📚 A-Level Physics: Formula Derivations from Unit 5 Jan 22 Paper | A-Level 物理：2022年1月单元5试卷公式推导
In the January 2022 A-Level Physics Unit 5 examination, candidates were challenged to derive and apply several fundamental equations from thermodynamics, nuclear physics and astrophysics. This article steps through those derivations with clarity, supporting students who wish to master the reasoning behind each relationship.
在2022年1月的A-Level物理单元5考试中，考生需要推导并应用热力学、核物理和天体物理中的多个基本方程。本文将通过清晰的步骤逐一拆解这些推导过程，帮助同学们彻底掌握每个关系式背后的逻辑。
1. Kinetic Theory Derivation of Ideal Gas Equation | 理想气体方程的分子动理论推导
Consider a cube of side L containing N identical particles, each of mass m. A single particle moves with velocity components (vₓ, v_y, v_z). Its x-component of momentum change when striking a wall is 2 m vₓ.
考虑一个边长为 L 的立方体，内有 N 个相同的粒子，每个质量为 m。一个粒子以速度分量 (vₓ, v_y, v_z) 运动。它撞击器壁时，动量的 x 分量变化为 2 m vₓ。
The time between successive collisions with the same wall is 2 L / vₓ, so the average force exerted by this particle on the wall is F = (2 m vₓ) / (2 L / vₓ) = m vₓ² / L.
同一壁面两次碰撞之间的时间为 2 L / vₓ，因此该粒子对器壁的平均作用力为 F = (2 m vₓ) / (2 L / vₓ) = m vₓ² / L。
Summing over all N particles and dividing by the wall area A = L² gives the pressure: P = (1 / L²) (m / L) Σ vₓ² = (m / V) Σ vₓ², where V = L³. Using the mean square speed ⟨v²⟩ = (vₓ² + v_y² + v_z²) and isotropy ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⅓ ⟨v²⟩, we obtain P = (1/3) (N m / V) ⟨v²⟩.
对所有 N 个粒子求和，并除以壁面积 A = L² 得到压强：P = (1 / L²) (m / L) Σ vₓ² = (m / V) Σ vₓ²，其中 V = L³。利用均方速率 ⟨v²⟩ = (vₓ² + v_y² + v_z²) 以及各向同性 ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⅓ ⟨v²⟩，可得 P = (1/3) (N m / V) ⟨v²⟩。
Introduce the average translational kinetic energy: ⟨Eₖ⟩ = ½ m ⟨v²⟩. The kinetic theory links this to absolute temperature via ⟨Eₖ⟩ = (3/2) k T, where k is the Boltzmann constant. Substituting gives P V = N k T, and with the mole concept N = n N_A, k N_A = R, we arrive at the ideal gas equation P V = n R T.
引入平均平动动能：⟨Eₖ⟩ = ½ m ⟨v²⟩。分子动理论将它与热力学温度联系起来，即 ⟨Eₖ⟩ = (3/2) k T，其中 k 为玻尔兹曼常数。代入得 P V = N k T，再利用摩尔概念 N = n N_A，k N_A = R，最终导出理想气体方程 P V = n R T。
2. Radioactive Decay Law and Half-Life Derivation | 放射性衰变定律与半衰期推导
The activity A of a sample is the number of decays per unit time, which is proportional to the number of undecayed nuclei N: A = -dN/dt = λ N, where λ is the decay constant.
样品的活度 A 是单位时间的衰变次数，与未衰变核的数目 N 成正比：A = -dN/dt = λ N，其中 λ 为衰变常量。
Separating variables and integrating from N₀ (at t = 0) to N gives ∫_{N₀}^{N} dN / N = -λ ∫_{0}^{t} dt, leading to ln (N / N₀) = -λ t. Rearranging yields the exponential decay law N = N₀ e^(-λ t).
分离变量并从 N₀（t = 0 时）积分到 N，可得 ∫_{N₀}^{N} dN / N = -λ ∫_{0}^{t} dt，由此得到 ln (N / N₀) = -λ t。整理后即得指数衰变律 N = N₀ e^(-λ t)。
Half-life T₁/₂ is the time when N = N₀ / 2. Substituting gives ½ = e^(-λ T₁/₂); taking natural logarithms produces T₁/₂ = ln 2 / λ. This shows the inverse relation between half-life and decay constant.
