — that is exactly what this item is testing. 你选了 D，把 (1) 也算作矛盾。(1) 属于粒子模型且与光电效应相符，这正是本题的区分点。
5 Vs–f graph遏止电压–频率图像
answerC
youA✗
The stopping potentials Vs of three metals are measured for different frequencies f. Which graph shows a possible experimental result?三种金属的遏止电压随频率变化，哪幅图可能是实验结果？
AfVs
BfVs
CfVs
DfVs
From eVs = hf − φ:
Vs = (h/e)·f − φ/e
  • Slope = h/e is the same for every metal ⇒ the three lines must be parallel;斜率恒为 h/e，与金属无关，三条线必平行。
  • Intercept f0 = φ/h differs, because a larger work function gives a higher threshold frequency;横截距 f₀=φ/h 不同，逸出功越大截距越大。
  • Vertical intercept = −φ/e is always negative.纵截距 −φ/e 必为负值。
f Vs P Q R slope = h/e for every metal ⇒ parallel lines
Correct graph: three parallel lines with different intercepts (dashed below the axis, where no photocurrent exists)正确图像：三条平行直线，横截距不同
  • A Different slopes meeting on the vertical axis — slopes must be equal.斜率不同却在纵轴相交，斜率必须相同。
  • B All lines through the origin, implying φ = 0.都过原点，意味着逸出功为零。
  • C Parallel lines with different frequency intercepts — correct.平行且横截距不同，正确。
  • D A common point on the f-axis (same threshold frequency) but different slopes — self-contradictory.交于 f 轴同一点却斜率不同，自相矛盾。
You chose A, probably because "lines converging downwards" looks like textbook figures. The single decisive test is: slope is always h/e, so the lines must be parallel. 你选了 A。唯一判据是：斜率恒为 h/e，三条线必须平行。
6 Certainties in the photoelectric effect必然结论
answerD
youC✗
Yellow light ejects electrons from a metal. Which statements must be correct?
(1) Electrons will also be emitted if the light is replaced by blue light. (2) Electrons are emitted almost immediately. (3) The maximum kinetic energy is independent of intensity.黄光已能使光电子逸出，下列哪（些）必然正确？
  • (1) True — blue light has a higher frequency, so if yellow already exceeds the threshold, blue certainly does.蓝光频率更高，黄光已超截止频率，蓝光必然也超。
  • (2) True — instantaneous emission (< 10−9 s) is one of the basic experimental facts of the photoelectric effect, and it is precisely what the wave model cannot explain.瞬时发射是光电效应的基本实验事实。
  • (3) True — Kmax = hf − φ depends on frequency and the metal only; intensity controls the number of electrons.最大动能只由频率与金属决定，强度只影响电子数目。
(1)(2)(3) → D. 三项皆对，选 D。
You chose C, doubting the instantaneous emission. The word must signals statements that follow from established facts whenever emission occurs. 你选了 C。题干中的 must 提示：只要发生发射，这些结论就必然成立。
Memory anchor: frequency sets the energy, intensity sets the number.记忆锚点：频率管动能，强度管数量。
7 Reverse voltage & threshold反向电压与截止频率
answerA
youD✗
With frequency f1 a current is detected. When the voltage is gradually raised to V the current just falls to zero. Keeping the voltage at V but switching to frequency f2, a current appears again. Which are correct?
(1) f2 > f1 (2) V is the stopping potential for f1 (3) f1 is the threshold frequency.初始 f₁ 有电流；升压到 V 时电流恰为零；保持 V 改用 f₂ 又有电流。哪（些）正确？

First identify the applied voltage: the battery's positive terminal is on the metal plate, so the voltage is retarding.

  • (2) True — the retarding voltage that just reduces the photocurrent to zero is the stopping potential, so eV = Kmax(f1).使光电流恰为零的反向电压就是遏止电压。
  • (1) True — at the same retarding voltage V, f2 still produces a current, so Kmax(f2) > eV = Kmax(f1), hence f2 > f1.同一遏止电压下 f₂ 仍有电流，故其最大动能更大，频率更高。
  • (3) False — with f1 and no applied voltage a current already exists, so f1 > f0. The threshold frequency is the frequency that just fails to eject electrons.f₁ 在零电压下已有电流，说明 f₁ 大于截止频率。
(1) and (2) → A. 结论：(1)(2)，选 A。
You chose D. Keep the three quantities strictly apart: threshold frequency f0 = φ/h (can electrons be emitted at all), incident frequency f (sets Kmax) and stopping potential Vs (the retarding voltage that kills the current). 你选了 D。务必区分：截止频率 f₀=φ/h、入射频率 f、遏止电压 Vs。
8 Photoelectric equation光电方程计算
answerD
youD✓
The work function of a metal is 5.12×10−19 J. The incident frequency is four times the threshold frequency. Find the maximum kinetic energy of the photoelectrons.逸出功 5.12×10⁻¹⁹ J，入射频率为截止频率的 4 倍，求最大动能。
Let the threshold frequency be f0; then φ = hf0 and f = 4f0.设截止频率 f₀，则逸出功 φ = hf₀，入射频率 f = 4f₀。
Photoelectric equation: Kmax = hf − φ = 4hf0 − hf0 = 3φ光电方程：最大动能等于入射频率倍数减一的差乘以逸出功。
Kmax = 3 × 5.12×10−19 = 1.536×10−18 J代入数值。
Convert: 1.536×10−18 ÷ 1.60×10−19 = 9.6 eV换算为电子伏特。
Answer D (9.60 eV). 答案 D。
There is no need to find f0 numerically: use Kmax = (n − 1)φ when the frequency is n times the threshold. 不必求 f₀ 的数值，直接用倍数关系 Kmax = (n−1)φ。
9 Atomic model diagram原子模型图
answerA
youA✓
Which diagram best represents J. J. Thomson's atomic model?哪幅图最能代表汤姆孙的原子模型？
A++++
B
C+
D−
Thomson's plum-pudding model (1904): positive charge is spread uniformly throughout the atom and the electrons are embedded in it. 汤姆孙"葡萄干布丁模型"：正电荷均匀分布在整个原子中，电子嵌于其中。
  • A A uniform positive sphere with embedded electron dots — correct.均匀正电球加嵌入的电子，正确。
  • B A random cluster of particles, not a coherent model.无规则堆积，不像一个模型。
  • C A concentrated nucleus with orbiting electrons — this is Rutherford's model.中心集中核加绕核电子，属卢瑟福核式模型。
  • D A negatively charged nucleus — wrong charge sign.核带负电，符号错误。
Answer A. 答案 A。
10 Alpha scatteringα 散射实验
answerC
youC✓
Which conclusion about electrons can be drawn from the alpha-particle scattering experiment?从 α 粒子散射实验可得出关于电子的哪一结论？

