A-Level Physics Unit 5 Formula Derivations (Jan20 Mark Scheme) | A-Level物理第五单元公式推导（Jan20评分方案）

📚 A-Level Physics Unit 5 Formula Derivations (Jan20 Mark Scheme) | A-Level物理第五单元公式推导（Jan20评分方案）

In A-Level Physics Unit 5 examinations, especially papers like the January 2020 session, mark schemes consistently reward clear, logical derivations of fundamental formulas. Students must show step‑by‑step mathematical reasoning, starting from basic definitions or laws. This article revisits the key derivations commonly examined, providing paired English–Chinese explanations that mirror the structure examiners expect. By mastering these derivations, you can confidently secure the method marks that often make the difference between grades.

在A-Level物理第五单元的考试中，尤其是像2020年1月那样的试卷，评分方案始终奖励对基本公式清晰、有逻辑的推导。考生必须展示从基本定义或定律出发、逐步推进的数学推理。本文重访常考的关键推导，提供与评分方案思路一致的中英文对照解释。掌握这些推导，你就能稳稳拿到那些往往决定等级差别的过程分。

1. Deriving the Exponential Decay Law | 推导指数衰减定律

Radioactive decay is a random process governed by the differential equation dN/dt = -λ N, where N is the number of undecayed nuclei, t is time, and λ is the decay constant. The negative sign indicates that the number of nuclei decreases with time.

放射性衰变是一个随机过程，由微分方程 dN/dt = -λ N 支配，其中 N 是未衰变原子核的数量，t 是时间，λ 是衰变常数。负号表示原子核数随时间减少。

Separate the variables: dN/N = -λ dt. Integrate both sides: ∫ dN/N = -λ ∫ dt, giving ln N = -λ t + C, where C is the constant of integration.

分离变量：dN/N = -λ dt。两边积分：∫ dN/N = -λ ∫ dt，得到 ln N = -λ t + C，其中 C 是积分常数。

At t = 0, N = N₀ (the initial number of nuclei). Substituting gives ln N₀ = C. Hence ln N = -λ t + ln N₀, which can be rewritten as ln(N/N₀) = -λ t.

当 t = 0 时，N = N₀（初始原子核数）。代入得 ln N₀ = C。因此 ln N = -λ t + ln N₀，可改写为 ln(N/N₀) = -λ t。

N = N₀ e⁻ λ t

This is the exponential decay law. The activity A = -dN/dt = λ N, so A also decreases exponentially: A = A₀ e⁻ λ t, where A₀ = λ N₀.

这就是指数衰减定律。活度 A = -dN/dt = λ N，因此 A 也呈指数衰减：A = A₀ e⁻ λ t，其中 A₀ = λ N₀。

2. Relationship Between Half-Life and Decay Constant | 半衰期与衰变常数的关系

The half‑life T₁/₂ is the time taken for the number of undecayed nuclei (or the activity) to fall to half its original value. Set N = N₀/2 at t = T₁/₂ in the decay law.

半衰期 T₁/₂ 是未衰变原子核数（或活度）降为原来一半所需的时间。在衰变定律中代入 t = T₁/₂ 时 N = N₀/2。

N₀/2 = N₀ e⁻ λ T₁/₂ → 1/2 = e⁻ λ T₁/₂

Take natural logarithms: ln(1/2) = -λ T₁/₂. Since ln(1/2) = -ln 2, we obtain -ln 2 = -λ T₁/₂.

取自然对数：ln(1/2) = -λ T₁/₂。因为 ln(1/2) = -ln 2，我们得到 -ln 2 = -λ T₁/₂。

T₁/₂ = ln 2 / λ

This relation is frequently used to determine λ from a measured half‑life. It also shows that a large decay constant corresponds to a short half‑life.

这个关系常用来从测得的半衰期求 λ。它也表明衰变常数大则半衰期短。

3. Mass Defect and Binding Energy | 质量亏损与结合能

The mass of a nucleus is always less than the sum of the masses of its individual protons and neutrons. This difference is the mass defect Δm.

原子核的质量总是小于其各质子与中子质量之和。这个差值就是质量亏损 Δm。

Δm = Z mₚ + (A – Z) mₙ – mₙᵤ꜀ₗₑᵤₛ

Here Z is the proton number, A the mass number, mₚ the proton mass, mₙ the neutron mass, and mₙᵤ꜀ₗₑᵤₛ the nuclear mass. Using atomic masses, electron masses must be accounted for carefully.