半衰期 T₁/₂ 是 N = N₀ / 2 的时刻。代入得 ½ = e^(-λ T₁/₂)，取自然对数后得到 T₁/₂ = ln 2 / λ。这表明半衰期与衰变常量成反比。
3. Gravitational Potential Energy and Escape Velocity | 引力势能与逃逸速度
The gravitational force between two point masses M and m separated by distance r is F = G M m / r². To bring m from infinity to a distance r against this force, work must be done.
两个质点 M 和 m 相距 r 时的引力为 F = G M m / r²。要克服此力将 m 从无穷远移动到距离 r 处，需要做功。
The work done by an external agent is W = ∫_{∞}^{r} (G M m / x²) dx = [- G M m / x]_{∞}^{r} = – G M m / r. This work is stored as gravitational potential energy U, so U = – G M m / r. The negative sign indicates a bound system.
外力做功为 W = ∫_{∞}^{r} (G M m / x²) dx = [- G M m / x]_{∞}^{r} = – G M m / r。此功储存为引力势能 U，故 U = – G M m / r。负号表示系统处于束缚态。
For an object to escape a planet’s gravity from its surface (radius R), its kinetic energy must equal the magnitude of the potential energy: ½ m vₑₛ꜀² = G M m / R. Cancelling m and solving gives escape velocity vₑₛ꜀ = √(2 G M / R).
物体要从行星表面（半径 R）脱离引力束缚，其动能必须等于势能的绝对值：½ m vₑₛ꜀² = G M m / R。消去 m 并求解得逃逸速度 vₑₛ꜀ = √(2 G M / R)。
4. Simple Harmonic Motion Displacement Equation | 简谐运动位移方程
Simple harmonic motion (SHM) arises when the restoring force is proportional to the displacement from equilibrium and directed opposite to it: F = – k x. Newton’s second law gives m a = – k x, so a = – (k/m) x.
当回复力与离开平衡位置的位移成正比且方向相反时，便产生简谐运动（SHM）：F = – k x。根据牛顿第二定律，m a = – k x，因此 a = – (k/m) x。
Defining the angular frequency ω = √(k/m), we have a = – ω² x. This is the defining equation of SHM. Its general solution can be written as x = A cos(ω t + φ), where A is the amplitude and φ the phase constant.
定义角频率 ω = √(k/m)，则有 a = – ω² x。这正是简谐运动的定义方程。其通解可写作 x = A cos(ω t + φ)，其中 A 为振幅，φ 为初相位。
Differentiating twice confirms the acceleration: v = dx/dt = – ω A sin(ω t + φ) and a = d²x/dt² = – ω² A cos(ω t + φ) = – ω² x. Thus the motion satisfies the SHM condition.
两次求导可验证加速度：v = dx/dt = – ω A sin(ω t + φ)，a = d²x/dt² = – ω² A cos(ω t + φ) = – ω² x。因此运动满足简谐运动条件。
5. Energy Transformations in Simple Harmonic Motion | 简谐运动中的能量转化
The kinetic energy of a mass–spring system in SHM is Eₖ = ½ m v² = ½ m ω² A² sin²(ω t + φ). The potential energy stored in the spring is Eₚ = ½ k x² = ½ m ω² A² cos²(ω t + φ), using k = m ω².
弹簧振子在做简谐运动时，动能为 Eₖ = ½ m v² = ½ m ω² A² sin²(ω t + φ)。弹簧的弹性势能为 Eₚ = ½ k x² = ½ m ω² A² cos²(ω t + φ)，其中利用了 k = m ω²。
The total mechanical energy is E_total = Eₖ + Eₚ = ½ m ω² A² [sin²(ω t + φ) + cos²(ω t + φ)] = ½ m ω² A². This shows that the total energy is constant and proportional to the square of the amplitude.
系统总机械能为 E_total = Eₖ + Eₚ = ½ m ω² A² [sin²(ω t + φ) + cos²(ω t + φ)] = ½ m ω² A²。可见总能量守恒，且与振幅的平方成正比。
At maximum displacement, energy is entirely potential; at equilibrium, energy is entirely kinetic. The continuous interchange illustrates energy conservation in an isolated oscillator.