Observations and inferences:

  • Most α-particles pass almost undeflected ⇒ most of the atom is empty space;绝大多数不偏转，说明原子内部大部分是空的。
  • A few are deflected through large angles ⇒ positive charge and nearly all the mass are concentrated in a tiny nucleus;极少数大角度散射，说明正电荷与质量集中在极小的核内。
  • An α-particle is about 7300 times heavier than an electron and is barely affected by electrons ⇒ electrons are very light compared with the nucleus.α 粒子比电子重约 7300 倍，几乎不受电子影响，说明电子质量远小于核。
  • A Electrons are negatively charged — from Thomson's cathode-ray work.电子带负电来自阴极射线实验。
  • B Electrons at the nucleus — simply wrong.电子在核内，错误。
  • C Electrons are very light compared with the nucleus — correct and inferred here.电子质量远小于核，正确。
  • D Electrons in circular orbits — the experiment gives no orbital information.该实验不能给出轨道信息。
Answer C. 答案 C。
11 Rutherford's model卢瑟福核式模型
answerD
youD✓
Which statements best describe Rutherford's atomic model?
(1) Most of the space occupied by an atom is empty. (2) All positive charge is concentrated in a small region at the centre. (3) Electrons orbit the nucleus.哪（些）最好地描述了卢瑟福的原子模型？

Rutherford's nuclear model (1911) contains exactly these three elements.

(1)(2)(3) → D. 三项皆为卢瑟福模型内容，选 D。
Contrast with Q14: "electrons move in circular orbits" does belong to Rutherford; "electrons occupy certain allowed orbits and do not radiate" is Bohr's new quantum assumption and does not. 与第 14 题对照："绕核圆周运动"属卢瑟福；"在允许轨道上且不辐射"是玻尔新增假设。
12 Hydrogen series氢原子谱系
answerC
youB✗
The Pfund series contains emission lines from transitions ending on the fourth excited state. Find the line of longest wavelength in this series.Pfund 系终止于第四激发态，求其中波长最长的谱线。
Count the levels: ground state n = 1, first excited state n = 2 … so the fourth excited state is n = 5.基态 n=1，故第四激发态为 n=5。
Longest wavelength ⇔ smallest energy gap ⇔ the adjacent transition 6 → 5.波长最长对应能量差最小，即相邻跃迁 6→5。
ΔE = 13.6(1/5² − 1/6²) = 13.6(0.0400 − 0.02778) = 0.1662 eV计算能级差。
λ = 1240 / 0.1662 ≈ 7.46×103 nm ≈ 7480 nm用 hc = 1240 eV nm 换算波长。
Answer C (7480 nm). 答案 C。
You chose B. Two traps: (i) the excited-state count (4th excited state = n = 5); (ii) "longest wavelength" means the transition closest to the final level (6 → 5), not the farthest. 你选了 B。两个陷阱：激发态编号；最长波长取最靠近终态的跃迁。
Infrared series to memorise: Paschen n→3, Brackett n→4, Pfund n→5. Higher series ⇒ longer wavelength. 红外谱系速记：Paschen→3、Brackett→4、Pfund→5，级数越大波长越长。
13 Energy levels能级与电离能
answerA
youA✓
A hydrogen atom is in the third excited state. Find the minimum energy needed to ionize it.氢原子处于第三激发态，求使其电离所需的最小能量。
The third excited state is n = k + 1 = 4.第 k 激发态对应 n = k+1，故第三激发态为 n=4。
E4 = −13.6 / 4² = −0.85 eV代入能级公式。
Ionisation energy = 0 − E4 = 0.85 eV电离能等于 0 减去该能级能量。
Answer A. 答案 A。
Ionisation energy = |En| = 13.6/n² eV. The same counting trap as Q12: k-th excited state ⇒ n = k + 1. 电离能 = 13.6/n² eV；与第 12 题同一类编号陷阱。
14 Bohr vs Rutherford玻尔与卢瑟福
answerA
youD✗
Which statements about Bohr's atomic model are consistent with Rutherford's model?
(1) The electron moves in a circular orbit. (2) The centripetal force is provided by the attraction of the nucleus. (3) The electron occupies certain allowed orbits without emitting EM radiation.哪（些）关于玻尔模型的陈述与卢瑟福模型相容？

Rutherford's model is a classical picture: the electron orbits the nucleus and the Coulomb attraction supplies the centripetal force. Bohr kept that picture and added quantisation.

