OCR物理 光电效应 德布罗意波 能级跃迁
OCR 的 A Level 物理 A（H556）把电子与光子的内容集中在模块 4，标题是 Electrons, waves and photons，其中的 4.5 节量子物理是学生普遍觉得最难拿分的一块。这部分题量不算大，却能把估算、单位换算和图像解释三类错误同时暴露出来。很多学生以为背下 hf = φ + KEmax 就算学完，真正丢分的地方在于说不清每个符号对应哪一次测量。下面按 OCR 考纲 4.5 节的顺序，把光子、光电效应、德布罗意波与能级四块内容拆开，每一步给出可以照着写的答题表述。
Quantum physics belongs to Module 4 of OCR A Level Physics A (H556), and it is examined in Component 02. Candidates often know the equations yet lose marks when they cannot attach each symbol to the measurement that produced it. This walk-through follows the order of topic 4.5 and keeps the phrasing that mark schemes reward.
4.5 在 H556 考卷上的位置与分值
模块 4 只出现在 Component 02，也就是 Exploring physics 那份卷子里，它覆盖模块 1、2、4、6，时长 2 小时 15 分、满分 100 分、占整个 A Level 的 37%。Section A 是 15 道选择题共 15 分，Section B 的 85 分以结构化题和扩展题为主，量子物理通常落在后者中段。Component 03 叫 Unified physics，1 小时 30 分、70 分、占 26%，会把 4.5 的内容和第六模块的粒子物理放进同一道题里问。按照 OCR 官网 2025 年 6 月发布的 specification 2.8 版，这份资格 2015 年 9 月首次教学、2017 年首次考试。阅卷时经常看到学生把 4.5 的公式写在电荷与电流的题上，说明模块内部的分节没有记清。
Module 4 is assessed only in Component 02, Exploring physics, which covers Modules 1, 2, 4 and 6 and runs for 2 hours 15 minutes. Section A carries 15 multiple-choice marks while the remaining 85 marks are structured and extended response, and Component 03 adds a 1 hour 30 minute synoptic paper worth 26%. The specification states first teaching in September 2015 with first assessment in 2017.
光子能量 E = hf 与波长换算
单个光子的能量由 E = hf 给出，h 取 6.63 × 10^-34 J s，这个值印在 OCR 的数据页上，答题时不必自己再推一遍量级。频率越高光子能量越大，红光 700 nm 约 2.8 × 10^-19 J，紫光 400 nm 约 5.0 × 10^-19 J，两者相差不到一倍，这个量级感在选择题里非常好用。换成电子伏就除以 1.60 × 10^-19，红光约 1.8 eV，紫光约 3.1 eV，正好横跨可见光区间的两端。题目给波长时，先用 c = fλ 换成频率或者直接用 E = hc/λ，少一步换算就少一次有效数字损失。这一步最容易错在把 nm 当作 m 直接代入，结果差九个数量级，写完之后先看数量级再检查一遍。
Photon energy follows E = hf, with h taken as 6.63 × 10^-34 J s from the OCR data sheet, and E = hc/λ is quicker whenever the wavelength is given. Red light at 700 nm carries about 2.8 × 10^-19 J, roughly 1.8 eV, while violet at 400 nm carries 5.0 × 10^-19 J, close to 3.1 eV; those two values bracket the visible range. Substituting nanometres as metres shifts the answer by nine orders of magnitude, so a sanity check on the power of ten is worth the extra second.
光强到底改变什么
光强描述的是单位时间通过单位面积的能量，不是单个光子的能量，这个区分决定了光电效应里哪些量会随光强变化。同一频率下把光强加倍，每秒到达的光子数加倍，单个光子的能量不变，所以逸出电子的最大动能也不变，只是单位时间逸出的电子数变多。饱和光电流正比于每秒逸出的光电子数，实验里光强加倍、电压足够大时电流也随之加倍。很多学生以为光强越大电子动能越大，这个直觉来自把光当成连续的波，而 4.5 的全部结论都建立在光子模型上。把这一点想通，后面几节的四条实验事实就不需要单独背了。
Intensity measures energy arriving per unit area per second, not the energy of an individual photon, and that distinction decides which quantities change in the photoelectric effect. Doubling the intensity at a fixed frequency doubles the number of photons per second while leaving the energy of each photon untouched, so the maximum kinetic energy stays put and only the photocurrent grows. The classical wave picture predicts the opposite outcome, which is precisely why the experiment carried so much weight.
