📚 Mock Test Walkthrough for OCR A Level Physics Module 6: Particles and Medical Physics | OCR A Level 物理 第六单元模拟测试卷解析
This walkthrough dissects a mock test designed for OCR A Level Physics Year 13, targeting Module 6: Particles and Medical Physics. The selected questions span capacitors, electric fields, electromagnetism, nuclear physics, and medical imaging. Each solution is presented with clear steps, key formulas, and exam-focused reasoning to deepen your understanding and refine problem-solving technique for the actual examination.
本文深入解析一份针对OCR A Level 物理 Year 13 第六单元（粒子与医学物理）的模拟测试卷。所选题目涵盖电容器、电场、电磁学、核物理和医学成像。每道题都以清晰的步骤、关键公式和应试策略呈现，帮助您加深理解、提升解题技巧，从容应对正式考试。
1. Capacitor Discharge Analysis | 电容器放电分析
Question: A 100 μF capacitor is charged to 6.0 V and then connected across a 20 kΩ resistor. Calculate (a) the time constant of the circuit, (b) the initial charge stored, and (c) the charge remaining after 2.0 s.
问题：一个100 μF电容器充电至6.0 V，然后连接到一个20 kΩ电阻上。计算(a) 电路的时间常数，(b) 初始储存的电荷量，(c) 2.0 s后剩余的电荷量。
The time constant τ for an RC circuit is given by τ = RC. Substituting the values: R = 20 kΩ = 20 × 10³ Ω, C = 100 μF = 100 × 10⁻⁶ F.
RC电路的时间常数 τ 由 τ = RC 给出。代入数值：R = 20 kΩ = 20 × 10³ Ω，C = 100 μF = 100 × 10⁻⁶ F。
τ = RC = (20 × 10³ Ω) × (100 × 10⁻⁶ F) = 2.0 s
The initial charge Q₀ is found from Q₀ = CV.
初始电荷 Q₀ 由 Q₀ = CV 求得。
Q₀ = (100 × 10⁻⁶ F) × (6.0 V) = 6.0 × 10⁻⁴ C
For discharge, the charge decays exponentially: Q = Q₀ e-t/τ. At t = 2.0 s, which equals exactly one time constant, we have Q = Q₀ e⁻¹.
放电过程中电荷呈指数衰减：Q = Q₀ e-t/τ。在 t = 2.0 s 时，恰好等于一个时间常数，因此 Q = Q₀ e⁻¹。
Q = 6.0 × 10⁻⁴ C × 0.3679 ≈ 2.2 × 10⁻⁴ C
This demonstrates that after one time constant, the charge drops to about 37% of its initial value – a useful checking point in both multiple-choice and structured questions.
这表明经过一个时间常数，电荷降至初始值的约37%——无论是选择题还是结构化题目，这都是一个实用的校验点。
2. Electric Force Between Point Charges | 点电荷之间的电场力
Question: Two point charges, +4.0 μC and –2.0 μC, are placed 30 cm apart in a vacuum. Calculate the magnitude of the electric force between them and state whether it is attractive or repulsive. (k = 8.99 × 10⁹ N m² C⁻²)
问题：两个点电荷，+4.0 μC 和 –2.0 μC，在真空中相距30 cm。计算两者之间电场力的大小，并说明是吸引力还是排斥力。(k = 8.99 × 10⁹ N m² C⁻²)
Coulomb’s law gives the force magnitude: F = k|Q₁Q₂| / r². The signs are opposite, so the force is attractive.
库仑定律给出力的大小：F = k|Q₁Q₂| / r²。电荷符号相反，因此力为吸引力。
|Q₁| = 4.0 × 10⁻⁶ C, |Q₂| = 2.0 × 10⁻⁶ C, r = 0.30 m
F = (8.99 × 10⁹) × (4.0 × 10⁻⁶) × (2.0 × 10⁻⁶) / (0.30)²
Carrying out the multiplication: (8.99 × 10⁹) × (8.0 × 10⁻¹²) = 7.192 × 10⁻²; dividing by 0.09 gives F ≈ 0.80 N. Always square the distance first to avoid a common error. The force is attractive because the charges have opposite signs; if asked for direction, you would state that the force on each charge points toward the other charge.
