📚 A-Level Physics Mark Scheme Unit 1 Jan21: Formula Derivation | A-Level物理评分方案单元1 2021年1月：公式推导
Understanding how to derive the photoelectric equation from experimental data is a core requirement in A-Level Physics Unit 1. The January 2021 mark scheme gives clear criteria for what gains marks: correct use of photon energy E = hf, the work function φ, and the link between stopping potential and maximum kinetic energy. This article walks through the derivation step by step, highlighting exactly what examiners look for.
理解如何从实验数据推导光电效应方程是A-Level物理单元1的核心要求。2021年1月的评分方案明确指出了得分点：正确使用光子能量E = hf、逸出功φ，以及截止电压与最大动能之间的关系。本文将一步步展示推导过程，并突出考官所关注的内容。
1. Introduction to the Photoelectric Effect | 光电效应简介
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency shines on it. In Unit 1, you are expected to explain why classical wave theory cannot account for the observations – namely, the existence of a threshold frequency, instantaneous emission, and kinetic energy dependence on frequency, not intensity.
光电效应是指当频率足够高的电磁辐射照射到金属表面时，电子从该表面逸出的现象。在单元1中，你需要解释为什么经典波动理论无法解释观察到的现象——即存在截止频率、瞬时发射以及动能依赖于频率而非光强。
2. Experimental Setup and Key Observations | 实验装置与关键观察
A typical experiment uses a vacuum phototube, a variable voltage supply, and a sensitive ammeter. Monochromatic light of known frequency f illuminates the cathode. By applying a reverse potential (stopping potential Vₛ), the photocurrent is reduced to zero. For each frequency, the stopping potential is recorded. The key observations are: no photoelectrons below a threshold frequency f₀, and above f₀, the stopping potential Vₛ increases linearly with f.
典型实验使用真空光电管、可调电压源和灵敏电流计。已知频率f的单色光照射阴极。通过施加反向电压（截止电压Vₛ），使光电流降至零。记录每个频率对应的截止电压。关键现象是：低于截止频率f₀时光电子数为零；在f₀以上，截止电压Vₛ随频率f线性增加。
3. Einstein’s Photoelectric Equation | 爱因斯坦光电效应方程
Einstein proposed that light consists of photons, each with energy E = hf. When a photon hits the metal, its energy is transferred to a single electron. The electron must use a minimum energy, the work function φ, to escape the surface. Any remaining energy appears as the electron’s maximum kinetic energy. This gives the famous equation: ½ m vₘₐₓ² = h f – φ.
爱因斯坦提出光由光子组成，每个光子能量为E = hf。光子撞击金属时，其能量传递给单个电子。电子必须消耗最小能量——逸出功φ，才能脱离表面。剩余能量表现为电子的最大动能。由此得到著名方程：½ m vₘₐₓ² = h f – φ。
4. Relating Stopping Potential to Maximum Kinetic Energy | 截止电压与最大动能的关系
When a stopping potential Vₛ is applied, even the most energetic electrons are repelled and just fail to reach the anode. At this point, the electrical work done e Vₛ equals the maximum kinetic energy. Hence: e Vₛ = ½ m vₘₐₓ². Substituting into Einstein’s equation gives: e Vₛ = h f – φ.
当施加截止电压Vₛ时，即使是动能最大的电子也会被排斥，恰好无法到达阳极。此时，电场做的功e Vₛ等于最大动能。因此：e Vₛ = ½ m vₘₐₓ²。代入爱因斯坦方程得到：e Vₛ = h f – φ。
5. Rearranging into the Form y = mx + c | 整理为 y = mx + c 形式
To determine Planck’s constant h, we rearrange the equation: Vₛ = (h/e) f – (φ/e). This is now a straight-line equation of the form y = mx + c, where y = Vₛ, x = f, the gradient m = h/e, and the intercept c = –φ/e. This linear relationship is central to the mark scheme.
为了测定普朗克常数h，我们重新整理方程：Vₛ = (h/e) f – (φ/e)。此时方程具有y = mx + c的直线形式，其中y = Vₛ，x = f，斜率m = h/e，截距c = –φ/e。这种线性关系是评分方案的重点。
6. Determining Planck’s Constant from the Graph | 从图像测定普朗克常数
Plot Vₛ on the y-axis and f on the x-axis. Draw a line of best fit through the data points above f₀. The gradient is ΔVₛ/Δf. Since gradient = h/e, we get h = e × gradient. The mark scheme rewards correct calculation of gradient and multiplication by the electronic charge e (1.60 × 10⁻¹⁹ C).
将Vₛ标在y轴，f标在x轴上作图。在f₀以上的数据点间画出最佳拟合线。斜率为ΔVₛ/Δf。由于斜率 = h/e，得到h = e × 斜率。评分方案认可正确计算斜率并乘以电子电荷e (1.60 × 10⁻¹⁹ C)。
7. Work Function and Threshold Frequency | 逸出功与截止频率
The work function φ is the minimum energy needed to eject an electron. From the graph, the x-intercept gives the threshold frequency f₀ where Vₛ = 0. At f₀, 0 = (h/e)f₀ – (φ/e), so φ = h f₀. Alternatively, using the y-intercept: c = –φ/e, giving φ = –e × intercept. Both methods are accepted in the mark scheme.