光电效应的四条实验观察
光电效应的四条观察是整个 4.5 节的骨架。低于某一频率时，无论光强多大都没有电子逸出；高于该频率时电子几乎立即逸出，延迟不超过 10^-9 s；逸出电子的最大动能随频率升高而增大；最大动能与光强无关。经典波动理论对前两条都解释不了，因为波的能量可以慢慢积累，金属总会攒够能量放出电子。OCR 考纲把瞬间逸出与能量与光强无关这两点写得很明确，答题时要把这两个意思都写出来，只写有电子逸出拿不到完整的分。阅卷时按点给分，漏掉一条观察就少一分。
Four observations define the effect: no emission below a threshold frequency regardless of intensity, emission within about 10^-9 s once that threshold is passed, a maximum kinetic energy that rises with frequency, and a maximum kinetic energy that does not respond to intensity. Wave theory fails on the first two because energy in a classical wave can accumulate over time. Mark schemes credit each observation separately, so all four have to be present in a full answer.
逸出功与截止频率的关系
逸出功是把一个电子从金属表面移出所需的最小能量，不同金属差别很大，铯约 2.1 eV，锌约 4.3 eV，铂接近 6 eV。截止频率由 φ = hf0 决定，铯的 f0 大约 5 × 10^14 Hz，对应波长约 600 nm，所以红光几乎打不出铯的电子，绿光就可以。这组数字解释了为什么教科书里的锌板实验要用紫外灯：锌的截止波长约 290 nm，正好落在紫外区。题目给的是逸出功而不是截止频率时，先算 f0 再判断有没有电子逸出，比直接比较波长可靠。反过来，若题目给的是波长，先换成频率再比较，别把两个量混着用。
The work function is the minimum energy needed to remove an electron from a metal surface, and it varies widely: roughly 2.1 eV for caesium, 4.3 eV for zinc and close to 6 eV for platinum. Threshold frequency follows from φ = hf0, giving about 5 × 10^14 Hz for caesium, which corresponds to roughly 600 nm, whereas zinc needs ultraviolet near 290 nm. Converting the work function into a threshold frequency first is far safer than comparing wavelengths by eye.
爱因斯坦方程 hf = φ + KEmax
爱因斯坦方程把入射光子的能量分成两份，一份用来越过逸出功，剩下的全部变成电子的最大动能。这个式子只在光子能量大于逸出功时才有物理意义，小于逸出功时右边会算出负的动能，说明没有电子逸出。做题时先判断频率有没有超过截止频率，再代方程，可以避免把零当成答案交上去。最大动能对应金属表面最外层的电子，它们离开时损失的能量最少，所以实验测到的是一段分布的上端而不是平均值。写答案时把符号 KEmax 说明为最大动能，不要含糊写成动能，否则表述分会被扣掉。
Einstein's equation splits the incoming photon energy into two parts, the work function needed to escape the metal and the maximum kinetic energy carried away. It only applies when the photon energy exceeds the work function, since a smaller photon would give a negative kinetic energy and no emission at all. The maximum refers to electrons at the very surface that lose the least energy on the way out, so the measured distribution has a sharp upper edge rather than a single value.
遏止电压与 h/e 斜率
遏止电压是把最快的光电子恰好拦下来所需的反向电压，满足 eVs = KEmax，因此电压表读数直接给出最大动能。实验里保持频率不变、只改光强，遏止电压不变而饱和电流随光强增大，这是区分光强效应与频率效应最干净的证据。改频率时 Vs 与 f 成直线关系，斜率等于 h/e，截距给出负的逸出功除以 e。这条直线的斜率测出来约 4.1 × 10^-15 V s，与普朗克常量除以电子电荷的公认值吻合。很多学生以为遏止电压靠调光强得到，其实它只跟频率和金属材料有关。
The stopping potential is the reverse voltage that just prevents the fastest photoelectrons from reaching the collector, so eVs = KEmax and a voltmeter reading yields the maximum kinetic energy directly. Holding the frequency fixed while raising intensity leaves that potential unchanged and increases only the saturation current. A graph of Vs against f gives a straight line of gradient h/e, about 4.1 × 10^-15 V s, which is the standard route to Planck's constant in this experiment.
德布罗意波长 λ = h/p 与电子衍射
德布罗意提出任何有动量的粒子都伴随一个波长，写成 λ = h/p，动量越大波长越短。对电子来说动量可以用 mv，也可以用 p = √(2mE)，后者在已知加速电压时更省事。50 V 加速电压下电子波长约 0.17 nm，100 V 约 0.12 nm，150 V 约 0.10 nm，与晶体里的原子间距同一量级，这正是电子衍射能够发生的原因。1927 年戴维孙与革末用镍晶体做出的衍射图样，把这个关系从假设变成了可测量的事实。考纲要求用电子衍射说明波粒二象性，答题要写出波长与原子间距相当这一句，光写电子是波拿不到分。
De Broglie proposed that every particle with momentum carries an associated wavelength, λ = h/p, so faster particles have shorter waves. For electrons the momentum can be written as p = √(2mE), which is convenient when the accelerating voltage is known. At 50 V the wavelength is about 0.17 nm and at 150 V about 0.10 nm, comparable with atomic spacings in a crystal; the nickel diffraction pattern reported by Davisson and Germer in 1927 turned the relation into a measurement.
