📚 Deriving Key Formulas from the PH05 International Physics Insert (Jan 2023) | 推导 PH05 国际物理插入页公式（2023年1月）
The Insert provided in the PH05 International Physics examination (January 2023) contains a wealth of equations, constants and data needed to tackle Unit 5 topics. Mastering the derivations behind these formulas not only helps you recall them accurately but also reveals the deep connections between the thermal, nuclear and oscillatory physics that shape our understanding of the universe. This article guides you through the step‑by‑step derivations of the most important relationships that appear in the Insert, using clear reasoning and accessible mathematics.
2023年1月PH05国际物理考试的插入页提供了丰富公式、常数和数据，用于解决第五单元的问题。掌握这些公式背后的推导，不仅能帮助你准确记忆它们，还能揭示热学、核物理和振动学之间的深层联系，这些联系构建了我们对宇宙的理解。本文将逐步推导插入页中最重要的一些关系式，采用清晰的推理和易懂的数学。
1. The Ideal Gas Equation and Its Constants | 理想气体方程及其常数
molecules, N, divided by the Avogadro constant Nₐ: n = N / Nₐ. Substituting into pV = nRT gives pV = (N / Nₐ) RT. Rearranging, pV = N (R / Nₐ) T. The ratio R / Nₐ is defined as the Boltzmann constant k. Therefore, we obtain pV = NkT and the essential relationship R = kNₐ. This shows that the molar gas constant is simply the gas constant per particle scaled up by one mole of particles.
插入页中的理想气体定律以 pV = nRT 和 pV = NkT 两种形式给出。为连接它们，考虑物质的量 n 等于气体分子总数 N 除以阿伏伽德罗常数 Nₐ：n = N / Nₐ。代入 pV = nRT 得到 pV = (N / Nₐ) RT。整理得 pV = N (R / Nₐ) T。比值 R / Nₐ 即定义为玻尔兹曼常数 k。因此，我们得到 pV = NkT 以及重要关系 R = kNₐ。这说明摩尔气体常数正是每粒子气体常数放大一摩尔粒子的结果。
2. Average Kinetic Energy of Gas Molecules | 气体分子的平均动能
theory model, the pressure exerted by an ideal gas on its container is given by pV = ⅓ N m ⟨c²⟩, where m is the mass of a single molecule and ⟨c²⟩ is the mean square speed. Equate this to the ideal gas equation pV = NkT: ⅓ N m ⟨c²⟩ = NkT. Cancel N and multiply through by 3 to get m ⟨c²⟩ = 3kT. Since the average translational kinetic energy of a molecule is Ek = ½ m ⟨c²⟩, substituting gives ½ m ⟨c²⟩ ctly links the microscopic motion of particles to the macroscopic property of temperature.
由分子动理论模型，理想气体对容器壁施加的压力可用 pV = ⅓ N m ⟨c²⟩ 表示，其中 m 是单个分子质量，⟨c²⟩ 是方均速率。令该式与理想气体方程 pV = NkT 相等：⅓ N m ⟨c²⟩ = NkT。消去 N 并乘以 3 得到 m ⟨c²⟩ = 3kT。由于分子的平均平动动能为 Ek = ½ m ⟨c²⟩，代入即得 ½ m ⟨c²⟩ = ³⁄₂ kT。这一优美结果将粒子的微观运动与温度的宏观性质直接联系起来。
3. Energy Changes with Heat: Specific and Latent | 热量引起的能量变化：比热容与潜热
e the temperature of 1 kg of a substance by 1 K. For a mass m undergoing a temperature change Δθ, the total energy transferred as heat is simply Q = mcΔθ. The second equation describes the energy needed to change the state of a mass m without a temperature change; L is the specific latent heat. Both are empirical relationships that, when combined with the first law of thermodynamics ΔU = Q − W, allow calculation of internal energy changes in thermal processes.
插入页中包含 Q = mcΔθ 和 Q = mL 两个方程。前者源自比热容 c 的定义：1 kg 物质温度升高 1 K 所需能量。对于质量 m、温度变化 Δθ 的过程，以热量形式传递的总能量即为 Q = mcΔθ。第二个方程描述了无温度变化时使质量 m 改变状态所需的能量；L 是比潜热。两者均为经验关系，与热力学第一定律 ΔU = Q − W 结合后，可用于计算热学过程中内能的变化。
4. The Law of Radioactive Decay | 放射性衰变定律
nstable nuclei present at time t, the rate of change of N is proportional to N: dN/dt = −λN. This differential equation can be solved by separation of variables: (1/N) dN = −λ dt. Integrating both sides gives ln N = −λt + constant. Exponentiating yields N = N₀ e−λt 0. This exponential decay law is a direct mathematical consequence of the constant probability assumption.
