A-Level Physics Unit 5 Jan 2019 Mark Scheme: Key Formula Derivations Explained | A-Level物理单元5 2019年1月评分方案：关键公式推导解析

📚 A-Level Physics Unit 5 Jan 2019 Mark Scheme: Key Formula Derivations Explained | A-Level物理单元5 2019年1月评分方案：关键公式推导解析

The January 2019 Edexcel A2 Physics Unit 5 (WPH05) examination rewarded students who could demonstrate rigorous formula derivations from first principles. This article revisits the most important derivations that feature in the mark scheme, explaining the logical steps, assumptions and final equations that A-Level learners must master for thermal physics, nuclear processes and astrophysics. Understanding these derivations not only secures marks in similar future questions but also builds a deeper intuition for the physical laws at work.

2019年1月爱德思A2物理单元5（WPH05）考试奖励了那些能够从基本原理出发展示严谨公式推导的考生。本文重温评分方案中最关键的推导，解释其中的逻辑步骤、假设与最终方程，这些都是A-Level学生必须掌握的热力学、核过程与天体物理内容。理解这些推导不仅能确保在类似试题中获得分数，也能为掌握背后的物理规律建立更深的直觉。

1. Deriving the Ideal Gas Pressure Equation | 推导理想气体压强公式

Consider a single particle of mass m moving with speed c inside a cubic container of side L. Its velocity component perpendicular to one wall is v_x. Between successive collisions with that wall the particle travels a distance 2L, taking time Δt = 2L / v_x.

考虑单个质量为 m、速率为 c 的粒子在边长为 L 的立方容器中运动。其垂直于某一壁面的速度分量为 v_x。两次连续碰撞之间，该粒子运动距离 2L，用时 Δt = 2L / v_x。

Each collision reverses the perpendicular component, giving a momentum change of 2mv_x. The average force exerted by this one particle on the wall is therefore F = (change in momentum) / (time between collisions) = (2mv_x) / (2L / v_x) = mv_x² / L.

每次碰撞使垂直分量反向，产生动量变化 2mv_x。因此这一个粒子对壁面施加的平均力为 F = (动量变化) / (碰撞时间间隔) = (2mv_x) / (2L / v_x) = mv_x² / L。

Summing over all N particles, the total force on the wall is F_total = (m / L) Σ v_x,i². The mean square velocity component ⟨v_x²⟩ is (1/N) Σ v_x,i², and because motion is random all three directions are equivalent: ⟨v_x²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = (1/3) ⟨c²⟩, where c is the speed of a particle.

对所有 N 个粒子求和，壁面受到的总力为 F_total = (m / L) Σ v_x,i²。速度分量平方平均值 ⟨v_x²⟩ = (1/N) Σ v_x,i²，由于运动是随机的，三个方向等效：⟨v_x²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = (1/3) ⟨c²⟩，其中 c 为粒子的速率。

Pressure p is force per unit area on the wall of area L²: p = F_total / L² = (m / L³) N ⟨v_x²⟩ = (m / V) N (1/3) ⟨c²⟩, where V = L³. This gives the celebrated result:

压强 p 是单位面积壁面上的力，壁面积为 L²：p = F_total / L² = (m / L³) N ⟨v_x²⟩ = (m / V) N (1/3) ⟨c²⟩，其中 V = L³。由此得到著名的结果：

pV = (1/3) N m ⟨c²⟩

In the January 2019 mark scheme, this step-by-step kinetic theory derivation was essential to secure full marks.

在2019年1月的评分方案中，这种逐步进行的气体动理论推导对获得满分至关重要。

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2. Connecting Mean Kinetic Energy to Absolute Temperature | 将平均动能与绝对温度联系起来

From the ideal gas equation pV = nRT and pV = (1/3) N m ⟨c²⟩, equate the right-hand sides. Since N = nN_A, where N_A is the Avogadro constant, we have (1/3) n N_A m ⟨c²⟩ = nRT.

从理想气体状态方程 pV = nRT 以及 pV = (1/3) N m ⟨c²⟩ 出发，使右边相等。因为 N = nN_A（N_A 为阿伏伽德罗常数），可得 (1/3) n N_A m ⟨c²⟩ = nRT。

Cancel n and multiply both sides by 3/2: (1/2) m ⟨c²⟩ = (3/2) (R / N_A) T. The ratio R/N_A is the Boltzmann constant k. Thus the average translational kinetic energy of a molecule is directly proportional to the absolute temperature:

约去 n 并将两边乘以 3/2：(1/2) m ⟨c²⟩ = (3/2) (R / N_A) T。比值 R/N_A 就是玻尔兹曼常量 k。因此一个分子的平均平动动能与绝对温度成正比：

(1/2) m ⟨c²⟩ = (3/2) kT

This derivation shows why temperature is a measure of the random kinetic energy of particles. The Jan 2019 paper expected candidates to link this to the kinetic model of gases.

