A-Level CIE Physics: Calculation Practice | A-Level CIE 物理：计算题专项训练

📚 A-Level CIE Physics: Calculation Practice | A-Level CIE 物理：计算题专项训练

Calculation questions form a substantial part of the CIE A-Level Physics examinations. They test not only your knowledge of physical principles but also your ability to manipulate equations, handle units, and apply mathematical reasoning to unfamiliar contexts. This article presents a systematic approach to tackling calculation problems, covering core topics from mechanics, waves, electricity, fields, thermal physics, nuclear physics, and oscillations. Each section includes key formulas, common pitfalls, and worked examples designed to build your confidence and precision.

计算题在 CIE A-Level 物理考试中占有相当大的比重。它们不仅考查你对物理原理的理解，还考查你对方程的运用、单位处理以及在陌生情境中应用数学推理的能力。本文提供了一个系统性的方法来应对计算题，涵盖了力学、波、电学、场、热物理、核物理以及振动等核心主题。每一小节都包含了关键公式、常见陷阱和设计好的例题，旨在提升你的信心和解题准确性。

1. Quantities, Units and Dimensional Homogeneity | 物理量、单位与量纲一致性

Always begin by ensuring every numerical value is expressed in SI base units. Convert masses to kilograms, lengths to metres, times to seconds, and so on. Checking the homogeneity of an equation by analysing base units can quickly reveal algebraic errors before you substitute numbers.

始终确保每个数值都用国际单位制基本单位表示。将质量换算为千克，长度换算为米，时间换算为秒，以此类推。通过分析基本单位来检验方程的量纲一致性，可以在代入数字之前快速揭示代数错误。

Write down the formula and rearrange it symbolically before inserting values. This prevents arithmetic mistakes and allows you to check units.

在代入数值之前，先写下公式并用符号进行移项整理。这可以防止算术错误，并让你检查单位。
Express all quantities in the standard form (e.g., 3.2 × 10⁻⁶ m, not 0.0000032 m) to avoid misplacing zeros.

将所有物理量用标准形式表示（例如 3.2 × 10⁻⁶ m，而不是 0.0000032 m），以免错放小数点。

Quantity	Base SI Units
Force (N)	kg m s⁻²
Pressure (Pa)	kg m⁻¹ s⁻²
Potential difference (V)	kg m² s⁻³ A⁻¹

Example: Verify that the period T = 2π√(l/g) for a simple pendulum is dimensionally consistent. [T] = s; [l] = m; [g] = m s⁻². The right-hand side gives √(m / (m s⁻²)) = √(s²) = s, matching the left-hand side.

示例：验证单摆周期公式 T = 2π√(l/g) 的量纲一致性。[T] = s；[l] = m；[g] = m s⁻²。右侧为 √(m / (m s⁻²)) = √(s²) = s，与左侧一致。

2. Kinematics and Projectile Motion | 运动学与抛体运动

The four SUVAT equations apply only when acceleration is constant. For projectile calculations, resolve the initial velocity into horizontal and vertical components. The horizontal motion has constant velocity, while the vertical motion has constant acceleration due to gravity g = 9.81 m s⁻².

四个 SUVAT 方程仅适用于加速度恒定的情况。对于抛体计算，将初速度分解为水平分量和竖直分量。水平方向为匀速运动，竖直方向则为重力加速度 g = 9.81 m s⁻² 作用下的匀加速运动。

Use v = u + at, s = ut + ½at², s = (u+v)t/2, v² = u² + 2as. Label the positive direction clearly to avoid sign errors.

使用 v = u + at、s = ut + ½at²、s = (u+v)t/2、v² = u² + 2as。清晰标记正方向以避免符号错误。
At maximum height, the vertical component of velocity is momentarily zero. The total time of flight for a symmetric trajectory is twice the time to reach peak height.

在最高点，速度的竖直分量瞬时为零。对称轨迹的总飞行时间是到达最高点时间的两倍。

Worked example: A ball is thrown at 20 m s⁻¹ at 30° above the horizontal from ground level. Find the range. vᵧ = 20 sin30° = 10 m s⁻¹. Time to peak: t = (vᵧ)/g = 10/9.81 ≈ 1.02 s. Total flight time = 2.04 s. Horizontal velocity vₓ = 20 cos30° ≈ 17.3 m s⁻¹. Range = vₓ × total time ≈ 17.3 × 2.04 ≈ 35.3 m.