从加速电压算波长的完整算例
给一道典型计算：电子经过 200 V 加速，求德布罗意波长。先用 eV = 1/2 mv^2 得 v = √(2eV/m)，代入 1.60 × 10^-19 C、200 V 和 9.11 × 10^-31 kg，得到速度约 8.4 × 10^6 m/s。动量 p = mv 约 7.6 × 10^-24 kg m/s，再除以 h 得 λ 约 8.7 × 10^-11 m，也就是 0.087 nm。这条链路有三处四舍五入，中间量保留三位有效数字，最后再压到两位，否则答案会偏出十分之一。若题目直接给动能而不给电压，跳过求速度那一步，直接用 λ = h/√(2mE) 更快也更不容易算错。
A typical question accelerates an electron through 200 V: the speed from eV = 1/2 mv^2 is about 8.4 × 10^6 m/s, the momentum is 7.6 × 10^-24 kg m/s, and the wavelength lands near 0.087 nm. Keeping three significant figures through the intermediate steps and rounding only at the end avoids the ten per cent drift that appears when every line is rounded early. When the kinetic energy is given directly, λ = h/√(2mE) removes one step and one source of error.
能级、跃迁与线状光谱
原子里的电子只能占据分立的能级，跃迁时吸收或放出的光子能量恰好等于能级差，写成 hf = E2 - E1。这一条解释了为什么气体放电管给出的是线状光谱而不是连续谱，每次跃迁对应的波长固定，谱线的位置就是能级结构的指纹。吸收谱与发射谱在同一位置分别出现暗线和亮线，说明同一批跃迁可以双向进行。氢原子从基态到电离需要 13.6 eV，从 n = 2 到 n = 1 的跃迁放出 10.2 eV 的光子，波长约 122 nm，落在紫外区，所以氢的可见谱线都来自更高的能级。阅卷时常见把能级差写成能级的答案，这种表述会直接判错。
Electrons in an atom occupy discrete levels, and a transition absorbs or emits a photon whose energy equals the gap, hf = E2 - E1. That is why a discharge tube produces a line spectrum instead of a continuum: each transition has a fixed wavelength, so the lines map the level structure directly. Hydrogen needs 13.6 eV to ionise from the ground state, and the n = 2 to n = 1 jump releases a 10.2 eV photon at about 122 nm, well inside the ultraviolet.
PAG6 用 LED 测定普朗克常量
PAG6 要求用 LED 或激光研究量子效应，最常见的做法是测每个二极管的阈值电压。正向电压刚让 LED 点亮时，电子获得的能量 eV 近似等于它发出的光子能量 hc/λ，于是 V 对 1/λ 作图得到一条直线，斜率等于 hc/e，乘上电子电荷再除以光速就得到 h。实验的关键在于判断哪一点算点亮，电流刚离开零点时读数最不稳定，通常取电流达到某一小值时的电压，或者把 I 对 V 的直线段外推到电流为零处。实验报告里要写清判断阈值的方法，只写点亮时的电压会被认为不可复现。用五种不同颜色的 LED 各取一个点，直线拟合的相关性往往比单点计算更可靠。
PAG6 asks for an experiment on a quantum effect, and the LED version measures the threshold voltage at which each diode starts to conduct. At that point the energy gained by an electron, eV, is roughly the photon energy hc/λ, so a graph of V against 1/λ is a straight line of gradient hc/e, from which h follows after multiplying by e and dividing by c. The weak point is deciding what counts as the threshold, because the current rises slowly near zero and readings there are unstable.
4.5 最常见的丢分点
最后把 4.5 最容易丢分的地方说清楚。单位要先用标准形式再代入，nm 转 m 乘 10^-9，eV 转 J 乘 1.60 × 10^-19，这两处换算每年都在卷面上出错。图像题要写清坐标轴的含义以及斜率对应哪个物理量，只画线不说明斜率等于什么，通常拿不到最后一分。解释类题按观察、光子模型解释、结论三段写，比如先说低于截止频率没有电子逸出，再说单个光子能量小于逸出功，最后说因此增加光强也不会改变结果。很多学生以为公式背熟就够了，实际上 4.5 的分数大半落在用词准确与单位一致上。
The recurring losses in 4.5 come from units and wording rather than from the physics: convert nanometres with 10^-9 and electronvolts with 1.60 × 10^-19 before substituting, and state what the gradient of a graph represents rather than drawing the line alone. Explanation questions score best when written as observation, photon-model explanation, then conclusion. Knowing the equation is rarely the limiting factor; saying precisely what each symbol means is.
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