WJEC物理 内阻 干涉 光电效应 半衰期
WJEC 的 A Level 物理里，电学、光学和核物理看起来是三块互不相干的内容，考卷却把它们按单元串在一条线上。这篇把 Unit 2 与 Unit 3 里最常考的公式、最常掉的单位换算、最常被误解的概念排在一起，给出能在考场上直接用的写法。文中所有卷面信息都取自 WJEC 官方 spec（课程代号 A Level 1420QS、AS 2420QS）与历年真题卷首的分值表，遇到具体数字可以自己回查核对。
Electricity, light and nuclear physics look like three unrelated blocks in WJEC A Level Physics, yet the specification links them through two examined units. This article gathers the equations that appear most often, the unit conversions that cost the most marks and the concepts candidates most often misunderstand, using the paper structure published in the WJEC specification (A Level code 1420QS, AS code 2420QS).
三张卷的位置与分值分布 / Where the Three Topic Areas Are Assessed
WJEC 把电学与光学放在 AS 的 Unit 2、把核物理放在 A2 的 Unit 3，复习前先把位置找准。Unit 2 Electricity and Light 考 1 小时 30 分钟，满分 80 分，占整个 A Level 的 20%，八个主题块从传导、电阻、直流电路一直排到波的性质、折射、光子和激光。Unit 3 Oscillations and Nuclei 考 2 小时 15 分钟，满分 100 分，占 25%，Section A 占 80 分、Section B 是 20 分的阅读理解，核衰变与核能排在最后两个主题块。Unit 5 的实验考试另占 10%，由 25 分实验任务与 25 分数据分析任务组成。2024 年 6 月的 Unit 2 卷（2420U20-1）九道题分值是 6、7、14、9、9、6、10、8、11，合计 80 分，QER 落在第 6 题。同年的 Unit 3 卷（1420U30-1）Section A 七道题是 10、10、11、23、7、7、12，Section B 一道 20 分，QER 在第 3(a) 题。这套 spec 从 2015 年 9 月起施教，AS 2016 年夏季首次认证、A Level 2017 年夏季首次认证。
WJEC splits electricity and light into AS Unit 2 and nuclear physics into A2 Unit 3, so the first task in revision is locating each area; Unit 2 runs 1 hour 30 minutes for 80 marks, worth 20% of the A Level, while Unit 3 runs 2 hours 15 minutes for 100 marks, worth 25%, with 80 marks of Section A and a 20-mark comprehension in Section B. In the June 2024 Unit 2 paper the nine questions carried 6, 7, 14, 9, 9, 6, 10, 8 and 11 marks, and Unit 3 split into 10, 10, 11, 23, 7, 7 and 12 marks plus a 20-mark Section B. Unit 5 adds a further 10% through two practical tasks of 25 marks each, and the specification has been taught since September 2015.
电流的微观图像与漂移速度 / Drift Velocity in a Metal
电流的定义式是 I = Q 除以 t，金属里真正说明问题的是它的微观版本 I = nAvq。n 是单位体积的自由电子数，铜约为 8.5 x 10^28 m^-3，A 是横截面积，v 是漂移速度，q 是元电荷 1.6 x 10^-19 C。把 1 A 与 1 mm^2 代进去，漂移速度只有 7 x 10^-5 m/s 这个量级，比蜗牛还慢。很多学生以为电子是从电池出发一路跑到灯泡里的，其实合上开关时整根导线里的电子几乎同时被电场推动，信号以接近光速的速度建立起来。这一步最容易丢分的是面积换算，1 mm^2 = 1 x 10^-6 m^2，漏掉这一步结果会差六个数量级。反过来问 n 时把式子移项成 n = I 除以 Avq，单位写成 m^-3。这类题常在 Unit 2 的传导电学部分出现，分值多在 3 到 4 分，评分看的是式子、代入、单位三步。
Current is defined by I = Q/t, and in a metal the useful version is I = nAvq, where n is the free-electron number density, A the cross-sectional area, v the drift velocity and q the elementary charge. Copper has roughly 8.5 x 10^28 free electrons per cubic metre, so 1 A through 1 mm^2 gives a drift velocity of only about 7 x 10^-5 m/s, far slower than a snail; all the electrons in the wire start moving almost together because the electric field is established at close to the speed of light. Marks are most often lost on the area conversion, since 1 mm^2 is 1 x 10^-6 m^2 and a missing factor of a million shifts the answer by six orders of magnitude.
