Interdisciplinary Integrated Question Practice for Year 12 Edexcel Physics | Year 12 Edexcel 物理：跨学科综合题型训练

📚 Interdisciplinary Integrated Question Practice for Year 12 Edexcel Physics | Year 12 Edexcel 物理：跨学科综合题型训练

In Edexcel A Level Physics, exam questions increasingly demand the ability to link topics across different modules and even across different subjects. This article presents a series of interdisciplinary question drills designed for Year 12 students, blending core physics ideas with mathematics, chemistry, biology, geology and engineering. Each section contains a worked example, helping you build confidence in tackling unfamiliar contexts and scoring top marks on Paper 1 and Paper 2.

在 Edexcel A Level 物理考试中，试题越来越强调跨模块、跨学科的知识综合。本文专为 Year 12 学生设计了一系列跨学科综合题型训练，将核心物理概念与数学、化学、生物、地质和工程相结合。每个小节均包含例题与详细解析，帮助你建立应对陌生情境的信心，并在试卷 1 和试卷 2 中夺取高分。

1. Mechanics Meets Mathematics: Vector Calculus in Projectile Motion | 力学与数学交融：抛体运动中的向量微积分

Question: A projectile is launched from ground level with an initial speed of 25 m s⁻¹ at an angle of 60° to the horizontal. Use vector methods and calculus to find: (a) the time to reach the highest point, (b) the maximum height, (c) the horizontal range. Assume g = 9.81 m s⁻² and neglect air resistance.

问题：一个抛体以初速 25 m s⁻¹ 与水平方向成 60° 角从地面发射。用向量方法和微积分求：(a) 到达最高点所需的时间，(b) 最大高度，(c) 水平射程。取 g = 9.81 m s⁻²，忽略空气阻力。

Analysis: Split initial velocity into components: uₓ = u cosθ, u_y = u sinθ. Use a = -g j. Integrate acceleration to obtain velocity, then integrate velocity for displacement. At the highest point, vertical velocity component is zero. The horizontal range is found when vertical displacement returns to zero.

分析：将初速度分解为两个分量：uₓ = u cosθ，u_y = u sinθ。加速度设为 a = -g j。对加速度积分得速度，再积分得位移。在最高点，垂直方向速度分量为零。当垂直位移回到零时，求得水平射程。

Solution: uₓ = 25 cos60° = 12.5 m s⁻¹, u_y = 25 sin60° ≈ 21.65 m s⁻¹. Velocity vector: v = (12.5) i + (21.65 – 9.81t) j. Setting v_y = 0 gives t = 21.65 / 9.81 ≈ 2.21 s. Maximum height: integrate y-component, y = 21.65t – 4.905t²; substitute t = 2.21 s → y_max ≈ 23.9 m. Range: time of flight T = 2 × 2.21 = 4.42 s, range = uₓ T ≈ 55.3 m.

解答：uₓ = 25 cos60° = 12.5 m s⁻¹，u_y = 25 sin60° ≈ 21.65 m s⁻¹。速度向量：v = (12.5) i + (21.65 – 9.81t) j。令 v_y = 0，得 t = 21.65 / 9.81 ≈ 2.21 s。最大高度：对 y 分量积分，y = 21.65t – 4.905t²；代入 t = 2.21 s → y_max ≈ 23.9 m。射程：全程时间 T = 2 × 2.21 = 4.42 s，射程 = uₓ T ≈ 55.3 m。

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2. Materials and Chemistry: Stress–Strain Curves and Intermolecular Bonds | 材料与化学：应力–应变曲线与分子间键

Question: A polymer sample of original length 100 mm and cross-sectional area 5.0 × 10⁻⁶ m² is stretched. The force–extension data give a Young modulus of 2.5 GPa up to a strain of 0.02, after which the curve becomes nonlinear and the sample breaks at a strain of 3.5. Explain the shape of the stress–strain graph using the chemistry of polymer chains, and calculate the breaking stress.

问题：一聚合物样品原长 100 mm，截面积 5.0 × 10⁻⁶ m²，被拉伸。力–伸长数据在应变小于 0.02 时给出杨氏模量 2.5 GPa，之后曲线呈非线性，最终在应变 3.5 时断裂。请用聚合物链的化学知识解释应力–应变图线的形状，并计算断裂应力。

Analysis: The linear region corresponds to elastic stretching of bonds and chain uncoiling. The nonlinear region involves chain slippage and alignment. Breaking occurs after extensive plastic deformation. Breaking stress is force at break divided by original area; force can be estimated from the stress–strain relationship if experimental data are given, but here we derive from the given Young modulus and breaking strain, assuming most of the deformation is plastic with little further stress increase. In a typical polymer, stress may plateau; the exact breaking stress would require a known breaking force. For simplicity, estimate breaking stress from the elastic modulus extended to breaking strain, which gives an overestimate but illustrates the calculation: stress = E × strain = 2.5 × 10⁹ Pa × 3.5 = 8.75 GPa. In reality it is lower due to plastic flow.