例题：一个球以 20 m s⁻¹ 的速度、与水平方向成 30° 角从地面抛出。求射程。vᵧ = 20 sin30° = 10 m s⁻¹。上升至最高点的时间：t = vᵧ/g = 10/9.81 ≈ 1.02 s。总飞行时间 = 2.04 s。水平速度 vₓ = 20 cos30° ≈ 17.3 m s⁻¹。射程 = vₓ × 总时间 ≈ 17.3 × 2.04 ≈ 35.3 m。

3. Forces, Newton’s Laws and Connected Bodies | 力、牛顿定律与连接体

Draw a clear free-body diagram for each mass. Resolve forces parallel and perpendicular to any inclined plane. Remember that tension in a light inextensible string is the same on both sides of a smooth pulley, whereas the normal reaction force is not necessarily equal to mg on an incline.

为每个物体画出清晰的受力图。将力沿斜面方向及其垂直方向分解。记住，轻质不可伸长绳在光滑滑轮两侧的张力大小相等，而斜面上的支持力不一定等于 mg。

Apply F = ma to the whole system first when objects move together as one, then consider individual bodies if internal forces (e.g., tension) are required.

当物体作为一个整体一起运动时，先对整个系统应用 F = ma；如果需要求内力（如张力），再单独考虑各个物体。
Frictional force is given by f = μR, where R is the normal reaction. Use μₛ for static friction (preventing motion) and μₖ for kinetic friction (during motion).

摩擦力由 f = μR 给出，其中 R 为支持力。用 μₛ 表示静摩擦（阻止运动），μₖ 表示动摩擦（运动过程中）。

Example: Two blocks of masses 3 kg and 5 kg are connected by a string over a smooth pulley at the top of a frictionless slope inclined at 30°. The 5 kg block hangs vertically. Determine the acceleration. Weight component of 3 kg down slope: 3g sin30° = 1.5g. Net force driving system = 5g − 1.5g = 3.5g. Total mass = 8 kg. a = 3.5g/8 ≈ (3.5 × 9.81)/8 ≈ 4.29 m s⁻².

示例：质量分别为 3 kg 和 5 kg 的两个物体通过轻绳相连，跨过一个位于倾角 30° 的光滑斜面顶部的光滑滑轮。5 kg 物体竖直悬挂。求加速度。3 kg 物体沿斜面向下的重力分量：3g sin30° = 1.5g。驱动系统的净力 = 5g − 1.5g = 3.5g。总质量 = 8 kg。a = 3.5g/8 ≈ (3.5 × 9.81)/8 ≈ 4.29 m s⁻²。

4. Work, Energy and Power | 功、能与功率

Distinguish clearly between kinetic energy (½mv²), gravitational potential energy (mgΔh), and elastic potential energy (½kx² for a spring). The work–energy theorem states that the net work done on an object equals its change in kinetic energy. When dissipative forces like friction are present, use the more general principle: initial energy = final energy + work done against friction.

清晰区分动能（½mv²）、重力势能（mgΔh）和弹性势能（弹簧的 ½kx²）。功能原理指出，对物体所做的净功等于其动能的变化量。当存在摩擦力等耗散力时，使用更普遍的原理：初始能量 = 最终能量 + 克服摩擦所做的功。

Power is the rate of doing work, P = W/t. For a force moving at constant speed v, useful power output is P = Fv.

功率是做功的速率，P = W/t。对于一个以恒定速度 v 运动的力，有用的输出功率为 P = Fv。
Efficiency = (useful output power or energy) / (total input power or energy) × 100%.

效率 = （有用的输出功率或能量）/（总的输入功率或能量）× 100%。

Calculation: A car of mass 1200 kg accelerates from rest to 20 m s⁻¹ up a hill gaining 15 m in height. If the average resistive force is 400 N and the distance travelled is 200 m, find the average power if the acceleration takes 8.0 s. Gain in KE = ½ × 1200 × 20² = 240 kJ. Gain in GPE = 1200 × 9.81 × 15 ≈ 176.6 kJ. Work against resistance = 400 × 200 = 80 kJ. Total work done = 240 + 176.6 + 80 = 496.6 kJ. Average power = 496.6 × 10³ / 8.0 ≈ 6.21 × 10⁴ W.