电阻率与温度对电阻的影响 / Resistivity and Temperature
电阻的定义式 R = V 除以 I 只是量出来的比值，真正描述材料的是电阻率，关系写成 R = ρL 除以 A。铜的电阻率约 1.7 x 10^-8 ohm m，镍铬合金约 1.1 x 10^-6 ohm m，差了六十多倍，所以电热丝用合金而不是铜。一根 2.0 m 长、横截面积 0.50 mm^2 的铜线，代入后 R = 1.7 x 10^-8 x 2.0 除以 5.0 x 10^-7 = 0.068 ohm。金属电阻随温度升高而增大，原因是晶格振动增强、电子碰撞更频繁；热敏电阻（NTC）正好相反，温度升高电阻下降，常接在分压电路里做温度传感。学生最常把电阻率与电阻混为一谈，题目表格给的是电阻率，代入时却当成电阻值用。画 I-V 图时要记住灯丝不是欧姆器件，曲线在电流增大后变平，这条曲线在 Unit 2 里会被要求用温度解释，写的时候把温度与电阻的因果顺序讲清楚，只写曲线变弯不给分。
Resistivity belongs to the material while resistance belongs to a particular piece of it, and the two are linked by R = ρL/A. Copper sits at about 1.7 x 10^-8 ohm m against roughly 1.1 x 10^-6 ohm m for nichrome, a factor of sixty that explains why heating elements are alloys; a 2.0 m copper wire of area 0.50 mm^2 comes out at 0.068 ohm. Metals gain resistance as they warm because the lattice vibrates more, whereas an NTC thermistor loses resistance, which is why it works as a temperature sensor inside a potential divider.
电动势、内阻与端电压图解 / EMF, Internal Resistance and the Terminal p.d. Graph
电源有两个参数，电动势 ε 与内阻 r，闭合电路的方程是 ε = I(R + r)。把 1.50 V、0.50 ohm 的电池接到 2.5 ohm 的电阻上，电流 I = 1.50 除以 3.0 = 0.50 A，端电压 V = IR = 1.25 V。内阻上消耗的功率是 I^2 r = 0.25 x 0.50 = 0.125 W，这部分不出现在外电路上。最常见的实验题是画端电压对电流的图，纵轴截距是 ε，斜率是负的 r，读数时要把内阻的影响算进去才能得到电池的电动势。很多学生以为端电压总等于电动势，其实只有在断路或内阻可以忽略时两者才相等。题目给一组 V 与 I 的数据时，用两点求斜率比用单点稳，用单点会把读数误差放大到整条线上。阅卷时看的是有没有先写出 ε = I(R + r)，只给一个数字答案通常拿不到过程分。
A cell is described by its emf and its internal resistance, and the closed-circuit equation ε = I(R + r) links them to the current. With 1.50 V and 0.50 ohm feeding a 2.5 ohm resistor the current is 0.50 A, the terminal p.d. is 1.25 V, and 0.125 W is lost inside the cell as I^2 r; plotting terminal p.d. against current gives the emf at the intercept and internal resistance as the negative gradient. Candidates who read the emf off a single data point rather than the intercept usually lose the accuracy mark.
