📚 A-Level Physics Unit 5 Mark Scheme Jan 22: Formula Derivations | A-Level 物理第五单元 2022年1月评分方案：公式推导
This article unpacks the key formula derivations that appeared in the A-Level Physics Unit 5 mark scheme for the January 2022 examination. Whether you are revising thermodynamics, nuclear physics, oscillations, or gravitational fields, understanding how to derive the essential equations is crucial for high marks. Each derivation is broken down step by step, with explanations in both English and Chinese to support bilingual learners. The content aligns with the approaches required by examiners and mirrors the logical flow expected in structured questions.
本文深度解析了 2022 年 1 月 A-Level 物理第五单元评分方案中出现的核心公式推导。无论你正在复习热力学、核物理、振荡还是引力场，掌握关键方程的推导过程对获取高分至关重要。每个推导都按步骤拆解，配以中英双语解释，助力双语学习者。内容贴合考官要求的思路，并反映了结构化问题中期望的逻辑流程。
1. Radioactive Decay Law Derivation | 放射衰变定律推导
The decay law arises from the assumption that the activity A is proportional to the number of undecayed nuclei N: A = λN. Since activity is the rate of decay, we write dN/dt = –λN, where the negative sign indicates a decrease in N. Separating variables gives dN/N = –λ dt. Integrating both sides yields ln N = –λt + constant. Applying the initial condition N = N₀ at t = 0 sets the constant to ln N₀. Hence ln(N/N₀) = –λt, which exponentiates to N = N₀ e⁻λᵗ.
衰变定律的推导基于一个假设：活度 A 与未衰变核数 N 成正比，即 A = λN。由于活度就是衰变率，因此可写为 dN/dt = –λN，其中负号表示 N 在减少。分离变量得到 dN/N = –λ dt。两边积分得 ln N = –λt + 常数。利用初始条件 t = 0 时 N = N₀，定出常数为 ln N₀。于是 ln(N/N₀) = –λt，取指数后即得 N = N₀ e⁻λᵗ。
This exponential relationship is fundamental to all nuclear decay calculations. In mark schemes, candidates must show clear separation of variables and correct handling of the integration constant.
这种指数关系是所有核衰变计算的基础。在评分方案中，考生必须清晰地展示分离变量以及正确处理积分常数。
2. Relationship Between Half-life and Decay Constant | 半衰期与衰变常数的关系
Half-life T₁/₂ is defined as the time for half the nuclei to decay. Substituting N = N₀/2 into N = N₀ e⁻λᵗ gives 1/2 = e⁻λT₁/₂. Taking natural logarithms: ln(1/2) = –λT₁/₂, so –ln 2 = –λT₁/₂. Therefore T₁/₂ = ln 2 / λ. This simple derivation must be presented logically in exam answers, often with a step explicitly stating that ln(1/2) = –ln 2.
半衰期 T₁/₂ 定义为半数核发生衰变所需的时间。将 N = N₀/2 代入 N = N₀ e⁻λᵗ 得 1/2 = e⁻λT₁/₂。取自然对数：ln(1/2) = –λT₁/₂，因此 –ln 2 = –λT₁/₂。故 T₁/₂ = ln 2 / λ。这个简短的推导在考试回答中必须逻辑清晰地呈现，通常需要明确写出 ln(1/2) = –ln 2 这一步骤。
The mark scheme often rewards both the algebraic manipulation and the correct interpretation of the half-life definition.
评分方案通常既奖励代数操作，也奖励对半衰期定义的正确理解。
3. Capacitor Discharge Equation | 电容器放电方程
For a capacitor discharging through a fixed resistor, the current I = dQ/dt is negative because the charge Q on the plates decreases. Using the definition of capacitance C = Q/V and Ohm’s law V = IR, the pd across the resistor equals the capacitor voltage: V = –IR (with sign conventions). Combining gives Q/C = –R dQ/dt. Rearranging: dQ/dt = –Q/(RC). This is of the same form as the radioactive decay equation, with time constant RC. The solution is Q = Q₀ e⁻ᵗ/ᴿᴳ.