计算：一辆质量为 1200 kg 的汽车从静止加速到 20 m s⁻¹，并爬上一个高度上升 15 m 的山坡。如果平均阻力为 400 N，行驶距离为 200 m，且加速过程用时 8.0 s，求平均功率。动能增加量 = ½ × 1200 × 20² = 240 kJ。重力势能增加量 = 1200 × 9.81 × 15 ≈ 176.6 kJ。克服阻力做功 = 400 × 200 = 80 kJ。总功 = 240 + 176.6 + 80 = 496.6 kJ。平均功率 = 496.6 × 10³ / 8.0 ≈ 6.21 × 10⁴ W。

5. Momentum, Collisions and Impulse | 动量、碰撞与冲量

Momentum is a vector quantity, p = mv. Impulse is the change in momentum, J = Δp = FΔt. In collisions, apply conservation of momentum in each axis independently. Whether kinetic energy is conserved distinguishes elastic from inelastic collisions.

动量是矢量，p = mv。冲量是动量的变化量，J = Δp = FΔt。在碰撞问题中，对每个轴向独立应用动量守恒。动能是否守恒是区分弹性碰撞与非弹性碰撞的关键。

For a perfectly elastic collision, relative speed of approach equals relative speed of separation, and kinetic energy is conserved.

对于完全弹性碰撞，接近的相对速度等于分离的相对速度，且动能守恒。
In an explosion or a gun-bullet problem, initial total momentum is zero, so final momenta must cancel vectorially.

在爆炸或枪-子弹问题中，初始总动量为零，因此最终动量必须在矢量上相互抵消。

Example: A 1500 kg car travelling at 20 m s⁻¹ collides and sticks to a stationary 2500 kg van. Find the speed after collision. Initial momentum = 1500 × 20 = 3.0 × 10⁴ kg m s⁻¹. Combined mass = 4000 kg. Final velocity = 3.0 × 10⁴ / 4000 = 7.5 m s⁻¹. Loss in KE = ½ × 1500 × 20² − ½ × 4000 × 7.5² = 300 kJ − 112.5 kJ = 187.5 kJ.

示例：一辆 1500 kg 的汽车以 20 m s⁻¹ 的速度行驶，与一辆静止的 2500 kg 货车碰撞后粘在一起。求碰撞后的速度。初始动量 = 1500 × 20 = 3.0 × 10⁴ kg m s⁻¹。总质量 = 4000 kg。末速度 = 3.0 × 10⁴ / 4000 = 7.5 m s⁻¹。动能损失 = ½ × 1500 × 20² − ½ × 4000 × 7.5² = 300 kJ − 112.5 kJ = 187.5 kJ。

6. Circular Motion and Gravitation | 圆周运动与万有引力

For an object moving in a circle of radius r at constant speed v, the centripetal acceleration is a = v²/r = ω²r, and the centripetal force is F = mv²/r. Do not treat centripetal force as an extra force; it is provided by tension, gravity, friction, or a normal reaction.

对于以恒定速率 v 在半径为 r 的圆周上运动的物体，向心加速度为 a = v²/r = ω²r，向心力为 F = mv²/r。不要将向心力当作一个额外的力；它由绳的张力、重力、摩擦力或支持力提供。

Combine circular motion with Newton’s law of gravitation: F = GMm/r². At the surface of a planet, g = GM/R².

将圆周运动与牛顿万有引力定律结合：F = GMm/r²。在行星表面，g = GM/R²。
For a satellite in orbit, centripetal force is supplied by gravity: GMm/r² = mv²/r, giving v² = GM/r. Use this to derive Kepler’s third law: T² ∝ r³.

对于轨道上的卫星，由万有引力提供向心力：GMm/r² = mv²/r，得出 v² = GM/r。利用此关系可推导开普勒第三定律：T² ∝ r³。

Calculation: A geostationary satellite has an orbital period of 24 hours. Given Earth’s mass M = 6.0 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻², find its orbital radius. T = 24 × 3600 = 8.64 × 10⁴ s. From T² = (4π²/GM) r³, r³ = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × (8.64 × 10⁴)²) / (4π²). Evaluate stepwise: GM = 4.002 × 10¹⁴, T² ≈ 7.46 × 10⁹, product ≈ 2.99 × 10²⁴. Divided by 4π² (≈39.5) gives r³ ≈ 7.57 × 10²², so r ≈ 4.23 × 10⁷ m.