  • (1) Consistent — circular orbits already appear in Rutherford's model.圆轨道本就是卢瑟福模型的内容。
  • (2) Consistent — in both treatments mv²/r = ke²/r².两种处理中库仑引力都提供向心力。
  • (3) Not consistent — this is Bohr's new postulate, introduced precisely to escape the classical prediction that a radiating electron would spiral into the nucleus. It contradicts, rather than extends, the Rutherford picture.这是玻尔为摆脱"辐射导致电子坠入核"的经典困难而新增的定态假设，与卢瑟福模型不相容。
(1) and (2) → A. 结论：(1)(2)，选 A。
You chose D. The test is always: was this already in Rutherford's model? Rutherford knew only "electrons orbit the nucleus" — he knew nothing about stationary states. 你选了 D。判据始终是：这一条卢瑟福模型里有没有？
15 Spectral lines氢原子光谱的性质
answerC
youC✓
Which statements about the spectral lines of hydrogen are correct?
(1) The line of highest frequency comes from the transition n = ∞ to n = 1. (2) Each spectral line arises from a transition in which one photon is emitted or absorbed. (3) An absorption spectrum is obtained when low-pressure hydrogen is bombarded by energetic electrons.哪（些）关于氢原子谱线的陈述正确？
  • (1) True — highest frequency ⇔ largest photon energy ⇔ largest level gap, i.e. ∞ → 1, the limit of the Lyman series.频率最大对应能量差最大，即 ∞→1，Lyman 系极限。
  • (2) True — one line, one photon absorbed or emitted; that is the definition of a spectral line.一条谱线对应一个光子的吸收或发射。
  • (3) False — electron bombardment excites the atoms, which then emit spontaneously, so you obtain an emission spectrum. An absorption spectrum requires white light to pass through the gas so that specific photons are removed.电子轰击是碰撞激发，随后自发辐射，得到发射谱；吸收谱需白光穿过气体。
(1) and (2) → C. 结论：(1)(2)，选 C。
Emission = bright lines on dark; absorption = dark lines on a continuous spectrum. "Electron bombardment" always means emission. 发射谱为暗底明线，吸收谱为连续谱上的暗线；电子轰击必为发射谱。
16 Lyman absorptionLyman 系吸收图
answerD
youC✗
Which energy-level diagram correctly represents the transitions of the Lyman series in the hydrogen absorption spectrum?哪幅能级图正确表示吸收光谱中 Lyman 系的跃迁？
AE1234∞
BE1234∞
CE1234∞
DE1234∞
Two steps. ① Lyman series: the transition endpoints are all at n = 1. ② Absorption: the atom starts low and is excited upward, so the arrows must point from n = 1 upwards. 两步判断：① Lyman 系终点为 n=1；② 吸收是低能级向高能级跃迁，箭头须由 n=1 向上。
E n = ∞ n = 4 n = 3 n = 2 n = 1 (ground) absorption — all arrows start from n = 1
Lyman absorption: n = 1 → n = 2, 3, 4, …, ∞Lyman 吸收：从 n=1 分别跃迁到更高能级
  • A Arrows point down into n = 1 — that is Lyman emission.箭头向下指向 n=1，属发射谱。
  • B Neither the directions nor the level combinations match the Lyman series.方向与能级组合都不符合 Lyman 系。
  • C Arrows start at n = 2 — that is Balmer absorption.箭头由 n=2 向上，属巴耳末系吸收。
  • D Arrows start at n = 1 and rise to 2, 3, 4, ∞ — correct.箭头由 n=1 向上，正确。
You chose C — the classic Lyman/Balmer mix-up. Remember Lyman ends at n = 1, Balmer at n = 2, Paschen at n = 3, then add "absorption upward, emission downward". 你选了 C，属 Lyman 与 Balmer 的典型混淆；记住各谱系终点再叠加箭头方向即可。
17 Evidence for matter waves物质波的实验证据
answerB
youB✓
Which phenomenon best demonstrates that electrons have wave properties?哪个实验现象最能说明电子具有波动性？
  • A The photoelectric effect demonstrates the particle nature of light.光电效应说明光的粒子性。
  • B The Davisson–Germer electron-diffraction experiment — electrons scattered by a crystal form a diffraction pattern, direct evidence for matter waves.戴维孙–革末电子衍射实验，电子波动性的直接证据。
  • C The discrete line spectrum reflects quantised energy levels, not electron wavelength.分立谱线反映能级量子化，不是电子波长。
  • D Alpha scattering established the nuclear model.α 散射确立核式结构。
Answer B. 答案 B。
de Broglie's hypothesis (1924) → confirmed by Davisson–Germer and G. P. Thomson (1927).德布罗意假说于 1927 年被电子衍射实验证实。
18 de Broglie wavelength德布罗意波长比较
answerD
youD✓
Under what condition can an electron have a shorter de Broglie wavelength than a neutron?什么条件下电子的德布罗意波长可比中子更短？
λ = h / p　　 and me ≈ mn/1836

A shorter λ needs a larger momentum. Because the electron is far lighter, at the same speed it has much less momentum, so it must travel much faster to beat the neutron.

