Formula Derivations for A-Level Physics Unit 5 (PH05) | A-Level 物理 PH05 公式推导

📚 Formula Derivations for A-Level Physics Unit 5 (PH05) | A-Level 物理 PH05 公式推导

Understanding how key equations are derived is essential for mastering A-Level Physics Unit 5 (PH05), which covers thermodynamics, radioactivity, gravitation, and oscillations. This article walks through the step-by-step derivations of the most important formulas, from the kinetic theory of gases to radioactive decay and gravitational potential, helping you grasp not just the results but the physical reasoning behind them.

理解关键方程是如何推导的，对于掌握A-Level物理单元5 (PH05) 至关重要，该单元涵盖热力学、放射性、引力和振动。本文将逐步推导最重要的公式，从气体分子运动论到放射性衰变和引力势，帮助你不仅掌握结果，更理解其背后的物理推理。

1. Kinetic Theory Derivation of Gas Pressure | 气体压强的动力学推导

Consider a single molecule of mass m moving with velocity v_x in the x‑direction inside a cubic container of side L. It collides elastically with a wall, reversing its x‑component of velocity while other components remain unchanged. The change in momentum is Δp = −2 m v_x (the wall gains +2 m v_x).

考虑一个质量为m的分子，在边长为L的立方形容器中以速度v_x沿x方向运动。它与壁面发生弹性碰撞，x方向的速度分量反向，其他分量不变。动量的变化为Δp = −2 m v_x（壁获得 +2 m v_x）。

The time between collisions on the same wall is the round‑trip time, Δt = 2L / v_x. The average force exerted by one molecule on that wall is therefore F = Δp / Δt = (2 m v_x) / (2L / v_x) = m v_x² / L.

与同一壁面两次碰撞的时间间隔为往返时间 Δt = 2L / v_x。因此，单个分子对壁施加的平均力为 F = Δp / Δt = (2 m v_x) / (2L / v_x) = m v_x² / L。

For N molecules moving with different speeds, the total force on one wall is F_total = (m / L) Σ v_x². Since the motion is random, the mean square speed in any direction is ⟨v_x²⟩ = ⟨v²⟩ / 3. Thus, total force becomes F_total = (N m ⟨v²⟩) / (3L). Pressure p = F / A, with area A = L², giving the fundamental kinetic theory equation:

对于N个以不同速度运动的分子，作用在一个壁上的总力为 F_total = (m / L) Σ v_x²。由于运动是无规则的，任何一个方向上的均方速度为 ⟨v_x²⟩ = ⟨v²⟩ / 3。于是总力变为 F_total = (N m ⟨v²⟩) / (3L)。压强 p = F / A，面积 A = L²，得到动力学理论的基本方程：

pV = ⅓ N m ⟨v²⟩

2. Linking Temperature to Kinetic Energy | 温度与动能的联系

From the experimental ideal gas law, pV = nRT, where n is the number of moles and R the gas constant. Introducing the Avogadro constant N_A and the total number of molecules N = n N_A, we can rewrite it as pV = N k T, where k = R / N_A is the Boltzmann constant.

由实验得出的理想气体定律 pV = nRT，其中n为摩尔数，R为气体常数。引入阿伏伽德罗常数 N_A 和总分子数 N = n N_A，可改写为 pV = N k T，其中 k = R / N_A 为玻尔兹曼常数。

Equating the two expressions for pV — one from kinetic theory and one from the experimental law — we obtain ⅓ N m ⟨v²⟩ = N k T. Multiplying both sides by 3/2 gives the average translational kinetic energy of a single molecule:

令动力学理论得出的 pV 表达式与实验定律的表达式相等，得到 ⅓ N m ⟨v²⟩ = N k T。两边乘以 3/2，得到单个分子的平均平动动能：

⟨KE⟩ = ½ m ⟨v²⟩ = (3/2) kT

3. Root‑Mean‑Square Speed from Temperature | 由温度求均方根速率

From the relation ½ m ⟨v²⟩ = (3/2) kT, we can solve for the root‑mean‑square speed c_rms = √⟨v²⟩. Multiplying the kinetic energy per molecule by N_A yields the average translational kinetic energy per mole: (3/2) RT. For a molar mass M (kg mol⁻¹), N_A m = M, so:

由关系式 ½ m ⟨v²⟩ = (3/2) kT 可解出均方根速率 c_rms = √⟨v²⟩。将单分子动能乘以 N_A 得到每摩尔平均平动动能：(3/2) RT。对于摩尔质量 M (kg mol⁻¹)，有 N_A m = M，因此：

c_rms = √(3RT / M)

This shows that, at a given temperature, lighter molecules move faster and the rms speed scales with the square root of the absolute temperature.

这表明在给定温度下，较轻的分子运动更快，且均方根速率与绝对温度的平方根成正比。

4. Deriving the Radioactive Decay Law | 放射性衰变定律的推导

Radioactive decay is a random process; the number of nuclei decaying per unit time is proportional to the number N of undecayed nuclei present. This gives the differential rate equation:

放射性衰变是一个随机过程；单位时间内衰变的原子核数与现存未衰变核数 N 成正比。由此得到微分速率方程：

dN/dt = −λ N

where λ is the decay constant. To solve, separate variables and integrate from an initial number N₀ at time t = 0:

其中 λ 为衰变常数。求解时，分离变量并从初始时刻 t=0、初始核数 N₀ 开始积分：

∫_N₀^N (1/N) dN = −λ ∫₀^t dt

Carrying out the integration yields ln(N / N₀) = −λ t, and exponentiating both sides gives the exponential decay law:

积分后得到 ln(N / N₀) = −λ t，两边取指数即得指数衰变定律：

N = N₀ e^{−λ t}

The activity A = |dN/dt| = λ N follows the same exponential form: A = A₀ e^{−λ t}.