直流电路的分压与功率 / Potential Dividers and Power in D.C. Circuits
串联电路电流相同、电压分配，并联电路电压相同、电流分配，这两条是后面所有计算的骨架。分压公式写成 Vout = Vin 乘 R2 除以 R1 加 R2，把 12 V 加在 100 ohm 与 220 ohm 的串联组合上，220 ohm 两端的电压是 12 x 220 除以 320 = 8.25 V。功率有三个等价写法 P = VI = I^2 R = V^2 除以 R，题目给了电压和电阻就直接用最后一个，免得先绕一圈算电流。电路图题最容易掉分的位置是电表接法，电流表要串在被测支路、内阻尽量小，电压表要并在元件两端、内阻尽量大。有的题目会问为什么实测值总比计算值略小，答案要落到电表内阻分流或者内阻上的压降，写实验误差拿不到分。这类解释题在 Unit 2 的直流电路部分通常值 2 到 3 分，一句现象加一句原因就够。
Series circuits share one current and split the voltage, while parallel circuits share one voltage and split the current, and every later calculation rests on those two statements. The divider rule gives 8.25 V across the 220 ohm resistor when 12 V is applied to 100 ohm and 220 ohm in series, and power can be written as VI, I^2 R or V^2/R, so choose the version that matches the quantities in the question. When a measured value comes out slightly below the calculated one, examiners want meter resistance or the drop across the internal resistance rather than a vague mention of experimental error.
波的叠加、相干与干涉条纹 / Superposition, Coherence and Interference
两列波在同一空间相遇时位移相加，这就是叠加原理；干涉要稳定，必须满足相干条件，也就是频率相同、相位差恒定。路径差是波长的整数倍时到达同相，出现亮纹；路径差是半波长的奇数倍时反相，出现暗纹。双缝实验的公式是 λ = ax 除以 D，a 是缝距、x 是相邻条纹间距、D 是屏距。取 a = 0.25 mm、D = 1.50 m、条纹间距 2.4 mm，得到 λ = 0.25 x 10^-3 x 2.4 x 10^-3 除以 1.50 = 4.0 x 10^-7 m，也就是 400 nm。偏振只能发生在横波上，这是光具有横波特性的直接证据，WJEC 2024 年夏季考试报告写明 Unit 2 里考偏振的 AO1 回忆题对多数考生偏难。驻波题要分清波节与波腹，绳子两端固定时基频波长等于两倍绳长，答案里带上这句话就是一个独立给分点。
Superposition adds displacements, and interference becomes steady only when the two sources are coherent, meaning equal frequency and a constant phase difference. Whole-wavelength path differences arrive in phase and give maxima, odd multiples of a half wavelength give minima, and for a = 0.25 mm, D = 1.50 m and a fringe spacing of 2.4 mm the wavelength comes out at 4.0 x 10^-7 m. Polarisation is the classic evidence that light is transverse, and WJEC's Summer 2024 report records that the Unit 2 recall question on polarisation proved too demanding for many candidates.
折射、全反射与光纤 / Refraction, Total Internal Reflection and Optical Fibres
折射定律写成 n1 sin θ1 = n2 sin θ2，光从空气以 45 度射入折射率 1.50 的玻璃，sin θ2 = sin 45 除以 1.50 = 0.471，得到 θ2 约 28 度，算完回头检查角度是变小还是变大。从玻璃射向空气时，临界角由 sin C = 1 除以 n 给出，n = 1.50 对应 C = 41.8 度，入射角超过这个值就发生全反射。光纤的芯折射率高于包层，光在芯与包层的界面上反复全反射前进，包层的作用就是把光约束在芯里。常见的错误是把临界角公式里的 n 写成两种介质的比值，正确写法是较密介质的折射率除以较疏介质的折射率。色散会让不同波长的光走不同路径，多模光纤里脉冲因此变宽，这是限制传输带宽的原因。阅卷时看到的失分大多是没有注明角度是与法线的夹角，题目里的角度常常故意画在与界面之间。
Refraction follows n1 sin θ1 = n2 sin θ2, so light entering glass of refractive index 1.50 at 45 degrees leaves at about 28 degrees with sin θ2 = 0.471. Travelling the other way, total internal reflection begins at the critical angle given by sin C = 1/n, which is 41.8 degrees for that glass; the fibre core carries the higher refractive index and the cladding keeps light inside by repeated reflection. Marks disappear when the angle is measured from the surface instead of the normal, and dispersion broadens pulses in multimode fibre because different wavelengths travel different paths.