  • A A narrow slit changes the diffraction pattern, not λ.狭缝宽度不影响波长。
  • B At rest p = 0 and λ → ∞.静止时动量零，波长无穷大。
  • C Slower electron ⇒ smaller p ⇒ longer λ.更慢则动量更小、波长更大。
  • D Much faster electron ⇒ larger p ⇒ shorter λ.快得多则动量大、波长更短。
Answer D. 答案 D。
19 Allotropes同素异形体
answerB
youB✓
Which statement about diamond and graphite is correct?关于金刚石与石墨，哪项正确？

Both are made of carbon atoms; they are allotropes and differ only in how the atoms are arranged — diamond is a three-dimensional network (each C bonded to four others, sp³) while graphite is layered hexagonal sheets (sp²).

  • A The atoms are identical (both carbon).两者都是碳原子。
  • B Their atoms are arranged differently — correct.原子排列方式不同，正确。
  • C / D Both are bulk forms, not nano forms; nano carbons are fullerenes, nanotubes and graphene.两者都是体相材料，纳米形态是富勒烯、碳纳米管、石墨烯。
Answer B. 答案 B。
20 Rayleigh criterion瑞利判据
answerC
youC✓
Two distant point sources form an image through a telescope. The intensity–displacement graph of the image is shown. Which statement is correct?两远处点光源经望远镜成像，强度–位置如图。哪项正确？
Ixpeakpeak
Rayleigh criterion: two images are just resolvable when the central maximum of one falls on the first minimum of the other, which leaves a dip of about 76% of the peak intensity. 瑞利判据：一个像的中央极大落在另一个像的第一极小上时为恰好可分辨，此时凹陷约为峰值的 76%。
  • The two main peaks in the figure are clearly separated and the dip between them falls far below 76% of the peak height;两主峰分离明显，中间凹陷远低于峰值的 76%。
  • Side maxima of each pattern are still individually visible, so the two patterns are largely independent;两侧仍能分辨各自的次极大，说明两图样基本独立。
  • Therefore the angular separation is greater than the Rayleigh limit.故角距离大于瑞利极限。
  • A Not resolvable — the dip would nearly vanish.不可分辨时凹陷几乎消失。
  • B Just resolvable — the dip would sit at about 76%.恰好可分辨时凹陷约在 76%。
  • C Resolvable — correct.可分辨，正确。
  • D The full intensity distribution is given, so a decision can be made.已有完整强度分布，可以判定。
Answer C. 答案 C。
21 Double-slit & matter waves双缝与物质波
answerA
youA✓
Two beams of electrons strike a double slit at a constant rate and form an interference pattern. If the kinetic energy of each electron increases, how does the pattern change?电子双缝干涉，若每个电子动能增大，图样如何变化？
λ = h / p = h / √(2mK)　⇒　K ↑ gives λ ↓
Fringe spacing　Δy = λ D / a　⇒　λ ↓ gives Δy ↓
  • A The fringes move closer together — correct.条纹间距变小、更密，正确。
  • B Fringes spread apart — opposite of what λ↓ gives.条纹变疏，与波长变小相反。
  • C / D Brightness is governed by the number of electrons, which the question does not change.亮度由电子数决定，题中未改变。
Answer A. 答案 A。
Chain to memorise: kinetic energy ↑ → momentum ↑ → wavelength ↓ → fringes closer.记忆链：动能↑ → 动量↑ → 波长↓ → 条纹变密。
22 Resolving power comparison分辨本领比较
answerA
youA✓
The apertures and de Broglie wavelengths of electrons in four transmission electron microscopes are given. If resolution is limited only by diffraction, which microscope has the smallest minimum resolvable angular separation?四台透射电镜的孔径与电子波长如表，哪台最小可分辨角最小？
θmin = 1.22 λ / D
MicroscopeAperture D / μmλ / nmθmin ≈ 1.22λ/D
P400.00401.22×10−7 rad
Q350.00501.74×10−7 rad
R140.00554.79×10−7 rad
S120.00656.61×10−7 rad
Smallest θmin ⇒ highest resolution ⇒ P, answer A. θmin 最小者分辨本领最高，为 P。
Convert μm and nm to the same unit (1 nm = 10−3 μm), or simply compare the ratio λ/D to avoid the unit trap entirely. 统一单位，或直接比较 λ/D 的比值。
23 TEM specimen透射电镜样品
answerB
youB✓
Which property of a specimen affects whether it can be observed under a transmission electron microscope?哪种样品性质会影响能否在透射电镜下观察？

A TEM forms its image from electrons that have passed through the specimen, so the specimen must be extremely thin (typically < 100 nm). A conducting specimen is also preferable to avoid charging.