活度 A = |dN/dt| = λ N 具有相同的指数形式：A = A₀ e^{−λ t}。

5. Half‑Life and Decay Constant | 半衰期与衰变常数

The half‑life T_½ is the time taken for the number of radioactive nuclei to fall to half its original value. Set N = N₀/2 and t = T_½ in the decay equation:

半衰期 T_½ 是放射性核数衰减到初始值一半所需的时间。在衰变方程中代入 N = N₀/2 和 t = T_½：

N₀/2 = N₀ e^{−λ T_½} → ½ = e^{−λ T_½}

Taking natural logarithms gives ln(½) = −λ T_½. Since ln(½) = −ln 2, we obtain the familiar relationship:

取自然对数得 ln(½) = −λ T_½。因为 ln(½) = −ln 2，于是得到熟悉的关系式：

T_½ = ln 2 / λ

This shows that a larger decay constant implies a shorter half‑life.

这表明衰变常数越大，半衰期越短。

6. Gravitational Potential Energy Between Two Point Masses | 两点质量间的引力势能

The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. The force of attraction is F = GMm / r², directed toward the central mass M. The work done against this force in moving from infinity to a distance r is:

引力势是将一个小的测试质量从无穷远处移至该点每单位质量所作的功。引力大小为 F = GMm / r²，方向指向中心质量 M。将质量从无穷远处移至距离 r 处克服引力所作的功为：

W = ∫_∞^r (GMm / r′²) dr′ = [−GMm / r′]_∞^r = −GMm / r

This work is stored as gravitational potential energy. Therefore, the potential energy of the two‑mass system is:

这个功储存为引力势能。因此，两质量系统的势能为：

U = −GMm / r

The gravitational potential V (J kg⁻¹) is simply U/m, giving V = −GM / r.

引力势 V (J kg⁻¹) 即为 U/m，故 V = −GM / r。

7. Escape Velocity from a Spherical Body | 球状天体的逃逸速度

To escape a planet’s gravitational field, an object must have kinetic energy equal to or greater than the magnitude of its gravitational potential energy at the surface. Using energy conservation, the threshold velocity v_esc is found by setting total mechanical energy to zero at infinity:

要挣脱行星引力场，物体的动能必须等于或大于其在行星表面引力势能的绝对值。利用能量守恒，将无穷远处总机械能设为零，即可求得临界速度 v_esc：

½ m v_esc² − G M m / R = 0 → v_esc = √(2GM / R)

Notice that the escape velocity does not depend on the object’s mass — it is a property of the planet’s mass M and radius R.

注意逃逸速度与物体质量无关 — 它取决于行星质量 M 和半径 R。

8. Period of a Simple Pendulum via SHM Approximation | 单摆周期的简谐运动近似推导

For a simple pendulum of length L displaced by a small angle θ, the restoring force along the arc is F = −mg sin θ. For small angles (θ ≪ 1 rad), sin θ ≈ θ, and the displacement s along the arc is s = Lθ. The tangential acceleration is a = d²s/dt² = L d²θ/dt². Newton’s second law gives:

对于长度为 L、偏移小角度 θ 的单摆，沿弧线的回复力为 F = −mg sin θ。对于小角度 (θ ≪ 1 rad)，sin θ ≈ θ，且沿弧线的位移 s = Lθ。切向加速度 a = d²s/dt² = L d²θ/dt²。由牛顿第二定律：

m L d²θ/dt² = −mg θ → d²θ/dt² = −(g / L) θ

This is the equation for simple harmonic motion with angular frequency ω = √(g / L). The period T is thus:

这是简谐运动的方程，角频率 ω = √(g / L)。因此周期 T 为：

T = 2π / ω = 2π √(L / g)

9. Mass–Spring System Period Derivation | 弹簧振子周期推导

For a mass m attached to a spring of force constant k, Hooke’s law gives the restoring force F = −k x, where x is the displacement from equilibrium. Newton’s second law yields:

对于连接在劲度系数为 k 的弹簧上的质量 m，胡克定律给出回复力 F = −k x，其中 x 为偏离平衡位置的位移。由牛顿第二定律：

m d²x/dt² = −k x → d²x/dt² = −(k / m) x

Comparing with the standard SHM form a = −ω² x identifies ω² = k / m. Hence the period is:

与标准简谐运动形式 a = −ω² x 比较，可得 ω² = k / m。因此周期为：

T = 2π √(m / k)

10. Combining Kinetic Theory with the Ideal Gas Equation | 动力学理论与理想气体方程的结合

Returning to the kinetic pressure equation pV = ⅓ N m ⟨v²⟩ and substituting the expression for average translational kinetic energy ⟨KE⟩ = ½ m ⟨v²⟩ = (3/2) kT, we obtain pV = N k T. Replacing N k by n R recovers the familiar ideal gas law pV = nRT. This three‑way link between macroscopic pressure, microscopic kinetic energy, and temperature lies at the heart of thermal physics.

回到动力论压强方程 pV = ⅓ N m ⟨v²⟩，代入平均平动动能表达式 ⟨KE⟩ = ½ m ⟨v²⟩ = (3/2) kT，得到 pV = N k T。用 n R 替换 N k，即恢复熟悉的理想气体定律 pV = nRT。这种宏观压强、微观动能和温度之间的三重联系正是热物理的核心。

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Formula Derivations for A-Level Physics Unit 5 (PH05) | A-Level 物理 PH05 公式推导