光子与光电效应 / Photons and the Photoelectric Effect
光子能量写成 E = hf = hc 除以 λ，普朗克常数 h = 6.63 x 10^-34 J s。波长 400 nm 的光子能量是 6.63 x 10^-34 x 3.0 x 10^8 除以 4.0 x 10^-7 = 4.97 x 10^-19 J，换算成电子伏约 3.1 eV。光电效应方程是 hf = φ + KEmax，若逸出功为 2.0 eV，最大动能就是 1.1 eV，遏止电压在数值上也是 1.1 V。三条实验规律要能直接写出来：存在阈值频率，光强只影响光电流大小，光电发射是瞬时的。很多学生以为增强光强就能打出动能更大的电子，其实光强只增加单位时间的光子数，单个光子的能量由频率决定。作图题里横轴用频率、纵轴用最大动能，斜率是 h，纵轴截距的绝对值是逸出功，这套读图法在历年卷子里反复出现。写解释时先比较光子能量与逸出功，再落到最大动能，顺序反了容易丢过程分。
A photon carries energy E = hf = hc/λ with h = 6.63 x 10^-34 J s, so light of 400 nm gives 4.97 x 10^-19 J, about 3.1 eV. The photoelectric equation hf = φ + KEmax means a 2.0 eV work function leaves 1.1 eV of kinetic energy, which is also the stopping potential in volts; intensity changes the number of photoelectrons per second but never their maximum energy. On a graph of maximum kinetic energy against frequency the gradient is h and the intercept on the energy axis is minus the work function, and that is where most of the marks sit.
能级、受激辐射与激光 / Energy Levels, Stimulated Emission and Lasers
原子只能停在分立的能级上，跃迁时吸收或发射的能量满足 hf = E2 减 E1，这就是线状光谱的来源。受激辐射是一个光子诱导激发态原子发出一个同频率、同相位、同方向的光子，两个变四个、四个变八个，形成放大。要让它持续发生，必须实现粒子数反转，也就是高能级上的原子数多于低能级，氦氖激光器的输出波长是 632.8 nm，属于红色可见光。激光的三个特征是单色性好、相干性好、方向性强，题目问为什么用激光做干涉实验时，答案要落到相干长度长，写亮度高不得分。常见的错误是把粒子数反转说成把原子激发到高能级就算完成，关键在分布反转，不是单个原子的能量高低。这一步最容易与自发辐射混淆，自发辐射不需要外来光子诱导，方向和相位都是随机的，两种辐射的比例决定激光的阈值条件。
Atoms occupy discrete energy levels, so absorption and emission occur at frequencies fixed by hf = E2 - E1 and produce line spectra rather than a continuum. Stimulated emission happens when a passing photon triggers an excited atom to emit an identical photon, and a laser needs a population inversion so that amplification beats absorption; the helium-neon laser emits at 632.8 nm. Coherence is the property that makes lasers useful in interference work, and an answer built on brightness instead of coherence, monochromaticity and directionality rarely scores the explanation marks.
核衰变、活度与半衰期 / Nuclear Decay, Activity and Half-Life
三种辐射要能分开说：alpha 粒子是氦核，电离本领最强、穿透力最弱，一张纸就能挡住；beta 粒子是核内中子转变时放出的电子，几毫米铝片可以吸收；gamma 是波长很短的电磁辐射，厚铅板只能把它明显减弱。WJEC 2023 年 6 月的 Unit 3 卷在第 3(d) 题用 6 分 QER 考过用不同吸收片以及电场或磁场区分三种辐射，写这类题要把偏移方向与电荷符号对应起来。衰变是随机过程，未衰变核的个数随时间按 N = N0 e^(-λt) 减少，活度 A = λN 的单位是贝可勒尔。半衰期与衰变常数的关系是 t1/2 = ln2 除以 λ，碳 14 的半衰期 5730 年对应 λ 约 1.2 x 10^-4 每年。经过三个半衰期还剩八分之一，也就是 12.5%，这类比例题先算半衰期个数再算剩余比例，比直接代指数式快。错得最多的是把活度当成剩余核个数，题目问活度时要先把 A = λN 写出来再算。
Alpha particles are helium nuclei with the strongest ionising power and the shortest range, beta particles are electrons released when a neutron turns into a proton, and gamma rays are short-wavelength electromagnetic radiation that thick lead only reduces. WJEC's June 2023 Unit 3 paper asked a six-mark QER on distinguishing the three with absorbers and with electric or magnetic fields, so deflection direction must be tied to the sign of the charge. Decay is random but the numbers follow N = N0 e^(-λt) with activity A = λN, and half-life is t1/2 = ln2/λ, so carbon-14 at 5730 years corresponds to λ of about 1.2 x 10^-4 per year.