  • A Colour — a TEM uses no visible light.电镜不用可见光，颜色无意义。
  • B Thickness — determines whether electrons can penetrate; correct.厚度决定电子能否穿透，正确。
  • C / D Conductivity and thermal conductivity matter for charging, but the decisive property here is thickness.导电性影响荷电效应，但决定性因素是厚度。
Answer B. 答案 B。
Compare with STM (B3): STM sees only the surface and needs a conducting sample; TEM sees the interior but needs a very thin sample. STM 只看表面、需导电；TEM 可看内部、需极薄。
24 Nanomaterials纳米材料
answerC
youC✓
Nanomaterials behave very differently from the bulk material because
(1) they are made of lighter elements; (2) quantum effects are significant at the nanoscale; (3) their properties are dominated by surface atoms.纳米材料性质与体相材料差异很大，原因是哪（些）？
  • (1) False — a nanomaterial has the same composition as the bulk material (gold is still gold); only the size differs.成分相同，仅尺寸不同。
  • (2) True — when the particle size approaches the electron wavelength the energy levels become discrete; quantum confinement effects appear (e.g. quantum dots change colour with size).尺寸接近电子波长时能级分立，量子限域效应显著。
  • (3) True — as size falls, the surface-area-to-volume ratio rises sharply and the low-coordination surface atoms dominate the physical and chemical behaviour.尺寸越小比表面积越大，表面原子主导材料性质。
(2) and (3) → C. 结论：(2)(3)，选 C。

B
Section B · Structured Questions结构题详解

Structured answers are marked on key terms + a complete logical chain + full working. Each item below gives the model answer first, then the marking points and a diagnosis of the marks you actually lost.结构题得分点是"关键词 + 逻辑链 + 完整运算"。每题先给范文，再给采分点与实际失分诊断。

B1　Photoelectric effect · 10 marks · scored 7光电效应

Given: green light λ = 495 nm (weak), threshold frequency f0 = 5.10×1014 Hz (corresponding to yellow light)已知：绿光波长 495 nm，金属截止频率 5.10×10¹⁴ Hz（对应黄光）

(a) Explain why the wave model of light cannot account for the threshold frequency.2 marks

Model answer

Under the wave model the energy of the incident light is spread out continuously over the wavefront, so an electron at the surface can absorb energy gradually and accumulate it until the work function is exceeded. The wave model therefore predicts that radiation of any frequency would eventually eject electrons, provided the intensity is high enough or the illumination lasts long enough — that is, emission would depend on intensity and there should be no sharp cut-off frequency.

Experimentally, however, no electrons are emitted at all when f < f0, however intense the light. This strong dependence on frequency rather than intensity is exactly what the wave model cannot explain.

波模型下光能连续分布在整个波前，表面电子可以逐渐吸收并累积能量，直到超过逸出功。因此波模型预言：只要强度足够大或照射时间足够长，任何频率的光最终都能打出电子，即发射只取决于强度，不应存在尖锐的截止频率。但实验中当 f < f₀ 时无论光多强都完全没有电子逸出——这一"只认频率、不认强度"的现象正是波模型无法解释的。
Marking points: ① wave energy is continuous and can be accumulated; ② hence any frequency would work given enough intensity/time; ③ this contradicts the observed absence of emission below f0. 采分点：① 波模型下能量连续、可累积；② 由此推出任何频率都行；③ 与"低于 f₀ 不发射"的实验事实矛盾。
You scored 0/2. Your answer only repeated the phrase "threshold frequency" without the causal chain about continuous accumulation of energy. An "Explain why" item is marked on mechanism words, not on restating the phenomenon. 你这一问得 0 分：只复述了"threshold frequency"，没有写出"能量连续累积"的因果链。Explain why 题必须出现机制性表述。

(b)(i) Calculate the maximum kinetic energy of the emitted photoelectrons.4 marks

Frequency of the incident light:
f = c/λ = (3.00×108) / (495×10−9) = 6.06×1014 Hz
先求入射光频率。
Photon energy:
E = hc/λ = (6.63×10−34 × 3.00×108) / (495×10−9) = 4.02×10−19 J
再求单个光子能量。
Work function:
φ = hf0 = 6.63×10−34 × 5.10×1014 = 3.38×10−19 J
由截止频率求逸出功。
Photoelectric equation:
Kmax = hf − φ = 4.02×10−19 − 3.38×10−19 = 6.4×10−20 J　(≈ 0.40 eV)
代入光电方程得最大动能。
Equivalent shorter route: Kmax = h(f − f0) = 6.63×10−34 × (6.06 − 5.10)×1014 = 6.4×10−20 J.更快的等价写法：直接用 K = h(f − f₀)。
Always state that f > f0 (green is higher in frequency than yellow) before concluding that emission occurs; a negative result would mean no emission at all. 先确认 f > f₀（绿光频率高于黄光）再下结论；若算出负值则表示没有电子逸出。

(b)(ii) Hence find the stopping potential.1 mark

eVs = Kmax　⇒　Vs = Kmax / e = (6.4×10−20) / (1.60×10−19) = 0.40 V
遏止电压由 eVs = Kmax 求得，答案 0.40 V。
Your score: full marks ✓ (your script gives 0.398 V, correct).你的得分：满分 ✓。

(c) Describe the change in stopping potential and maximum photocurrent when the plate is illuminated by (i) blue light of the same intensity and (ii) green light of the same wavelength but higher intensity.4 marks