对于通过固定电阻放电的电容器，电流 I = dQ/dt 取负值，因为极板上的电荷 Q 在减少。利用电容定义 C = Q/V 和欧姆定律 V = IR，电阻两端的电压等于电容电压：V = –IR（注意符号规则）。联立得 Q/C = –R dQ/dt。整理后为 dQ/dt = –Q/(RC)。这与放射性衰变方程形式相同，时间常数为 RC。解为 Q = Q₀ e⁻ᵗ/ᴿᴳ。
Examiners expect a derivation that carefully addresses the negative sign; otherwise the exponential decay constant loses its physical meaning. The voltage and current equations follow naturally by substituting V = Q/C and I = dQ/dt.
考官期望推导演算时仔细处理负号，否则指数衰减常数会失去物理意义。将 V = Q/C 和 I = dQ/dt 代入后，电压和电流方程自然得出。
4. Energy Stored in a Charged Capacitor | 电容器储存的能量
The work done to add a small charge dq when the potential difference is v is dW = v dq. Since v = q/C, we have dW = (q/C) dq. Integrating from q = 0 to Q gives total stored energy W = ∫₀̄ᴼ (q/C) dq = [q²/(2C)]₀̄ᴼ = ½ Q²/C. Using Q = CV, alternative forms are W = ½ CV² and W = ½ QV. This derivation is frequently assessed in structured questions; the definite integral must be shown explicitly.
当电势差为 v 时，转移微小电荷 dq 所做的功为 dW = v dq。由于 v = q/C，可得 dW = (q/C) dq。从 q = 0 到 Q 积分，得到储存的总能量 W = ∫₀̄ᴼ (q/C) dq = [q²/(2C)]₀̄ᴼ = ½ Q²/C。利用 Q = CV，也可写出 W = ½ CV² 和 W = ½ QV。该推导在结构化问题中经常考查，定积分必须明确展示。
Mark schemes often require recognising that the area under a voltage–charge graph represents energy, reinforcing the integral derivation.
评分方案常要求认识到电压-电荷图下方面积代表能量，从而印证积分推导。
5. Simple Harmonic Motion Acceleration Equation | 简谐运动加速度方程
SHM is defined by a restoring force proportional to displacement and directed towards equilibrium: F = –kx. Applying Newton’s second law, F = ma, gives ma = –kx, hence a = –(k/m)x. Defining ω² = k/m yields the standard SHM acceleration equation a = –ω²x. The period T can then be derived from ω = 2π/T. This definition-based derivation is a key starting point in exam mark schemes.
简谐运动的定义是恢复力与位移成正比且指向平衡位置：F = –kx。应用牛顿第二定律 F = ma，得 ma = –kx，故 a = –(k/m)x。定义 ω² = k/m，即得到标准的简谐运动加速度方程 a = –ω²x。周期 T 可由 ω = 2π/T 推出。这个基于定义的推导是考试评分方案中的关键起点。
Examiners look for the correct linking of force, acceleration and the ω² substitution, as well as the application to mass-spring and simple pendulum systems.
考官期望看到力、加速度和 ω² 代换之间的正确联系，以及将其应用于弹簧振子和单摆系统。
6. Velocity–Displacement Relation in SHM | 简谐运动速度与位移关系
Starting from a = d²x/dt² = –ω²x, we can derive the velocity equation by using the chain rule: a = dv/dt = dv/dx · dx/dt = v dv/dx. Thus v dv/dx = –ω²x. Separating variables: ∫ v dv = –ω² ∫ x dx, which integrates to ½ v² = –½ ω² x² + constant. Using the condition that v = 0 when x = A (amplitude), the constant becomes ½ ω² A². Hence v² = ω² (A² – x²), or v = ± ω√(A² – x²).
从 a = d²x/dt² = –ω²x 出发，利用链式法则可以推导速度方程：a = dv/dt = dv/dx · dx/dt = v dv/dx。因此 v dv/dx = –ω²x。分离变量得 ∫ v dv = –ω² ∫ x dx，积分后为 ½ v² = –½ ω² x² + 常数。利用 v = 0 时 x = A（振幅）的条件，定出常数为 ½ ω² A²。最终得到 v² = ω² (A² – x²)，或 v = ± ω√(A² – x²)。
This derivation is often tested as a multi-step structured question. The mark scheme typically splits marks for separating variables, setting limits, and interpreting the constant correctly.