在最大位移处，能量全部为势能；在平衡位置，能量全部为动能。这种持续的转化展示了孤立振动系统的能量守恒。
6. Thermal Energy Transfer and Specific Latent Heat | 热能传递与比潜热
When a substance changes temperature without changing phase, the energy transferred ΔQ is related to the temperature change Δθ by ΔQ = m c Δθ, where c is the specific heat capacity. This is a direct consequence of the definition of c.
当物质温度变化而物态不变时，传递的能量 ΔQ 与温度变化 Δθ 的关系为 ΔQ = m c Δθ，其中 c 为比热容。这是比热容定义的直接结果。
During a phase change at constant temperature, the energy supplied goes into breaking intermolecular bonds rather than raising kinetic energy. The energy needed per unit mass is the specific latent heat L, so ΔQ = m L.
在恒定温度下发生相变时，所提供能量用于打破分子间键，而非增加动能。单位质量所需的能量称为比潜热 L，因此 ΔQ = m L。
These relations can be combined with power P = ΔQ / Δt or electrical methods (P = V I) to determine c or L experimentally. In a typical exam problem, students might derive L from a cooling curve or from the gradient of a temperature–time graph.
这些关系式可与功率 P = ΔQ / Δt 或电学方法 (P = V I) 结合，通过实验测定 c 或 L。典型考题中，学生可能需要从冷却曲线或温度–时间图的斜率推导 L。
7. Stellar Luminosity and the Stefan–Boltzmann Law | 恒星光度与斯特藩–玻尔兹曼定律
A star radiates energy from its photosphere, which can be treated as a black body. The Stefan–Boltzmann law states that the power emitted per unit area is J = σ T⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and T is the surface temperature.
恒星从光球层辐射能量，可视作黑体。斯特藩–玻尔兹曼定律表明，单位面积发射功率为 J = σ T⁴，其中 σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴，T 为表面温度。
If the star has radius R, its surface area is 4πR², so its luminosity L (total power output) is L = 4πR² σ T⁴. This equation links measurable quantities and allows astronomers to estimate stellar radii if luminosity and temperature are known.
若恒星半径为 R，表面积为 4πR²，则其光度 L（总输出功率）为 L = 4πR² σ T⁴。该方程将可测量量联系起来，若已知光度和温度，天文学家便可估算恒星半径。
A derivation of peak wavelength from Wien’s displacement law λ_max T = constant is often paired with the Stefan–Boltzmann law in astrophysics questions. No explicit derivation of the T⁴ dependence is required at A-Level, but students should be able to use L ∝ R² T⁴ in proportional reasoning.
维恩位移定律 λ_max T = 常数 常与斯特藩–玻尔兹曼定律一同出现在天体物理题中。A-Level不要求推导 T⁴ 依赖关系，但学生应能运用 L ∝ R² T⁴ 进行比例推理。
8. Nuclear Binding Energy and Mass Defect | 核结合能与质量亏损
Experimental measurements show that the mass of an atomic nucleus is less than the sum of the masses of its separate protons and neutrons. This difference is the mass defect Δm.
实验测量显示，原子核的质量小于其独立质子和中子质量之和。这一差值即为质量亏损 Δm。
According to Einstein’s mass–energy equivalence, the binding energy that holds the nucleus together is E_binding = Δm c², where c is the speed of light in vacuum. This energy represents the work required to separate the nucleus into its individual nucleons.
根据爱因斯坦质能等价关系，束缚原子核的结合能为 E_binding = Δm c²，其中 c 为真空光速。该能量代表将原子核拆分成单个核子所需的功。
To find the binding energy per nucleon, divide E_binding by the mass number A. The curve of binding energy per nucleon against A peaks around iron-56, explaining the stability of nuclei and the release of energy in fusion and fission.
要计算每个核子的结合能，可将 E_binding 除以核子数 A。每个核子的结合能随 A 变化的曲线在铁-56附近达到峰值，这解释了原子核的稳定性以及聚变和裂变中能量的释放。
When a nucleus undergoes decay or reaction, the difference in total binding energy between products and reactants is released as kinetic energy of the products or as photons. This underpins the energy calculations in nuclear physics exam questions.
当原子核发生衰变或反应时，产物与反应物之间总结合能的差值以产物动能或光子形式释放。这是核物理考题中能量计算的基础。
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