计算：一颗地球同步卫星的轨道周期为 24 小时。已知地球质量 M = 6.0 × 10²⁴ kg，G = 6.67 × 10⁻¹¹ N m² kg⁻²，求其轨道半径。T = 24 × 3600 = 8.64 × 10⁴ s。由 T² = (4π²/GM) r³ 得 r³ = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × (8.64 × 10⁴)²) / (4π²)。逐步计算：GM = 4.002 × 10¹⁴，T² ≈ 7.46 × 10⁹，乘积 ≈ 2.99 × 10²⁴。除以 4π² (≈39.5) 得 r³ ≈ 7.57 × 10²²，因此 r ≈ 4.23 × 10⁷ m。

7. Simple Harmonic Motion (SHM) | 简谐运动

SHM is characterised by a ∝ −x, with the defining equation a = −ω²x. The displacement–time graph is a sinusoid: x = x₀ sin(ωt) or x = x₀ cos(ωt). Velocity is given by v = ±ω√(x₀² − x²), and maximum speed is ωx₀ at the equilibrium position.

简谐运动的特征是 a ∝ −x，其定义方程为 a = −ω²x。位移-时间图像是正弦曲线：x = x₀ sin(ωt) 或 x = x₀ cos(ωt)。速度由 v = ±ω√(x₀² − x²) 给出，最大速度出现在平衡位置，大小为 ωx₀。

For a mass–spring system, T = 2π√(m/k). For a simple pendulum, T = 2π√(l/g). Know how to derive these from first principles.

对于质量-弹簧系统，T = 2π√(m/k)；对于单摆，T = 2π√(l/g)。要懂得如何从基本原理推导这些公式。
Energy in SHM swaps between kinetic (½mv²) and potential (½kx² for spring, or ½mω²x² in general). Total energy E = ½mω²x₀².

简谐运动中的能量在动能（½mv²）和势能（弹簧为 ½kx²，或一般形式为 ½mω²x²）之间转换。总能量 E = ½mω²x₀²。

Example: A 0.50 kg mass attached to a spring of negligible mass oscillates with amplitude 0.080 m and period 1.2 s. Find the maximum speed and the total energy. ω = 2π/T = 2π/1.2 ≈ 5.24 rad s⁻¹. v_max = ωx₀ = 5.24 × 0.080 ≈ 0.419 m s⁻¹. Total energy = ½mω²x₀² = 0.5 × 0.50 × (5.24)² × (0.080)² ≈ 0.0439 J.

示例：一个 0.50 kg 的物体连接到质量可忽略的弹簧上，以振幅 0.080 m 和周期 1.2 s 振动。求最大速度和总能量。ω = 2π/T = 2π/1.2 ≈ 5.24 rad s⁻¹。v_max = ωx₀ = 5.24 × 0.080 ≈ 0.419 m s⁻¹。总能量 = ½mω²x₀² = 0.5 × 0.50 × (5.24)² × (0.080)² ≈ 0.0439 J。

8. Electric Fields, Potential and Capacitance | 电场、电势与电容

For a uniform electric field between parallel plates, E = V/d. The force on a charge q is F = qE. Electric potential energy in a uniform field is U = qEd (parallel to field). In a radial field due to a point charge Q, E = kQ/r² and V = kQ/r (with k = 1/(4πε₀)).

对于平行板间的匀强电场，E = V/d。电荷 q 所受的力为 F = qE。匀强电场中的电势能为 U = qEd（平行于电场）。在点电荷 Q 产生的径向电场中，E = kQ/r²，V = kQ/r（其中 k = 1/(4πε₀)）。

Capacitance C = Q/V. For a parallel-plate capacitor, C = ε₀A/d. Energy stored in a capacitor: U = ½QV = ½CV² = ½Q²/C.

电容 C = Q/V。对于平行板电容器，C = ε₀A/d。电容器储存的能量：U = ½QV = ½CV² = ½Q²/C。
When capacitors are combined, series: 1/C_total = Σ 1/C_i; parallel: C_total = Σ C_i. Pay attention to charge conservation in switched circuits.

电容器组合时，串联：1/C_total = Σ 1/C_i；并联：C_total = Σ C_i。注意开关切换电路中的电荷守恒。

Calculation: A 470 μF capacitor is charged to 12 V. It is then connected across an uncharged 220 μF capacitor. Find the final voltage. Initial charge Q = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C. Total capacitance in parallel = (470+220) μF = 690 μF. Final voltage V = Q_total / C_total = 5.64 × 10⁻³ / (690 × 10⁻⁶) ≈ 8.17 V.