此处 Z 是质子数，A 是质量数，mₚ 是质子质量，mₙ 是中子质量，mₙᵤ꜀ₗₑᵤₛ 是核的质量。若使用原子质量，则须谨慎考虑电子质量。

Binding energy is the energy equivalent of the mass defect: E_b = Δm c². In nuclear physics, it is convenient to use unified atomic mass units: 1 u = 1.66 × 10⁻²⁷ kg, and 1 u c² = 931.5 MeV.

结合能是质量亏损的能量当量：E_b = Δm c²。在核物理中，使用原子质量单位为便：1 u = 1.66 × 10⁻²⁷ kg，且 1 u c² = 931.5 MeV。

To calculate binding energy in MeV, find Δm in atomic mass units and multiply by 931.5. Binding energy per nucleon is E_b / A, a measure of nuclear stability.

要计算以 MeV 为单位的结合能，先求得以原子质量单位表示的 Δm，再乘以 931.5。每个核子的结合能为 E_b / A，这是核稳定性的量度。

4. Energy Released in Nuclear Reactions | 核反应中的能量释放

In both fission and fusion, the total mass of the products is less than the total mass of the reactants. The energy released Q is given by Q = (Δm) c², where Δm = total initial mass – total final mass.

在裂变和聚变中，产物的总质量小于反应物的总质量。释放的能量 Q 由 Q = (Δm) c² 给出，其中 Δm = 初态总质量 – 末态总质量。

For example, in a typical fission reaction of uranium‑235 induced by a neutron:

²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3¹n

Using precise atomic masses, compute the mass difference. Then convert to MeV using 1 u ≡ 931.5 MeV. The energy appears mainly as kinetic energy of the product nuclei and neutrons.

使用精确的原子质量计算质量差，再利用 1 u ≡ 931.5 MeV 转换为能量。能量主要以产物核与中子的动能形式出现。

Mark schemes expect candidates to show the mass calculation clearly, to state the mass difference, and to apply the conversion factor accurately.

评分方案期望考生清晰地列出质量计算、写明质量差并准确运用转换因子。

5. Assumptions of the Kinetic Theory of Gases | 分子动理论的假设

To derive the pressure of an ideal gas, the kinetic theory makes several simplifying assumptions about the gas molecules.

为推导理想气体的压强，分子动理论对气体分子作出若干简化假设。

The gas consists of a very large number of identical molecules, each moving in rapid, random motion.

气体由极大量完全相同的分子组成，每个分子都在作快速而杂乱无章的运动。
The volume of the molecules themselves is negligible compared with the volume of the container.

分子本身的体积与容器的容积相比可忽略不计。
Collisions between molecules and with the walls are perfectly elastic.

分子间的碰撞以及分子与器壁的碰撞都是完全弹性的。
No intermolecular forces act except during collisions; between collisions, molecules move in straight lines at constant speed.

除碰撞瞬间外，分子之间无力作用；在两次碰撞之间，分子沿直线匀速运动。
The duration of a collision is negligible compared to the time between collisions, so the impulse is instantaneous.

碰撞持续时间与两次碰撞之间的时间相比可忽略不计，因此冲量可视为瞬时作用。
Newtonian mechanics applies to the motion of the molecules.

分子的运动遵从牛顿力学。

These assumptions allow us to focus on momentum changes at the walls, leading to a simple expression for pressure.

这些假设使我们能够专注于器壁处的动量变化，从而导出压强的简洁表达式。

6. Derivation of Pressure of an Ideal Gas | 理想气体压强公式推导

Consider a single molecule of mass m moving with velocity component vₓ perpendicular to one wall of a cubic container of side L. When it collides elastically with the wall, its velocity reverses from +vₓ to -vₓ, so the change in momentum is Δp = -2 m vₓ. The magnitude of change transferred to the wall is 2 m vₓ.