逸出功φ是逸出电子所需的最小能量。从图中看，x轴截距给出截止频率f₀，此时Vₛ = 0。在f₀处，0 = (h/e)f₀ – (φ/e)，因此φ = h f₀。另一种方法是利用y轴截距：c = –φ/e，得到φ = –e × 截距。评分方案两种情况都接受。
8. Step-by-Step Derivation with Mark Scheme Points | 结合评分方案的分步推导
Below is a breakdown of how marks are typically allocated. (Based on the Jan 2021 unit 1 mark scheme logic.)
以下是典型得分点分解。（基于2021年1月单元1评分方案逻辑。）
- Step 1: State photon energy E = hf. (1 mark)
步骤1： 写出光子能量E = hf。（1分） - Step 2: Write energy conservation: photon energy = work function + maximum kinetic energy. (1 mark)
步骤2： 写出能量守恒：光子能量 = 逸出功 + 最大动能。（1分） - Step 3: Express maximum kinetic energy as ½ m vₘₐₓ². Accept Eₖ(max). (1 mark)
步骤3： 将最大动能表达为½ m vₘₐₓ²，也接受Eₖ(max)。（1分） - Step 4: Relate to stopping potential: e Vₛ = ½ m vₘₐₓ². (1 mark)
步骤4： 关联截止电压：e Vₛ = ½ m vₘₐₓ²。（1分） - Step 5: Combine to get e Vₛ = h f – φ. (1 mark)
步骤5： 合并得到e Vₛ = h f – φ。（1分） - Step 6: Rearrange to Vₛ = (h/e) f – (φ/e) and identify gradient. (1 mark)
步骤6： 整理为Vₛ = (h/e) f – (φ/e)并识别斜率。（1分） - Step 7: Calculate h from gradient × e. (1 mark)
步骤7： 由斜率 × e计算h。（1分）
Examiners also look for correct units: Vₛ in V, f in Hz, h in J s, φ in J. Marks may be deducted for missing unit conversions.
考官还要求单位正确：Vₛ用V，f用Hz，h用J s，φ用J。缺少单位换算可能被扣分。
9. Example Data and Graph Interpretation | 示例数据与图像解读
| Frequency f / 10¹⁴ Hz | Stopping potential Vₛ / V |
|---|---|
| 5.5 | 0.35 |
| 6.0 | 0.58 |
| 6.5 | 0.80 |
| 7.0 | 1.02 |
| 7.5 | 1.25 |
Plotting these points yields a straight line with equation Vₛ = 0.41 f – 1.9 (approximately). Here h/e ≈ 0.41 × 10⁻¹⁴ V s, so h ≈ 0.41 × 10⁻¹⁴ × 1.6 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ J s. The x-intercept (Vₛ=0) gives f₀ ≈ 4.6 × 10¹⁴ Hz, hence φ = h f₀ ≈ 3.0 × 10⁻¹⁹ J.
描绘这些点可得一条直线，近似方程为Vₛ = 0.41 f – 1.9。这里h/e ≈ 0.41 × 10⁻¹⁴ V s，因此h ≈ 0.41 × 10⁻¹⁴ × 1码.6 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ J s。x轴截距（Vₛ=0）给出f₀ ≈ 4.6 × 10¹⁴ Hz，所以φ = h f₀ ≈ 3.0 × 10⁻¹⁹ J。
10. Common Pitfalls and Examiner Tips | 常见错误与考官提示
One frequent mistake is confusing intensity with frequency. The mark scheme penalises using intensity to explain kinetic energy variations. Another pitfall is forgetting to convert frequency units (e.g., from 10¹⁴ Hz to Hz) when calculating gradient. Also, ensure you draw the line of best fit only for points above f₀; points below f₀ are not part of the linear relationship. Examiners reward clear labelling of axes and stating the gradient value with its unit.
一个常见错误是混淆光强与频率。评分方案对用光强解释动能变化会扣分。另一个易错点是计算斜率时忘记频率单位换算（例如从10¹⁴ Hz换算到Hz）。此外，务必仅在f₀以上的点画最佳拟合线；f₀以下的点不属于线性关系。考官喜欢坐标轴清晰标注，并写出斜率值及单位。
11. Why This Derivation Matters Beyond the Exam | 推导在考试之外的意义
The photoelectric equation derivation is not just a mark-scoring exercise. It confirms the particle nature of light and was pivotal in the development of quantum mechanics. The ability to extract Planck’s constant from a simple voltage-frequency graph illustrates how fundamental constants can be measured with relatively basic equipment. In real-world applications, photoelectric cells are used in solar panels, light meters, and imaging sensors, all relying on the same principles.
光电方程推导不仅是得分练习，它证实了光的粒子性，并在量子力学发展中起到关键作用。从简单的电压-频率图像中提取普朗克常数，展示了如何用相对基础的设备测量基本常数。在现实应用中，光电池被用于太阳能板、测光表和图像传感器，全都依赖相同的原理。
12. Summary and Final Revision Checklist | 总结与最终复习清单
The core of the Unit 1 Jan21 mark scheme is the logical chain: E = hf → hf = φ + KEₘₐₓ → KEₘₐₓ = e Vₛ → e Vₛ = hf – φ → Vₛ = (h/e)f – (φ/e). Master this chain, and you secure nearly all marks in the derivation question. Be ready to interpret the graph, calculate h, and determine the work function. Practice with different numbers of significant figures to match the data precision.
单元1 2021年1月评分方案的核心是逻辑链：E = hf → hf = φ + KEₘₐₓ → KEₘₐₓ = e Vₛ → e Vₛ = hf – φ → Vₛ = (h/e)f – (φ/e)。掌握这一链条，你几乎能拿到推导题的全部分数。准备好解读图像，计算h，并确定逸出功。练习使用不同有效数字以匹配数据精度。
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