放射性衰变是一种随机过程，单个核子每单位时间衰变的概率为衰变常数 λ。若 N 为 t 时刻存在的未衰变核子数，则 N 的变化率正比于 N：dN/dt = −λN。这个微分方程可通过分离变量法求解：(1/N) dN = −λ dt。两边积分得 ln N = −λt + 常数。取指数得到 N = N₀ e−λt，其中 N₀ 是 t = 0 时的核子数。这个指数衰变定律是恒定概率假设的直接数学结果。
5. Half-Life and the Decay Constant | 半衰期与衰变常数
The half‑ r half of the original radioactive nuclei to decay. Set N = N₀/2 in the decay law N = N₀ e−λt: N₀/2 = N₀ e−λ T₁/₂ nce ln(½) = −ln 2, we obtain −ln 2 = −λ T₁/₂, so T₁/₂ = ln 2 / λ. This shows that the half‑life and the decay constant are inversely proportional. A larger λ means a shorter half‑life and a more rapid decay.
半衰期 T₁/₂ 定义为原有放射性核子衰变一半所需的时间。令 N = N₀/2 代入衰变定律 N = N₀ e−λt：N₀/2 = N₀ e−λ T₁/₂。消去 N₀ 并取自然对数：ln(½) = −λ T₁/₂。由于 ln(½) = −ln 2，得到 −ln 2 = −λ T₁/₂，所以 T₁/₂ = ln 2 / λ。这表明半衰期与衰变常数成反比。λ 越大意味着半衰期越短，衰变得越快。
6. Activity of a Radioactive Sample | 放射性样品的活度
r of decays per unit time, which is equal to the magnitude of the rate of change of N: A = |dN/dt|. From the decay law, dN/dt = −λN, so A = λN. Substituting N = N₀ e−λt gives A = λN₀ e−λt. Since λN₀ is the initial activity A₀, we can also write A = A₀ e−λt. Therefore, activity follows the same exponential decay pattern as the number of parent nuclei, a fact frequently used in radiometric dating.
活度 A 定义为单位时间内的衰变次数，等于 N 的变化率的绝对值：A = |dN/dt|。由衰变定律 dN/dt = −λN，可得 A = λN。代入 N = N₀ e−λt 给出 A = λN₀ e−λt。由于 λN₀ 即为初始活度 A₀，我们也可以写成 A = A₀ e−λt。可见，活度与母核数量遵循相同的指数衰减规律，这一事实常用于放射性定年。
7. Mass Defect and Binding Energy | 质量亏损与结合能
a nucleus is understood through the concept of binding energy, which derives from Einstein's mass– m = Z mₚ + N mₙ − M_nucleus, where Z is the proton number, N the neutron number, mₚ the proton mass and mₙ the neutron mass. The binding energy Eb that holds the nucleus together is then Eb = Δm c². Dividing Eb Z + N gives the average binding energy per nucleon, a key indicator of nuclear stability.
原子核的稳定性可通过结合能的概念来理解，这源自爱因斯坦的质能等价关系 E = mc²。一个原子核的质量总是小于其各个独立质子和中子质量之和。这个质量亏损为 Δm = Z mₚ + N mₙ − M_nucleus，其中 Z 为质子数，N 为中子数，mₚ 为质子质量，mₙ 为中子质量。将核子聚合在一起的结合能 Eb 则为 Eb = Δm c²。将 Eb 除以核子数 A = Z + N 即得到每个核子的平均结合能，这是衡量核稳定性的关键指标。
8. Equations of Simple Harmonic Motion | 简谐运动方程
force is proportional to displacement from equilibrium and directed towards it. For a mass‑spring system, Hooke's law gives F = −kx. Newton' where ω is the angular frequency. Equating the two expressions yields ω² = k/m, hence ω = √k/m. The period T = 2π/ω then becomes T = 2π √m/k. For a simple pendulum, using the small‑ (x/L), leading to a = −(g/L) x. Comparing to a = −ω² x gives ω² = g/L and therefore T = 2π √L/g.
当回复力与相对平衡位置的位移成正比且指向平衡位置时，物体做简谐运动。对于质量‑弹簧系统，胡克定律给出 F = −kx。牛顿第二定律 F = ma，因此 ma = −kx，加速度 a = −(k/m) x。在简谐运动中，加速度写为 a = −ω² x，其中 ω 为角频率。对比两式可得 ω² = k/m，即 ω = √k/m。周期 T = 2π/ω 则变为 T = 2π √m/k。对于单摆，利用小角度近似 sin θ ≈ θ，回复力为 −mg sin θ ≈ −mg (x/L)，从而 a = −(g/L) x。与 a = −ω² x 比较得 ω² = g/L，因此 T = 2π √L/g。
9. Gravitational Field Strength from Potential | 从势求引力场强
potential at a distance r is V = −GM/r. Differentiating with respect to r gives dV/dr = GM/r²
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