ConditionStopping potential VsMaximum photocurrentReasoning
(i) Blue light, same intensity蓝光，强度相同 Increases增大 Decreases减小 Blue has a shorter wavelength and therefore a higher frequency, so each photon carries more energy and Kmax = hf − φ is larger; hence Vs = Kmax/e increases. The same intensity means the same total energy per second, but each photon is more energetic, so the number of photons per second falls, fewer photoelectrons are released per second and the maximum photocurrent is smaller. 蓝光波长更短、频率更高，单光子能量更大，最大动能更大，故遏止电压增大。强度相同意味着每秒总能量相同，而单光子能量更大，因此每秒光子数减少，光电子数减少，最大光电流减小。
(ii) Green light, same wavelength, higher intensity绿光，波长相同，强度更高 Unchanged不变 Increases增大 Same wavelength means the same photon energy, so Kmax and therefore Vs are unchanged. Higher intensity means more photons per second, so more photoelectrons per second and a larger current. 波长（频率）不变，单光子能量不变，最大动能与遏止电压不变；强度增大使每秒光子数增多，光电子数增多，电流增大。
One-line summary: frequency decides how energetic each electron is (Vs); intensity decides how many there are (current). Raising the frequency at fixed intensity actually lowers the current, because the photons become fewer. 一句话总结：频率决定单个电子的能量（遏止电压），强度决定电子数目（电流）；同强度下提高频率反而使电流下降。
You scored half of (i): "Vs increases" was right, but you wrote that the maximum photocurrent increases. That happens if intensity is confused with photon number — intensity is energy per unit area per unit time = (number of photons) × (energy per photon). 你 (i) 只答对一半：Vs 增大正确，但写成光电流增大。把强度当成光子数就会出错。

B2　Hydrogen-like ions and the Rydberg formula · 10 marks · scored 5类氢离子与里德伯公式

Given: En = Z²(−13.6/n²) eV for a hydrogen-like ion, where EG is the ground-state energy of hydrogen题给：类氢离子能级 En = Z²(−13.6/n²) eV

(a) Compare a hydrogen atom in the ground state with a He+ ion in the ground state. Which has the smaller ionisation energy? Explain briefly.4 marks

Model answer

Hydrogen has the smaller ionisation energy.

H (Z = 1)：E1 = −13.6 eV ⇒ ionisation energy = 13.6 eV
He+ (Z = 2)：E1 = 2²×(−13.6) = −54.4 eV ⇒ ionisation energy = 54.4 eV = 4 × 13.6 eV

The energy of a hydrogen-like ion is proportional to Z², so the electron in He+ feels an attractive Coulomb force from a +2e nucleus that is four times stronger. It is therefore bound far more tightly and four times as much energy is needed to remove it.

氢原子的电离能更小。类氢离子能量正比于 Z²，He⁺ 的电子受 +2e 核的库仑引力是氢的 4 倍，被束缚得更紧，需要 4 倍能量才能移出（13.6 eV 对比 54.4 eV）。
Marking points: ① H gives 13.6 eV; ② correct substitution with Z = 2 gives 54.4 eV; ③ conclusion that hydrogen is smaller; ④ the reason — E ∝ Z², larger nuclear charge, tighter binding. 采分点：① 算出 13.6 eV；② 正确代 Z=2 得 54.4 eV；③ 结论"氢更小"；④ 用 E ∝ Z² 解释。
You scored 2/4. The value 13.6 eV was right, but the Z² factor was replaced by something like "×0.2²" instead of Z² = 4, and the required one-line physical reason (larger nuclear charge ⇒ tighter binding) was missing. 你得 2/4：13.6 eV 正确，但 Z² 未正确代入（写成 0.2² 之类），也缺少"核电荷大、束缚更紧"的物理解释句。

(b)(i) Prove the Rydberg formula from the energy levels, and express R in terms of EG, h and c.3 marks

Write the hydrogen levels using the magnitude of the ground-state energy EG = 13.6 eV:
En = − EG / n²
用基态能量大小表示氢原子能级。
For a transition from nhigher down to nlower the emitted photon has energy
hc/λ = Ehigher − Elower = −EG/nhigher² + EG/nlower²
跃迁释放的光子能量等于两能级之差。
Rearranging:
hc/λ = EG ( 1/nlower² − 1/nhigher² )
⇒　1/λ = (EG/hc) ( 1/nlower² − 1/nhigher² )
整理成 1/λ 的形式。
Compare term by term with 1/λ = R(1/nlower² − 1/nhigher²):
R = EG / (hc)
与里德伯公式逐项比较得 R = EG/(hc)。
Check: R = (13.6 × 1.60×10−19) / (6.63×10−34 × 3.00×108) ≈ 1.10×107 m−1, in agreement with the accepted 1.097×107 m−1.数值校验与公认值一致。
You scored 2/3: the derivation chain and R = EG/hc were present, but the final result line was crossed out, so the conclusion was not cleanly stated. In an exam, write the required result once more, boxed and on its own line. 你得 2/3：推导与 R = EG/hc 都写到了，但结论行被划掉。考试时把结论单独写成一行。

(b)(ii) Can the formula be applied directly to He+ without modification? If not, modify it.3 marks

Model answer

No. The formula was derived for a nucleus of charge +e (Z = 1). For He+ the nuclear charge is +2e, so every energy level is multiplied by Z² = 4:

En(He+) = Z²(−EG/n²) = −4EG/n²
hc/λ = 4EG(1/nlower² − 1/nhigher²)