考虑一个质量为 m 的单一分子，以垂直于边长为 L 的立方形容器某一壁的速度分量 vₓ 运动。当它与器壁发生弹性碰撞时，其速度从 +vₓ 反向为 -vₓ，故动量变化量为 Δp = -2 m vₓ。传递给器壁的动量大小为 2 m vₓ。

The time between successive collisions on the same wall is the round‑trip time: Δt = 2L / |vₓ|. Hence, the average force exerted by one molecule on that wall is F = Δp / Δt = (2 m vₓ) / (2L/vₓ) = m vₓ² / L.

同一分子连续撞击同一器壁的时间间隔为往复时间：Δt = 2L / |vₓ|。因此，单个分子对该器壁施加的平均力为 F = Δp / Δt = (2 m vₓ) / (2L/vₓ) = m vₓ² / L。

Pressure p is force per unit area. The area of the wall is L², so p from one molecule = (m vₓ² / L) / L² = m vₓ² / L³ = m vₓ² / V, where V = L³ is the volume of the cube.

压强 p 是单位面积上的力。器壁面积为 L²，故由单个分子产生的压强为 p = (m vₓ² / L) / L² = m vₓ² / L³ = m vₓ² / V，其中 V = L³ 是立方体的体积。

For N molecules, the total pressure is p = (m / V) Σ vₓ,ᵢ² = N m ⟨vₓ²⟩ / V, where ⟨vₓ²⟩ is the mean square velocity component in the x‑direction.

对 N 个分子，总压强为 p = (m / V) Σ vₓ,ᵢ² = N m ⟨vₓ²⟩ / V，其中 ⟨vₓ²⟩ 是 x 方向上的均方速度分量。

Due to random motion, ⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨vᶻ²⟩ = (1/3) ⟨v²⟩, where ⟨v²⟩ is the mean square speed. Therefore, p = (N m / V) × (1/3) ⟨v²⟩.

由于运动的无规性，⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨vᶻ²⟩ = (1/3) ⟨v²⟩，其中 ⟨v²⟩ 是均方速率。因此，p = (N m / V) × (1/3) ⟨v²⟩。

pV = ⅓ N m ⟨v²⟩

Defining the root‑mean‑square speed cᵣₘₛ = √⟨v²⟩, we obtain the standard kinetic theory equation:

定义方均根速率 cᵣₘₛ = √⟨v²⟩，即得标准的分子动理论方程：

pV = ⅓ N m cᵣₘₛ²

7. Relating Absolute Temperature to Mean Kinetic Energy | 联系绝对温度与平均动能

From the ideal gas equation, pV = nRT = N k T, where n is the number of moles, R is the molar gas constant, N is the number of molecules, and k is Boltzmann’s constant (k = R / Nₐ).

由理想气体状态方程，pV = nRT = N k T，其中 n 是摩尔数，R 是摩尔气体常数，N 是分子数，k 是玻尔兹曼常数（k = R / Nₐ）。

Equate this with the kinetic theory result: N k T = ⅓ N m cᵣₘₛ². Cancel N (for N > 0) and multiply both sides by 3/2:

将此式与分子动理论结果联立：N k T = ⅓ N m cᵣₘₛ²。消去 N（N > 0）并两边乘以 3/2：

½ m cᵣₘₛ² = (3/2) k T

The left‑hand side is the mean translational kinetic energy of a single molecule. Thus, the absolute temperature of an ideal gas is directly proportional to the average kinetic energy of its molecules.

等式左边是单个分子的平均平动动能。因此，理想气体的绝对温度与其分子的平均动能成正比。

This result links the macroscopic quantity temperature to the microscopic motion of particles. It also explains the concept of absolute zero, where particle kinetic energy would theoretically become zero.

这一结果将宏观量温度与微观粒子运动联系起来，也解释了绝对零度的概念——理论上此时粒子的动能将变为零。

8. Deriving the Gas Laws from Kinetic Theory | 从分子动理论推导气体定律

Using pV = ⅓ N m cᵣₘₛ² and T ∝ ⟨Eₖ⟩ = ½ m cᵣₘₛ², we can explain the empirical gas laws.

利用 pV = ⅓ N m cᵣₘₛ² 和 T ∝ ⟨Eₖ⟩ = ½ m cᵣₘₛ²，我们可以解释经验气体定律。

Boyle’s law (T constant): For a fixed mass of gas at constant temperature, N, m and cᵣₘₛ remain constant. Therefore pV = constant, or p ∝ 1/V.

