题型概述：光子与光电效应Question type: photons & the photoelectric effect
Top GoldSection 1 几乎每年都有一道光子 / 光电效应小题（3–6 分），套路固定：先算一个光子的能量，再用爱因斯坦光电方程，最后按题目要求换算成电势、速度、时间或光子个数。失分几乎都来自单位（nm、eV、mW、mm）和对"最大动能"含义的误解。Almost every Section 1 paper has a short photon / photoelectric question (3–6 marks) with a fixed pattern: find the energy of one photon, apply Einstein's photoelectric equation, then convert to whatever is asked — a potential, a speed, a time or a number of photons. Marks are lost almost entirely to units (nm, eV, mW, mm) and to misreading what "maximum kinetic energy" means.
本模块统一使用：Constants used throughout: h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m/s, e = 1.60×10⁻¹⁹ C, me = 9.11×10⁻³¹ kg, ε₀ = 8.85×10⁻¹² F/m。答案可用科学计数法输入，例如 7.16e-19 或 7.16×10^-19。. Answers can be typed in scientific notation, e.g. 7.16e-19 or 7.16×10^-19.
1光子能量：E = hf = hc/λPhoton energy: E = hf = hc/λ
光由一份一份的光子组成，每个光子的能量只由频率决定，与光的亮暗无关。题目通常给波长，用 c = fλ 换成频率。原子尺度的能量非常小，所以常用电子伏特（eV）：1 eV 是一个电子穿过 1 V 电势差时获得的能量。Light comes in packets called photons, and the energy of each one depends only on its frequency, not on how bright the light is. Questions usually give the wavelength, so use c = fλ. Atomic-scale energies are tiny, so the electronvolt is common: 1 eV is the energy an electron gains crossing a potential difference of 1 V.
E = hf = hc/λ 1 eV = 1.60×10⁻¹⁹ J速算：E(eV) ≈ 1240 / λ(nm)。例如 400 nm 紫光 ≈ 3.1 eV，150 nm 紫外 ≈ 8.3 eV，639 nm 红光 ≈ 1.9 eV。用它检查计算结果的数量级。Quick check: E(eV) ≈ 1240 / λ(nm). So 400 nm violet ≈ 3.1 eV, 150 nm UV ≈ 8.3 eV, 639 nm red ≈ 1.9 eV. Use it to sanity-check the order of magnitude.
2爱因斯坦光电方程与阈值Einstein's equation and the threshold
一个光子只能把能量交给一个电子。电子先要"付"一笔功函数 φ（也写作 W）才能离开金属表面，剩下的变成动能。束缚最弱的电子剩得最多，所以方程给出的是最大动能 Kmax；其他电子在内部多损失了一些能量，动能在 0 到 Kmax 之间。One photon gives all its energy to one electron. The electron must first "pay" the work function φ (also written W) to leave the metal surface; what is left becomes kinetic energy. The least-bound electrons keep the most, so the equation gives the maximum kinetic energy Kmax; other electrons lose a little more inside the metal and have anything from 0 to Kmax.
hf = φ + Kmax 阈值threshold: f₀ = φ/h, λ₀ = hc/φ光强只决定每秒到达多少个光子（从而决定光电流大小），不改变每个光子的能量，所以 Kmax 与光强无关。若 hf < φ（λ > λ₀），单个光子不够付"出门费"，再强的光也打不出电子。Intensity only sets how many photons arrive per second (and so the size of the photocurrent); it does not change the energy of each photon, so Kmax does not depend on intensity. If hf < φ (λ > λ₀) a single photon cannot pay the "exit fee", and no amount of intensity will release electrons.
试一试：① 保持波长，调光强——电子变多（电流变大），但 Kmax 不变；② 把波长调到 λ₀ 以上——即使光强 100%，也一个电子都没有。Try it: ① keep λ fixed and change the intensity — more electrons (bigger current) but the same Kmax; ② push λ beyond λ₀ — even at 100% intensity, not a single electron.
3遏止电势与"孤立导体会充电"Stopping potential and the charging isolated conductor
让收集极比发射极低 V（反向/遏止电压），电子穿过间隙时要"爬坡"，动能减少 eV。把 V 调大到连最快的电子也到不了收集极时，光电流为零，这时的电压就是遏止电势 Vs。Make the collector lower in potential than the emitter by V (a retarding or stopping voltage): electrons crossing the gap "climb a hill" and lose kinetic energy eV. Raise V until even the fastest electron cannot reach the collector; the current drops to zero and that voltage is the stopping potential Vs.
eVs = Kmax = hf − φ孤立导体（不接地的金属板或金属球）自己就会造出这个"坡"：电子跑掉后导体带正电、电势升高，后来的电子要克服这个电势才能逃走。电势升到 eV = Kmax 时连最快的电子也被拉回，电荷不再改变。对球体，再用孤立球的电势公式求电荷和电子数。An isolated conductor (an unearthed plate or sphere) builds this hill by itself: each electron that leaves makes it more positive, so later electrons must overcome its potential to escape. Once eV = Kmax even the fastest electron is pulled back and the charge stops changing. For a sphere, the isolated-sphere potential then gives the charge and the number of electrons.
Vmax = Kmax/e, Q = 4πε₀rVmax, n = Q/e数值技巧：Kmax 用 eV 表示时，Vmax 的伏特数和它相同（3.79 eV ⇒ 3.79 V）。Numerical shortcut: with Kmax in eV, Vmax in volts is the same number (3.79 eV ⇒ 3.79 V).
| 子题型Sub-type | 识别特征How to spot it | 核心步骤Core step | 真题Past paper |
|---|---|---|---|
| 求功函数Find the work function | 给 λ 和 Kmaxλ and Kmax given | W = hc/λ − Kmax | 2016 Q1(j) |
| 孤立导体的最大电势Maximum potential of an isolated conductor | "acquire a positive potential", "isolated sphere" | eV = hf − φ；球：Q = 4πε₀rV，n = Q/eeV = hf − φ; sphere: Q = 4πε₀rV, n = Q/e | 2012 Q1(f), 2019 Q1(p) |
| 电子飞行时间Electron time of flight | "electrode placed … m away", "stopping potential applied" | Kmax → v，再用运动学（匀速 / 匀减速）Kmax → v, then kinematics (uniform / uniform deceleration) | 2021 Q1(p) |
| 光子计数Counting photons | 给功率、时间、波长power, time and wavelength given | N = Pt / (hc/λ) | 2017 Q1(j) |