这个推导常以多步结构化问题的形式考查。评分方案通常将分数分配给分离变量、设定边界条件以及正确解释常数这几个环节。
7. Kinetic Theory Derivation of pV = NkT | 用分子动理论推导 pV = NkT
Consider N particles of gas in a cube of side L. A particle moving with velocity component vₓ hits the wall and rebounds elastically, changing momentum by 2mvₓ. The time between collisions with one wall is 2L/vₓ, so the average force from one particle is Δp/Δt = (2mvₓ) / (2L/vₓ) = mvₓ²/L. Summing over all particles, the total force F = Σ mvₓ²/L = (m/L) Σ vₓ². Pressure p = F/L² = (m/L³) Σ vₓ² = (m/V) Σ vₓ². Using the root-mean-square speed, Σ vₓ² = N⟨vₓ²⟩, and by isotropy ⟨vₓ²⟩ = ⅓⟨v²⟩. Hence p = (m/V) · N · ⅓⟨v²⟩ = ⅓ (N/V) m⟨v²⟩. Finally, recognising that ½ m⟨v²⟩ = ³⁄₂ kT gives pV = NkT.
考虑一个边长为 L 的立方体容器内有 N 个气体粒子。一个粒子以速度分量 vₓ 运动，与器壁弹性碰撞后动量变化为 2mvₓ。与同一器壁两次碰撞的时间间隔为 2L/vₓ，因此一个粒子的平均作用力为 Δp/Δt = (2mvₓ) / (2L/vₓ) = mvₓ²/L。对所有粒子求和，总力 F = Σ mvₓ²/L = (m/L) Σ vₓ²。压强 p = F/L² = (m/L³) Σ vₓ² = (m/V) Σ vₓ²。引入方均根速率，Σ vₓ² = N⟨vₓ²⟩，且由各向同性有 ⟨vₓ²⟩ = ⅓⟨v²⟩。于是 p = (m/V) · N · ⅓⟨v²⟩ = ⅓ (N/V) m⟨v²⟩。最后利用 ½ m⟨v²⟩ = ³⁄₂ kT，即得 pV = NkT。
This derivation is a cornerstone of Unit 5 thermodynamics. The mark scheme emphasises clear steps: momentum change, time between collisions, summing forces, introducing mean square speed and linking to temperature.
这个推导是 Unit 5 热力学的基石。评分方案强调清晰的步骤：动量变化、碰撞时间间隔、力的求和、引入均方速率以及与温度的联系。
8. Internal Energy and the First Law of Thermodynamics | 内能与热力学第一定律
The first law states that the increase in internal energy ΔU of a system equals the net heat energy added Q minus the net work done by the system W: ΔU = Q – W. For an ideal gas, internal energy depends only on temperature, and for a reversible expansion at constant pressure, work done is W = p ΔV. Derivations involving isothermal or adiabatic processes start from these relations. For an adiabatic change (Q = 0), ΔU = –W, and using U ∝ T leads to the pV^γ = constant relation.
热力学第一定律指出，系统内能的增量 ΔU 等于净增加的热量 Q 减去系统对外做的净功 W：ΔU = Q – W。对于理想气体，内能只与温度有关；在等压可逆膨胀中，功为 W = p ΔV。涉及等温或绝热过程的推导都从这些关系出发。对于绝热变化（Q = 0），ΔU = –W，再利用 U ∝ T 即可推出 pV^γ = 常数的关系。
The mark scheme often requires careful sign conventions and the ability to apply the first law to specific gas processes, as seen in January 2022 paper questions.
评分方案通常要求仔细处理符号规则，并能够将第一定律应用于具体气体过程，这在 2022 年 1 月的试卷问题中也有所体现。
9. Gravitational Potential Derivation | 引力势的推导
Gravitational potential V at a point in a radial field is defined as the work done per unit mass to bring a test mass from infinity to that point. The gravitational force on a mass m is F = GM m/r². Work done against this force over a small displacement dr is dW = –F dr = –(GM m/r²) dr. Integrating from r = ∞ to r gives V = W/m = –GM ∫ₒᵣ (1/r²) dr = –GM [–1/r]ₒᵣ = –GM(1/∞ – 1/r) = –GM/r. The negative sign indicates that work is done by the field when a mass moves towards the source.