计算过程：(8.99 × 10⁹) × (8.0 × 10⁻¹²) = 7.192 × 10⁻²，再除以0.09得到 F ≈ 0.80 N。务必先对距离进行平方，以避免常见错误。由于两电荷异号，力为吸引力；若要说明方向，每个电荷所受的力均指向另一电荷。
3. Cyclotron: Charged Particle in Magnetic Field | 回旋加速器：磁场中的带电粒子
Question: A proton moves perpendicular to a uniform magnetic field of 0.60 T in a cyclotron. Calculate the radius of its circular path when its speed is 3.0 × 10⁷ m s⁻¹. Explain why the radius increases as the proton gains energy. (mₚ = 1.67 × 10⁻²⁷ kg, e = 1.60 × 10⁻¹⁹ C)
问题：一个质子在回旋加速器中垂直于0.60 T的匀强磁场运动。当速度为3.0 × 10⁷ m s⁻¹时，计算其圆形轨迹的半径。解释为何随着质子能量增大，半径会增大。(mₚ = 1.67 × 10⁻²⁷ kg，e = 1.60 × 10⁻¹⁹ C)
The magnetic force provides the centripetal force: Bev = mₚv² / r, which rearranges to r = mₚv / (Be).
磁场力提供向心力：Bev = mₚv² / r，整理得 r = mₚv / (Be)。
r = (1.67 × 10⁻²⁷ × 3.0 × 10⁷) / (0.60 × 1.60 × 10⁻¹⁹) ≈ 0.522 m
The radius is directly proportional to the proton’s momentum mₚv. As the proton is repeatedly accelerated by the electric field in the ‘dee’ gaps, its speed and kinetic energy increase, so the momentum rises. Consequently, the radius grows each half-cycle, producing an outward spiral until the proton reaches the extraction radius. Notice that the cyclotron frequency f = Be/(2πmₚ) stays constant as long as the mass doesn’t change relativistically.
半径与质子的动量 mₚv 成正比。每当质子在D形盒间隙被电场加速，其速度和动能增加，动量也随之增大。因此每个半周期半径都会增大，形成向外扩展的螺旋线，直到质子到达引出半径。注意，只要质量没有相对论性变化，回旋频率 f = Be/(2πmₚ) 保持不变。
4. Alpha Decay Equation and Conservation Laws | α衰变方程与守恒定律
Question: Uranium-238 decays by alpha emission. Write the complete decay equation and identify the daughter nucleus.
问题：铀-238发生α衰变。写出完整的衰变方程，并指出子核。
In alpha decay, the parent nucleus emits a helium nucleus ⁴₂He. Mass number (A) decreases by 4 and atomic number (Z) decreases by 2.
在α衰变中，母核放出一个氦核 ⁴₂He。质量数 (A) 减4，原子序数 (Z) 减2。
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
The daughter nucleus has A = 234, Z = 90, which corresponds to thorium-234. Conservation of mass–energy and momentum also apply, but at A-level we mainly verify the conservation of nucleon number and charge. Always check that the total A on the right equals 238 (234+4) and total Z equals 92 (90+2).
子核的质量数为234，原子序数为90，即钍-234。质能守恒和动量守恒同样适用，但在A Level中我们主要验证核子数守恒和电荷守恒。务必检查右侧总A为238（234+4），总Z为92（90+2）。
5. Radioactive Decay and Half-Life Calculation | 放射性衰变与半衰期计算
Question: A radioactive source has an initial activity of 2400 Bq and a half-life of 4.0 hours. Calculate its activity after 12 hours.
问题：某放射源的初始活度为2400 Bq，半衰期为4.0小时。计算12小时后的活度。
The number of half-lives elapsed is n = total time / T½ = 12 h / 4.0 h = 3.
经过的半衰期数目 n = 总时间 / T½ = 12 h / 4.0 h = 3。
Activity decreases by a factor of (1/2)ⁿ: A = A₀ (½)ⁿ.
活度按 (1/2)ⁿ 倍数衰减：A = A₀ (½)ⁿ。
A = 2400 Bq × (½)³ = 2400 / 8 = 300 Bq
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