质量亏损、结合能与核能 / Mass Defect, Binding Energy and Nuclear Energy
核反应前后质量不守恒，差出来的质量亏损按 E = Δm 乘 c^2 变成能量，1 个原子质量单位对应 931.5 MeV。结合能除以核子数得到每个核子的结合能，曲线在铁附近最高，约 8.8 MeV，这就是中等质量核最稳定的原因。铀 235 裂变一次释放约 200 MeV，换成焦耳是 200 x 10^6 x 1.6 x 10^-19 = 3.2 x 10^-11 J，单次数值很小，但一克铀里的原子数极大。重核裂变与轻核聚变都能放能，原因是两者都在朝结合能曲线的高峰靠近，方向相反但终点都在铁附近。常见的错误是把质量亏损说成质量消失，正确说法是质量转化成了能量，能量总量仍然守恒。计算题里单位最容易出错，用 MeV 时要与电子伏换算配套，用焦耳时把光速写成 3.0 x 10^8 m/s，两套单位不要混在一道题里。
A nuclear reaction does not conserve mass, and the missing mass reappears as energy through E = Δm c^2, with one atomic mass unit equal to 931.5 MeV. Binding energy per nucleon peaks near iron at about 8.8 MeV, which is why medium-mass nuclei are the most stable, and one uranium-235 fission releases roughly 200 MeV, or 3.2 x 10^-11 J. Fission of heavy nuclei and fusion of light nuclei both move towards that peak, and answering that the mass disappears rather than that mass is converted into energy costs the explanation mark.
四周复习排期与 QER 作答顺序 / Four-Week Revision and QER Technique
四周的节奏可以按卷子切：第一周过 Unit 2 的传导、电阻与直流电路计算题，第二周做波与光子，第三周处理 Unit 3 的核衰变与核能，第四周整卷限时。WJEC 2024 年夏季考试报告指出 AO1 回忆型题目继续失分较多，而给出数据的 AO2 应用题得分偏高，复习重心因此要放在把式子用起来，而不是只背结论。QER 是质量扩展回答，2024 年 6 月的 Unit 2 放在第 6 题、Unit 3 放在第 3(a) 题，通常值 6 分。写 QER 的顺序是先给定义，再用一个具体例子或数字支撑，最后回到题目情境，三句话对应三档分。阅卷时看到的空白回答里，很多学生以为写满就有分，其实评分看的是论点数量与物理术语的准确性，同一句话重复三遍不加分。实验方法类的 QER 在 Unit 3 里得分情况相对较好，因为可以写吸收片、计数率、本底扣除这些具体做法，复习时把这几个模板句准备好就够用。
A four-week plan works well split by paper: calculations from Unit 2 electricity first, waves and photons next, then Unit 3 nuclear decay and nuclear energy, finishing with a timed paper. WJEC's Summer 2024 report states that AO1 recall questions continue to cause problems while AO2 questions with data provided score well, so practice should use the equations rather than recite conclusions. QER items carry six marks, sitting in question 6 of the June 2024 Unit 2 paper and question 3(a) of Unit 3, and three clear points with accurate terminology beat a long answer that repeats one idea.
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