这个推导表明温度是粒子无规动能的量度。2019年1月试卷期望考生将此与气体动理论模型联系起来。

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3. Radioactive Decay Law: Exponential Derivation | 放射性衰变定律的指数推导

Radioactive decay is a random process. The activity A is the number of nuclei decaying per unit time, which is proportional to the number N of undecayed nuclei: A = –dN/dt = λ N, where λ is the decay constant.

放射性衰变是一种随机过程。活度 A 是单位时间内衰变的原子核数，它与未衰变的原子核数 N 成正比：A = –dN/dt = λ N，λ 是衰变常量。

Separating variables: dN / N = –λ dt. Integrating both sides from N₀ (at t = 0) to N (at time t) gives ln(N) – ln(N₀) = –λ t.

分离变量：dN / N = –λ dt。从 N₀（t=0 时）到 N（t 时刻）积分两边，得到 ln(N) – ln(N₀) = –λ t。

Taking exponentials yields the well-known exponential decay law:

取指数得到熟知的指数衰变定律：

N = N₀ e^{–λ t}

The January 2019 mark scheme gave credit for correctly setting up the differential equation and showing the integration steps.

2019年1月评分方案对正确设立微分方程并展示积分步骤给予分数。

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4. Half-Life and Its Relationship to Decay Constant | 半衰期与衰变常数的关系

The half-life T_½ is the time after which half the original nuclei remain: N = N₀/2. Substituting into the decay law: N₀/2 = N₀ e^{–λ T_½}.

半衰期 T_½ 是原有原子核剩下半数所经历的时间：N = N₀/2。代入衰变定律：N₀/2 = N₀ e^{–λ T_½}。

Cancel N₀ and take natural logarithms: ln(1/2) = –λ T_½. Since ln(1/2) = –ln 2, we obtain:

约去 N₀ 并取自然对数：ln(1/2) = –λ T_½。因为 ln(1/2) = –ln 2，我们得到：

T_½ = ln 2 / λ

This simple derivation reinforces the inverse relationship between half-life and decay constant. The Jan 2019 Unit 5 paper frequently required students to calculate one from the other.

这个简单的推导强化了半衰期与衰变常量的反比关系。2019年1月单元5试卷经常要求学生由其中一个量计算另一个量。

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5. Mass-Energy Equivalence in Nuclear Reactions | 核反应中的质能等价

Einstein’s mass–energy relation states that a mass Δm is equivalent to an energy E = Δm c². In nuclear physics, the mass defect of a nucleus is the difference between the total mass of its separate nucleons and the mass of the nucleus itself. This lost mass is released as binding energy.

爱因斯坦的质能关系表明质量 Δm 等效于能量 E = Δm c²。在核物理中，一个原子核的质量亏损是其独立核子的总质量与原子核本身质量之差。这些损失的质量以结合能的形式释放。

For a reaction a + X → Y + b, the Q-value is Q = (m_a + m_X – m_Y – m_b) c². If Q > 0 the reaction is exothermic; if Q < 0 it is endothermic. The Jan 2019 mark scheme rewarded careful conversion between atomic mass units (u) and MeV: 1 u = 931.5 MeV/c².

对于反应 a + X → Y + b，Q 值为 Q = (m_a + m_X – m_Y – m_b) c²。若 Q > 0 则反应放热；若 Q < 0 则吸热。2019年1月评分方案奖励准确使用原子质量单位 (u) 与 MeV 之间的转换：1 u = 931.5 MeV/c²。

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6. Binding Energy and Mass Defect Calculation | 结合能与质量亏损计算

The binding energy B of a nucleus with Z protons and N neutrons is B = (Z m_p + N m_n – m_nucleus) c². To find the binding energy per nucleon, divide by the mass number A = Z + N.

一个含 Z 个质子和 N 个中子的原子核的结合能为 B = (Z m_p + N m_n – m_nucleus) c²。要计算比结合能，除以质量数 A = Z + N。

In many exam problems, masses are given in atomic mass units and the binding energy must be expressed in MeV. The conversion factor 931.5 MeV/u is used. A graph of binding energy per nucleon against mass number shows a maximum around iron-56, illustrating stability. The Jan 2019 paper tested this understanding in the context of fusion and fission.

在许多考题中，质量以原子质量单位给出，结合能需以 MeV 表达，用到转换因子 931.5 MeV/u。比结合能随质量数变化的曲线在铁-56附近出现最大值，体现了稳定性。2019年1月的试卷在聚变与裂变的情境中考查了这一理解。

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7. Stellar Evolution: Deriving Luminosity from Stefan-Boltzmann | 恒星演化：从斯特藩-玻尔兹曼推导光度

A star can be treated approximately as a black body. The power radiated per unit area from its surface is given by the Stefan-Boltzmann law: F = σ T⁴, where σ = 5.67 × 10^–8 W m^–2 K^–4.