玻意耳定律（T 恒定）：对一定质量的气体，在温度不变时，N、m 和 cᵣₘₛ 保持不变。因此 pV = 常数，即 p ∝ 1/V。
Charles’s law (p constant): At constant pressure, pV = constant × T, so V ∝ T, provided N and m are fixed. (Since T ∝ cᵣₘₛ², a rise in T increases cᵣₘₛ² and thus V must increase to keep p unchanged.)

查理定律（p 恒定）：在压强不变时，pV = 常数 × T，因此 V ∝ T，前提是 N 和 m 固定。（因 T ∝ cᵣₘₛ²，T 升高会增大 cᵣₘₛ²，故 V 必须增大以保持 p 不变。）
Pressure law (V constant): For a fixed volume, pV = constant × T, so p ∝ T.

压强定律（V 恒定）：体积固定时，pV = 常数 × T，因此 p ∝ T。

These derivations demonstrate the power of the kinetic model in unifying macroscopic observations with microscopic behaviour.

这些推导展示了分子动模型在统一宏观观测与微观行为方面的力量。

9. SHM Displacement, Velocity and Acceleration Equations | 简谐运动的位移、速度和加速度方程

Simple harmonic motion (SHM) is defined by a restoring force proportional to displacement: F = -k x. Applying Newton’s second law gives a = – (k/m) x. Defining ω² = k/m, we obtain the fundamental differential equation:

简谐运动由与位移成正比的回复力定义：F = -k x。应用牛顿第二定律得 a = – (k/m) x。定义 ω² = k/m，即得基本微分方程：

a = – ω² x or d²x/dt² = – ω² x

A solution to this equation is sinusoidal: x = A sin(ω t) or x = A cos(ω t), where A is the amplitude. The choice depends on initial conditions. If at t = 0, x = 0 and velocity is positive, we use x = A sin(ω t).

该方程的一个解为正弦函数：x = A sin(ω t) 或 x = A cos(ω t)，其中 A 是振幅。选择取决于初始条件。若 t = 0 时 x = 0 且速度为正，则使用 x = A sin(ω t)。

Velocity is the first derivative: v = dx/dt = A ω cos(ω t) or, using the cosine solution, v = -A ω sin(ω t). The maximum speed is v_max = ω A.

速度为一阶导数：v = dx/dt = A ω cos(ω t)，或对余弦解 v = -A ω sin(ω t)。最大速率为 v_max = ω A。

Acceleration is the second derivative: a = d²x/dt² = -A ω² sin(ω t) = -ω² x, which satisfies the defining equation. The maximum acceleration is a_max = ω² A.

加速度为二阶导数：a = d²x/dt² = -A ω² sin(ω t) = -ω² x，满足定义方程。最大加速度为 a_max = ω² A。

These relationships are summarised in mark schemes using the standard forms and are essential for solving pendulum or mass‑spring problems.

评分方案中使用标准形式总结这些关系，它们对解决摆或质块‑弹簧问题至关重要。

10. Energy in Simple Harmonic Motion | 简谐运动中的能量

For a mass‑spring system in SHM, the total mechanical energy is constant and interchanges between kinetic energy (K) and potential energy (U).

对处于简谐运动的质块‑弹簧系统，总机械能守恒，在动能 (K) 和势能 (U) 之间相互转换。

K = ½ m v² = ½ m ω² (A² – x²)

U = ½ k x² = ½ m ω² x²

The total energy E_total = K + U = ½ m ω² A² = ½ k A², which is independent of displacement. This expression is often required in Unit 5 questions to find the amplitude from given energy values.

总能量 E_total = K + U = ½ m ω² A² = ½ k A²，与位移无关。第五单元的问题常要求在给定能量值的情况下用此式求振幅。

Similarly, for a simple pendulum, potential energy is gravitational, but the same principle of energy conservation applies, with maximum speed at the equilibrium position and maximum potential energy at the extremes.

类似地，对单摆，势能为重力势能，但能量守恒原理相同：平衡位置速率最大，极端点势能最大。

Mark schemes award marks for stating either the kinetic or potential energy formula correctly and for equating total energy to ½ k A² at the point of maximum displacement.

评分方案对正确写出动能或

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A-Level Physics Unit 5 Formula Derivations (Jan20 Mark Scheme) | A-Level物理第五单元公式推导（Jan20评分方案）