计算：一个 470 μF 的电容器充电至 12 V，然后与一个未充电的 220 μF 电容器并联。求最终电压。初始电荷 Q = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C。并联总电容 = (470+220) μF = 690 μF。最终电压 V = Q_total / C_total = 5.64 × 10⁻³ / (690 × 10⁻⁶) ≈ 8.17 V。

9. DC Circuits, Kirchhoff’s Laws and Potential Dividers | 直流电路、基尔霍夫定律与分压器

Apply Kirchhoff’s first law (current sum at a junction is zero) and second law (sum of e.m.f.s equals sum of p.d.s around any closed loop). Write equations systematically and solve simultaneously for unknown currents.

应用基尔霍夫第一定律（节点处电流代数和为零）和第二定律（沿任一闭合回路，电动势之和等于电势差之和）。系统地写出方程，并联立求解未知电流。

The potential divider equation: V_out = V_in × (R₂/(R₁+R₂)) for two resistors in series. For a sensor circuit, think how R changes with temperature or light.

分压器方程：对于两个串联电阻，V_out = V_in × (R₂/(R₁+R₂))。对于传感器电路，要考虑电阻如何随温度或光线变化。
Internal resistance r of a cell: terminal p.d. V = E − Ir. Plotting V against I gives a gradient of −r and y-intercept E.

电池的内阻 r：路端电压 V = E − Ir。绘制 V 对 I 的图像，斜率为 −r，y 轴截距为 E。

Example: A 9.0 V battery with internal resistance 0.80 Ω is connected to a 5.0 Ω and a 12.0 Ω resistor in parallel. Find the terminal voltage. External resistance: 1/R = 1/5.0 + 1/12.0 = 0.283… , R ≈ 3.53 Ω. Total circuit resistance = 3.53 + 0.80 = 4.33 Ω. Current I = 9.0/4.33 ≈ 2.08 A. Terminal p.d. = 9.0 − (2.08×0.80) ≈ 7.34 V.

示例：一个 9.0 V 的电池，内阻为 0.80 Ω，连接到一个 5.0 Ω 和一个 12.0 Ω 的并联电阻。求电池的路端电压。外电阻：1/R = 1/5.0 + 1/12.0 = 0.283…，R ≈ 3.53 Ω。电路总电阻 = 3.53 + 0.80 = 4.33 Ω。电流 I = 9.0/4.33 ≈ 2.08 A。路端电压 = 9.0 − (2.08×0.80) ≈ 7.34 V。

10. Magnetic Fields, Force on a Conductor and Faraday’s Law | 磁场、通电导体受力与法拉第定律

The force on a current-carrying conductor in a magnetic field is given by F = BIL sinθ, where θ is the angle between the conductor and the field. For a moving charge, F = Bqv sinθ. Use Fleming’s left-hand rule for direction.

磁场中通电导体所受的力由 F = BIL sinθ 给出，其中 θ 是导体与磁场之间的夹角。对于运动电荷，F = Bqv sinθ。用弗莱明左手定则判定方向。

Magnetic flux φ = BA cosθ; flux linkage = Nφ. Faraday’s law: induced e.m.f. E = −d(Nφ)/dt. Lenz’s law gives the direction of the induced e.m.f.

磁通量 φ = BA cosθ；磁链 = Nφ。法拉第定律：感应电动势 E = −d(Nφ)/dt。楞次定律给出感应电动势的方向。
For a conductor of length L moving at velocity v perpendicular to field B, E = BLv.

对于长度为 L 的导体以速度 v 垂直于磁场 B 运动的情况，E = BLv。

Calculation: A coil of 500 turns with area 4.0 × 10⁻³ m² is rotated through 90° in a uniform 0.25 T field in 0.10 s. Initial flux linkage = NBA cos0° = 500 × 0.25 × 4.0×10⁻³ = 0.50 Wb. Final flux linkage at 90° = 0. Change in flux linkage = −0.50 Wb. Average induced e.m.f. magnitude = |Δ(Nφ)/Δt| = 0.50/0.10 = 5.0 V.

计算：一个 500 匝的线圈，面积为 4.0 × 10⁻³ m²，在 0.10 s 内在 0.25 T 的匀强磁场中转过 90°。初始磁链 = NBA cos0° = 500 × 0.25 × 4.0×10⁻³ = 0.50 Wb。90° 时的末磁链 = 0。磁链变化量 = −0.50 Wb。平均感应电动势大小 = |Δ(Nφ)/Δt| = 0.50/0.10 = 5.0 V。

11. Thermal Physics and Ideal Gases | 热物理与理想气体

Temperature must be in kelvin for gas law calculations. The ideal gas equation is pV = nRT, where R = 8.31 J mol⁻¹ K⁻¹, and n = mass / molar mass. For combined gas laws, use p₁V₁/T₁ = p₂V₂/T₂ for a fixed mass of ideal gas.