恒星可近似视为黑体。其表面单位面积辐射的功率由斯特藩-玻尔兹曼定律给出：F = σ T⁴，其中 σ = 5.67 × 10^–8 W m^–2 K^–4。

The total luminosity L is the power radiated through the whole surface area of the star. For a sphere of radius R:

总光度 L 是恒星整个表面积辐射的功率。对于一个半径为 R 的球体：

L = 4πR² σ T⁴

In the Jan 2019 Unit 5 paper, this relation was used to compare two stars on the Hertzsprung-Russell diagram, requiring rearrangement to find radius or temperature. Students had to derive one quantity from the other with given data.

在2019年1月单元5试卷中，该关系被用于比较赫罗图上的两颗恒星，需要通过移项求半径或温度。学生需要由给定数据推导出一个量。

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8. Hubble’s Law and the Expanding Universe | 哈勃定律与膨胀宇宙

Hubble’s law states that the recessional speed v of a galaxy is proportional to its distance d: v = H₀ d, where H₀ is the Hubble constant. This empirical law was used in the Jan 2019 exam to estimate the age of the Universe.

哈勃定律指出星系的退行速度 v 与其距离 d 成正比：v = H₀ d，H₀ 为哈勃常数。这一经验规律在2019年1月考试中被用于估算宇宙的年龄。

If the expansion rate has been constant, the time since the Big Bang is approximately t = d / v = 1 / H₀. Converting H₀ to SI units (e.g. 2.2 × 10^–18 s^–1) yields an estimate of about 14 billion years, though precise values depend on cosmological models.

若膨胀速率恒定，大爆炸以来经历的时间近似为 t = d / v = 1 / H₀。将 H₀ 转换为国际单位（例如 2.2 × 10^–18 s^–1）得到约140亿年的估算值，但准确数值依赖于宇宙学模型。

The derivation t = 1/H₀ was a favourite in mark schemes: candidates had to show that v = d/t, equate with H₀d, cancel d and rearrange.

推导 t = 1/H₀ 是评分方案中的常客：考生须写出 v = d/t，与 H₀d 相等，约去 d 并整理。

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9. Gravitational Potential in Radial Fields | 径向场中的引力势

Gravitational potential V at a point is defined as the work done per unit mass to bring a test mass from infinity to that point. The gravitational force is F = –GMm / r², so the work done against this force is W = ∫_∞^r (GMm / r²) dr.

引力势 V 定义为将单位质量的检验物体从无穷远处移至该点所做的功。引力为 F = –GMm / r²，因此克服该力所做的功为 W = ∫_∞^r (GMm / r²) dr。

Evaluating the integral: W = m [ –GM / r ]_∞^r = –m GM / r – 0 = –GMm / r. Dividing by the test mass m gives the potential:

计算积分：W = m [ –GM / r ]_∞^r = –m GM / r – 0 = –GMm / r。除以检验质量 m 得到势：

V = –GM / r

The January 2019 mark scheme expected students to recognise that potential is negative because work is done by the field as masses approach, and that equipotential surfaces are spheres. This derivation was fundamental to understanding satellite motion and escape velocity.

2019年1月评分方案期望学生认识到势为负值是因为质量靠近时场做正功，且等势面为球面。这一推导是理解卫星运动与逃逸速度的基础。

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10. Escape Velocity: A Derivation from Energy Conservation | 逃逸速度：能量守恒推导

To escape a planet’s gravitational pull, a projectile must have enough kinetic energy to reach infinity where the gravitational potential energy becomes zero. Using energy conservation at the surface (radius R):

要摆脱行星的引力束缚，抛射体必须具有足够的动能以到达无穷远处，那里的引力势能为零。在表面（半径为 R）使用能量守恒：

½ m v_esc² + ( – GMm / R ) = 0

Here, the total mechanical energy at infinity is taken as zero. Solve for v_esc:

这里无穷远处的总机械能被取为零。解出 v_esc：

v_esc = √(2GM / R)

This expression does not depend on the mass of the escaping object, only on the planet’s mass and radius. In the Jan 2019 Unit 5 paper, candidates were required to derive this equation and then calculate escape speed for a given astronomical body.

该表达式与逃脱物体的质量无关，仅取决于行星的质量和半径。在2019年1月单元5试卷中，要求考生推导该方程并计算特定天体的逃逸速度。

The derivation illustrates the close link between gravitational potential and kinetic energy. Students who could clearly set up the initial and final energy balance secured the marks allocated in the mark scheme.

该推导展示了引力势与动能之间的紧密联系。能够清晰设立初态与末态能量平衡的学生获得了评分方案中所分配的分数。

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