The modified formula is therefore

1/λ = 4R ( 1/nlower² − 1/nhigher² )　，i.e. replace R by Z²R
不能直接套用：该公式按核电荷 +e（Z=1）导出，而 He⁺ 核电荷为 +2e，能级整体乘 Z²=4。修改后即把 R 换成 Z²R = 4R。
Marking points: ① answer No; ② reason — nuclear charge +2e, energy contains Z²; ③ the corrected formula with Z² = 4. 采分点：① 答 No；② 原因（Z=2，含 Z² 因子）；③ 写出带 Z²=4 的修正式。
You scored 1/3: only "No." was written, with neither the reason nor the modified formula — almost half the question left unanswered. Even when unsure, always write 1/λ = Z²R(1/n1² − 1/n2²). 你得 1/3：只写了 "No."，既无原因也无修正公式。即使不确定，也应写出 1/λ = Z²R(...)。

B3　Scanning tunnelling microscope (STM) · 10 marks · scored 5扫描隧道显微镜

A probe scans across the sample surface at a constant height.探针在样品表面上方以恒定高度扫描。

(a) Name the electric current and the effect that produces it.2 marks

Model answer

The current is a tunnelling current, produced by the quantum tunnelling effect — electrons cross the vacuum gap between tip and sample even though they classically lack the energy to do so.

该电流是隧道电流，由量子隧穿效应产生：电子在经典上能量不足的情况下仍能穿过针尖与样品之间的真空势垒。
Two marking points: the current (tunnelling current) and the effect (quantum tunnelling effect). "Electric current" earns nothing.两个采分点：隧道电流 + 量子隧穿效应。
You scored 1/2: "tunnelling current" was given, but the effect was written as "a tunnelling effect" without "quantum", and part of the line was struck out. Write the noun phrase first, then add one explanatory clause. 你得 1/2：电流名称对了，但效应未点明"量子"，且卷面有涂划。先写名词短语再补解释。

(b) Describe how the STM maps the surface of the sample.3 marks

Model answer

A very sharp probe is scanned across the sample at a (nominally) constant height, with a small voltage applied between tip and sample so that electrons tunnel across the gap. The tunnelling current depends exponentially on the tip–sample separation d:

I ∝ e−2κd

so a tiny change in separation produces a large change in current. By recording the current at every position (x, y) while scanning — or by using a feedback loop that moves the tip up and down to keep the current constant and recording that height — a three-dimensional topographical map of the surface is built up, ultimately to atomic resolution.

用极细的探针在样品上方以（近似）恒定高度扫描，并在针尖与样品间加小电压使电子隧穿。隧道电流随间距 d 指数变化，间距的极小改变会引起电流的显著变化。逐点记录扫描中各位置的隧道电流（或用反馈回路调节针尖高度以保持电流恒定并记录该高度），即可重建表面三维形貌，最高可达原子分辨。
Marking points: ① the probe is scanned over the surface; ② the tunnelling current varies exponentially with the tip–sample gap; ③ recording current (or feedback height) at each point builds the surface image. 采分点：① 探针扫描表面；② 隧道电流随间距指数变化；③ 逐点记录（或反馈调高）重建表面图像。
You scored 0/3. Your script reads "It maps its topography by electrical effect / the surface of the sample need be reachable to electromagnetic effects / the electric current on the current and the probe create tunneling" — fragments with no main line. Answer description questions in three moves: operation → physical quantity changes → how an image is produced. 你得 0/3：答卷是零碎短语，缺少主线。描述题按"操作 → 物理量变化 → 成像"三步组织。

(c)(i) State one major limitation of the STM. (ii) Hence, is the STM suitable for the internal structure of a slice of metallic specimen?1 + 1 mark

Model answer

(i) It can probe only the surface (the outermost atomic layer) and reveals nothing about the interior. Another accepted limitation is that the sample must be electrically conducting.

(ii) No. The tunnelling current depends only on the gap between the tip and the topmost surface atoms, so information about the interior of the slice cannot be obtained.

(i) 只能探测表面（最外层原子），无法反映内部结构（另一可接受答案：样品必须导电）。(ii) 不合适：隧道电流只取决于针尖与最外层表面原子的间距。
Your score: 2/2 ✓你的得分：满分 ✓

(d) State another usage of the STM apart from imaging.1 mark

Model answer

Atomic manipulation — the tip can pick up, drag or position individual atoms and molecules on a surface (for example building atomic-scale patterns or writing information), or it can measure the local electronic density of states.

原子操纵：用针尖搬运、移动或放置单个原子/分子，构建原子尺度的图案或写入信息；也可测量表面局域电子态密度。
You scored 0/1 with the single word "modification", which is too vague to award. One mark questions need one precise term: atomic manipulation. 你得 0/1，只写了 "modification"，过于笼统。1 分题要给出准确术语。

(e) Is the resolving power of the STM limited by the Rayleigh criterion? Explain.2 marks

Model answer

No. The Rayleigh criterion applies to instruments that form an image from waves passing through a lens or aperture, where the resolution is set by diffraction, θmin = 1.22λ/D.

An STM involves no lens and no wave-imaging process. Its resolution is determined by the sharpness of the tip and by the exponential dependence of the tunnelling current on the tip–sample distance, so an extremely small change in separation produces a measurable change in current. Atomic resolution is therefore achieved without any diffraction limit.