在气体定律计算中，温度必须使用开尔文。理想气体状态方程为 pV = nRT，其中 R = 8.31 J mol⁻¹ K⁻¹，n = 质量 / 摩尔质量。对于组合气体定律，对于一定质量的理想气体，使用 p₁V₁/T₁ = p₂V₂/T₂。

Kinetic theory: pV = ⅓ N m ⟨c²⟩, and average kinetic energy per molecule = (3/2) kT. Root-mean-square speed c_rms = √(3kT/m).

分子动理论：pV = ⅓ N m ⟨c²⟩，每个分子的平均动能 = (3/2) kT。方均根速率 c_rms = √(3kT/m)。
Specific heat capacity Q = mcΔθ; specific latent heat Q = mL. Conservation of energy is essential in calorimetry: heat lost = heat gained.

比热容 Q = mcΔθ；比潜热 Q = mL。在量热学中，能量守恒至关重要：失热 = 得热。

Example: A 2.0 mol sample of an ideal gas occupies 0.050 m³ at 300 K. Calculate its pressure and the r.m.s. speed of its molecules if the molar mass is 0.028 kg mol⁻¹. p = nRT/V = 2.0 × 8.31 × 300 / 0.050 ≈ 9.97 × 10⁴ Pa. Mass of one molecule m = 0.028 / (6.02 × 10²³) ≈ 4.65 × 10⁻²⁶ kg. c_rms = √(3 × 1.38×10⁻²³ × 300 / 4.65×10⁻²⁶) ≈ √(2.67×10⁵) ≈ 517 m s⁻¹.

示例：2.0 mol 的理想气体在 300 K 时占据 0.050 m³ 的体积。计算其压强以及分子的方均根速率（摩尔质量为 0.028 kg mol⁻¹）。p = nRT/V = 2.0 × 8.31 × 300 / 0.050 ≈ 9.97 × 10⁴ Pa。单个分子质量 m = 0.028 / (6.02 × 10²³) ≈ 4.65 × 10⁻²⁶ kg。c_rms = √(3 × 1.38×10⁻²³ × 300 / 4.65×10⁻²⁶) ≈ √(2.67×10⁵) ≈ 517 m s⁻¹。

12. Nuclear Physics and Radioactive Decay | 核物理与放射性衰变

The decay law is N = N₀ e^(−λt) or A = A₀ e^(−λt), where λ is the decay constant. The half-life t₁/₂ = ln2/λ. Always convert activities to becquerels (s⁻¹) when using the formula.

衰变规律为 N = N₀ e^(−λt) 或 A = A₀ e^(−λt)，其中 λ 为衰变常数。半衰期 t₁/₂ = ln2/λ。使用公式时，务必将活度转换为贝克勒尔（s⁻¹）。

Binding energy per nucleon = (total mass defect × c²) / number of nucleons. 1 u = 931.5 MeV.

平均结合能（每个核子的结合能）= （总质量亏损 × c²）/ 核子数。1 u = 931.5 MeV。
In radioactive dating, the ratio of remaining nuclei to a stable daughter product allows determination of age.

在放射性测年中，剩余核与稳定子体的比例可用于确定年代。

Calculation: A sample of wood contains 6.0 × 10¹⁰ atoms of carbon‑14 and has an activity of 0.23 Bq. Find the half-life of carbon‑14. Activity A = λN, so λ = A/N = 0.23 / (6.0×10¹⁰) ≈ 3.83 × 10⁻¹² s⁻¹. t₁/₂ = ln2/λ = 0.693 / (3.83×10⁻¹²) ≈ 1.81 × 10¹¹ s, which converts to about 5740 years.

计算：一块木材样本含有 6.0 × 10¹⁰ 个碳‑14 原子，活度为 0.23 Bq。求碳‑14 的半衰期。活度 A = λN，所以 λ = A/N = 0.23 / (6.0×10¹⁰) ≈ 3.83 × 10⁻¹² s⁻¹。t₁/₂ = ln2/λ = 0.693 / (3.83×10⁻¹²) ≈ 1.81 × 10¹¹ s，换算后约为 5740 年。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

A-Level CIE Physics: Calculation Practice | A-Level CIE 物理：计算题专项训练