分析：线性区域对应键的弹性拉伸和分子链的解卷曲。非线性区域涉及链的滑移和定向排列。断裂发生在大量塑性变形之后。断裂应力为断裂力除以原面积；力可由应力–应变关系估算，但如果没有实验数据，我们根据给定的杨氏模量和断裂应变来估算，虽然大部分变形为塑性，应力可能不再按线性增加，典型聚合物会出现应力平台。为简化，用弹性模量乘以断裂应变得到应力上限：应力 = E × 应变 = 2.5 × 10⁹ Pa × 3.5 = 8.75 GPa，实际会因塑性流动而低于此值。

Solution: The initial linear part reflects stretching of covalent bonds and uncoiling of cross-linked chains, requiring high force. Once the chains start to slide past one another, the material yields and the curve flattens or rises slowly. Finally, chains are fully aligned and break. Breaking stress ~ 8.75 × 10⁹ Pa (upper bound). Exact value depends on the actual fracture force.

解答：初始线性段反映共价键的拉伸和交联链的解卷曲，需要较大力。一旦链开始相对滑动，材料屈服，曲线趋于平缓或缓慢上升。最终链完全定向排列并断裂。断裂应力约 8.75 × 10⁹ Pa（上限值）。准确值取决于实际断裂力。

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3. Waves and Biology: Ultrasound Imaging and Acoustic Impedance | 波与生物学：超声成像与声阻抗

Question: Medical ultrasound uses a frequency of 3.5 MHz. The speed of sound in soft tissue is 1540 m s⁻¹. Calculate the wavelength. Explain why a gel is applied to the skin during the scan, referencing the idea of acoustic impedance and the density–velocity product. The acoustic impedance of air is about 430 kg m⁻² s⁻¹ and that of soft tissue is 1.63 × 10⁶ kg m⁻² s⁻¹. Determine the percentage of ultrasound intensity reflected at an air–tissue interface without gel.

问题：医学超声使用频率 3.5 MHz。声音在软组织中的速度为 1540 m s⁻¹。计算波长。解释为什么扫描时要在皮肤上涂抹耦合剂，并引用声阻抗及密度–速度乘积的概念。空气的声阻抗约为 430 kg m⁻² s⁻¹，软组织的声阻抗为 1.63 × 10⁶ kg m⁻² s⁻¹。求在没有耦合剂时，空气–组织界面反射的超声强度百分比。

Analysis: Wavelength λ = v / f. Gel replaces air, matching the acoustic impedance of tissue and reducing reflection. The intensity reflection coefficient for normal incidence is given by R = ((Z₂ – Z₁) / (Z₂ + Z₁))². Apply to air (Z₁) and tissue (Z₂).

分析：波长 λ = v / f。耦合剂取代空气，使声阻抗与组织匹配，减少反射。正入射时的强度反射系数为 R = ((Z₂ – Z₁) / (Z₂ + Z₁))²。代入空气 (Z₁) 和组织 (Z₂) 计算。

Solution: λ = 1540 / (3.5 × 10⁶) = 4.4 × 10⁻⁴ m = 0.44 mm. Without gel, R = ((1.63×10⁶ – 430) / (1.63×10⁶ + 430))² ≈ (1.62957×10⁶ / 1.63043×10⁶)² ≈ 0.99947² ≈ 0.9989. So about 99.9% of intensity is reflected, leaving almost nothing transmitted. Gel (Z ≈ 1.5×10⁶ kg m⁻² s⁻¹) reduces this to <1% reflection, making imaging possible.

解答：λ = 1540 / (3.5 × 10⁶) = 4.4 × 10⁻⁴ m = 0.44 mm。无耦合剂时，R = ((1.63×10⁶ – 430) / (1.63×10⁶ + 430))² ≈ (1.62957×10⁶ / 1.63043×10⁶)² ≈ 0.99947² ≈ 0.9989。约 99.9% 的强度被反射，几乎没有透射。耦合剂 (Z ≈ 1.5×10⁶ kg m⁻² s⁻¹) 可将反射降至低于 1%，使成像成为可能。

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4. Electric Circuits and Geology: Resistivity Surveying for Groundwater | 电路与地质学：地下水电阻率勘探

Question: Geologists inject current into the ground through two electrodes and measure the potential difference across two other electrodes. In a homogeneous layer, the resistance R between the potential electrodes is linked to the ground resistivity ρ by R = ρ / (2πa) for a Wenner array with electrode spacing a. If a = 5.0 m and the measured R = 120 Ω, calculate the ground resistivity. Suggest how this value changes if the subsurface contains saline water rather than dry rock.