不受限。瑞利判据适用于以波经透镜/孔径成像、分辨率受衍射限制的仪器。STM 没有透镜成像过程，其分辨率由针尖尖锐程度以及隧道电流对间距的指数依赖决定，间距的极小变化即可引起可测的电流变化，故可在无衍射限制的情况下达到原子分辨。
Your score: 2/2 ✓ — your phrase "no diffraction and light processing in STM" already captured the key idea, though the wording could be tighter. 你得 2/2 ✓，已抓住"无衍射成像"这一关键。
Contrast with the TEM: its resolution is diffraction-limited, θmin = 1.22λ/D with the electron wavelength. The STM/TEM contrast is a frequent exam question. 对比：TEM 的分辨率确实受衍射限制；TEM 与 STM 的对比是常见考点。

C
Formulae and Constants公式与常数速查

Quantity / lawFormula or value
Planck constanth = 6.63×10−34 J s   (hc = 1240 eV nm)普朗克常数
Speed of light · elementary charge · electron massc = 3.00×108 m s−1; e = 1.60×10−19 C; me = 9.11×10−31 kg光速、元电荷、电子质量
Photon energyE = hf = hc/λ光子能量
Photoelectric equationhf = φ + Kmax, with φ = hf0光电方程
Stopping potentialeVs = Kmax ⇒ Vs = (h/e)f − φ/e遏止电压
Hydrogen energy levelsEn = −13.6/n² eV氢原子能级
Hydrogen-like ion levelsEn = Z²(−13.6/n²) eV类氢离子能级
Rydberg formula1/λ = R(1/n1² − 1/n2²), R = 1.097×107 m−1 = EG/(hc); for hydrogen-like ions use Z²R里德伯公式
Series limitsLyman → n=1; Balmer → n=2; Paschen → n=3; Brackett → n=4; Pfund → n=5各谱系终点
de Broglie wavelengthλ = h/p = h/√(2mK) = h/√(2meV)德布罗意波长
Rayleigh criterionθmin = 1.22 λ / D瑞利判据
Double-slit fringe spacingΔy = λD/a双缝条纹间距

D
Top Ten Pitfalls in This Paper本次考试的十大易错点（按失分排序）

1. Intensity vs photon number

Intensity = energy per unit area per unit time = (photons) × (energy per photon). Higher frequency at the same intensity means fewer photons, hence a smaller current. (B1(c)(i))强度 = 光子数 × 单光子能量；同强度下提高频率反而使电流减小。

2. Threshold / incident frequency / stopping potential

f0 = φ/h decides whether emission is possible; f sets Kmax; Vs is the retarding voltage that kills the current. (Q7)截止频率、入射频率、遏止电压三者不可互代。

3. Write the equation before reading a graph

I–V graph: intercept ⇔ frequency, plateau ⇔ intensity. Vs–f graph: slope = h/e, so the lines must be parallel. (Q3, Q5)图像题先写关系式：截距看频率、平台看强度、斜率恒为 h/e。

4. Excited-state counting

k-th excited state ⇒ n = k + 1. Third excited state is n = 4; fourth is n = 5. (Q12, Q13)第 k 激发态对应 n = k+1。

5. Direction of the Lyman series

Ends on n = 1; absorption arrows rise from n = 1, emission arrows fall into n = 1. (Q16)Lyman 终点 n=1；吸收向上、发射向下。

6. Rutherford vs Bohr

"Circular orbits" and "Coulomb force as centripetal force" belong to Rutherford; "stationary states with no radiation" is Bohr's new postulate. (Q14)圆轨道与库仑力属卢瑟福；定态不辐射属玻尔。

7. Emission vs absorption spectrum

Electron bombardment excites atoms that then emit ⇒ emission spectrum. White light through low-pressure gas ⇒ absorption spectrum. (Q15)电子轰击得发射谱；白光穿低压气体得吸收谱。

8. "Explain" needs a mechanism

For the threshold-frequency question, the phrase "energy is accumulated continuously" must appear. (B1(a))Explain 题必须写出机制性语句。

9. Resolution chains

λ↓ ⇒ diffraction limit↓ ⇒ resolution↑, but the energy change must be traced through the photoelectric and matter-wave relations. (Q21, Q22)分辨率链条要联立光电方程与物质波公式。

10. STM vs TEM

STM: surface only, needs a conducting sample, not diffraction-limited. TEM: interior, needs a very thin sample, diffraction-limited. (B3, Q23)STM 看表面、需导电、不受衍射限制；TEM 看内部、需极薄、受衍射限制。

How to revise from here下一步复习建议

  1. Redo the eight wrong MCQs and write one "decision sentence" for each, e.g. "slope = h/e, so the lines must be parallel". These templates transfer to unseen questions.重做 8 道错题，每题写一句判据句，形成可迁移的思维模板。
  2. Close the three half-answered gaps: the wave-model failure argument in B1(a), the modified formula in B2(b)(ii) and the imaging mechanism in B3(b) — about 8 marks in total, the cheapest marks to recover.补齐三处半道题，共约 8 分，最容易收复。
  3. Build the English terminology: tunnelling current, quantum tunnelling effect, atomic manipulation, threshold frequency, stopping potential, work function, continuous transfer of energy. Marks are often lost when a Chinese idea cannot be put into the accepted English phrase.建立术语英文表达，避免思路对但用词不达标而失分。
  4. Drill four calculation families: the photoelectric equation, level transitions, the Rydberg formula and the Rayleigh criterion — five questions each, always in the form "formula → substitution → result with units".四类计算各练 5 题，固定写成"公式 → 代入 → 带单位结果"。
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