径向场中某点的引力势 V 定义为将单位质量检验物体从无穷远移至该点外力所做的功。作用在质量 m 上的引力为 F = GM m/r²。克服此力移动微小位移 dr 所做的功为 dW = –F dr = –(GM m/r²) dr。从 r = ∞ 积分到 r，得 V = W/m = –GM ∫ₒᵣ (1/r²) dr = –GM [–1/r]ₒᵣ = –GM(1/∞ – 1/r) = –GM/r。负号表明当质量向场源移动时，场做正功。
This derivation is frequently examined in the context of gravitational fields. The mark scheme expects a clear integral set-up with correct limits and an explanation of the negative sign.
这个推导在引力场情境下常被考查。评分方案期望清晰的积分设置、正确的积分限以及对负号的解释。
10. Wien’s Displacement Law and Peak Wavelength | 维恩位移定律与峰值波长
Wien’s law states that the wavelength λₘₐₓ at which a black-body radiation curve peaks is inversely proportional to its absolute temperature: λₘₐₓ T = constant ≈ 2.898 × 10⁻³ m K. While the full quantum derivation requires Planck’s law, the examination often tests the conceptual understanding that higher temperature shifts the peak to shorter wavelengths. In a mark scheme, candidates are expected to interpret the peak of a given spectral curve and apply the proportionality.
维恩定律指出，黑体辐射曲线峰值对应的波长 λₘₐₓ 与其绝对温度成反比：λₘₐₓ T = 常数 ≈ 2.898 × 10⁻³ m·K。虽然完整的量子推导需要借助普朗克定律，但考试常考查概念理解：温度越高，峰值波长越短。在评分方案中，考生应能解读给定的光谱曲线并应用比例关系。
Derivations may involve recognising that the product λₘₐₓ T remains unchanged for a given source, allowing calculation of temperature or peak wavelength when one is known.
推导可能涉及认识到对于给定源，λₘₐₓ T 乘积保持不变，从而在已知一个量时计算温度或峰值波长。
11. Stefan–Boltzmann Law and Luminosity | 斯特藩–玻尔兹曼定律与光度
The Stefan–Boltzmann law relates the total power radiated per unit area of a black body to its temperature: L = σ A T⁴, where σ is the Stefan–Boltzmann constant. For a spherical star of radius R, the luminosity becomes L = 4πR² σ T⁴. Although the full law stems from integrating Planck’s curve, students are required to use it to compare luminosities and temperatures. A typical mark scheme expects the equation to be rearranged and values substituted correctly, often in a ratio form to eliminate constants.
斯特藩–玻尔兹曼定律将黑体单位面积的总辐射功率与其温度联系起来：L = σ A T⁴，其中 σ 为斯特藩–玻尔兹曼常数。对于半径为 R 的球形恒星，光度变为 L = 4πR² σ T⁴。尽管完整的定律来自对普朗克曲线的积分，但学生需运用它来比较光度和温度。典型的评分方案期望对方程进行变形并正确代入数值，常以比值形式约去常数。
Understanding this derivation helps solve problems on stellar brightness and black-body radiation, which appear regularly in the astrophysics section of Unit 5.
理解这个推导有助于解决关于恒星亮度和黑体辐射的问题，这些问题在 Unit 5 的天体物理部分经常出现。
12. Linking the Derivations to the Jan 22 Mark Scheme | 将推导关联至 2022 年 1 月评分方案
The January 2022 Unit 5 paper included structured questions that required robust derivations of several of the above equations. Mark scheme annotations highlighted the importance of stating initial assumptions, showing clear integration limits, and correctly handling negative signs in exponential relations. Reviewing these derivations in the context of the mark scheme reinforces exam technique: partial marks are awarded for method, even if the final answer has a minor slip. Practising these derivations bilingual ensures you can meet the examiner’s expectations with confidence.
2022 年 1 月的 Unit 5 试卷包含结构化问题，要求对上述多个方程进行严谨推导。评分方案的批注强调了陈述初始假设、清晰展示积分限以及正确处理指数关系中负号的重要性。结合评分方案复习这些推导能强化考试技巧：即使最终答案有小错误，方法步骤也能获得部分分数。用中英双语练习这些推导，可确保你能自信地满足考官的期望。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导