问题：地质学家通过两个电极向大地注入电流，并用另两个电极测量电势差。在均匀地层中，对于电极间距为 a 的温纳排列，电势电极间电阻 R 与大地电阻率 ρ 的关系为 R = ρ / (2πa)。若 a = 5.0 m，测得 R = 120 Ω，计算大地电阻率。若地下含盐水而非干燥岩石，该值将如何变化？

Analysis: Rearrange the formula to ρ = 2πa R. The presence of saline water, rich in ions, drastically lowers resistivity. This is a direct application of Ohm’s law in a continuous medium and links to chemistry (ion mobility).

分析：整理公式得 ρ = 2πa R。富含离子的盐水会显著降低电阻率。这是欧姆定律在连续介质中的直接应用，并与化学（离子迁移率）相联系。

Solution: ρ = 2π × 5.0 m × 120 Ω ≈ 3.77 × 10³ Ω m. For dry rock, resistivity can be 10³–10⁶ Ω m; for saline water saturated rock, it can drop to 1–100 Ω m. Hence a low resistivity survey anomaly suggests groundwater or saline intrusion.

解答：ρ = 2π × 5.0 m × 120 Ω ≈ 3.77 × 10³ Ω m。干燥岩石的电阻率在 10³–10⁶ Ω m 之间；含盐水岩石可降至 1–100 Ω m。因此，低电阻率异常指示地下水或盐水入侵。

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5. Particle Physics and Astrophysics: Cosmic Ray Muons and Time Dilation | 粒子物理与天体物理：宇宙线 μ 子与时间膨胀

Question: Muons are created in the upper atmosphere at an altitude of about 15 km, moving at 0.995c. Their rest half-life is 1.52 µs. In classical physics, how far would they travel before half decay? In special relativity, why are they detected at sea level? Calculate the distance in the Earth frame using time dilation and verify that it matches the flight distance.

问题：μ 子产生于约 15 km 高空，以 0.995c 朝地面运动。静止半衰期为 1.52 µs。按经典物理，它们在半衰期内能飞多远？在狭义相对论框架下，为什么它们能在海平面被探测到？利用时间膨胀计算地球参考系中的飞行距离，并验证其与路径长度一致。

Analysis: Classical distance = v × t₀. With time dilation, the observed half-life in Earth frame is γ t₀ where γ = 1/√(1 – v²/c²). Then distance d’ = v × γ t₀. Compare to 15 km.

分析：经典距离 = v × t₀。考虑时间膨胀后，地球系中观测到的半衰期为 γ t₀，其中 γ = 1/√(1 – v²/c²)。于是距离 d’ = v × γ t₀。与 15 km 比较。

Solution: v = 0.995c, c = 3.0×10⁸ m s⁻¹, t₀ = 1.52×10⁻⁶ s. Classical d = 0.995×3.0×10⁸ × 1.52×10⁻⁶ ≈ 454 m. γ = 1/√(1 – 0.995²) ≈ 10.0. Dilated half-life = 15.2 µs, d’ = 0.995×3.0×10⁸ × 15.2×10⁻⁶ ≈ 4.54×10⁴ m = 45.4 km. This is well above 15 km, so many muons survive to sea level, consistent with relativistic predictions and experimental evidence.

解答：v = 0.995c，c = 3.0×10⁸ m s⁻¹，t₀ = 1.52×10⁻⁶ s。经典 d = 0.995×3.0×10⁸ × 1.52×10⁻⁶ ≈ 454 m。γ = 1/√(1 – 0.995²) ≈ 10.0。膨胀后半衰期 = 15.2 µs，d’ = 0.995×3.0×10⁸ × 15.2×10⁻⁶ ≈ 4.54×10⁴ m = 45.4 km。远大于 15 km，因此大量 μ 子能到达海平面，与相对论预言和实验证据一致。

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6. Mechanics and Biomechanics: Lever Systems in the Human Arm | 力学与生物力学：人体手臂中的杠杆系统

Question: The biceps muscle exerts a force Fₘ at a distance of 4.0 cm from the elbow joint. The forearm and hand (weight 15 N) act at 15 cm from the elbow, and a load of 50 N is held in the hand at 35 cm from the elbow. Assuming equilibrium, calculate Fₘ and the reaction force at the elbow joint. Classify this lever system.

问题：肱二头肌在距肘关节 4.0 cm 处施加力 Fₘ。前臂和手（重量 15 N）作用在距肘 15 cm 处，手中握有 50 N 负载，距离肘 35 cm。设系统平衡，计算 Fₘ 及肘关节处的反作用力。并判断此杠杆类型。

Analysis: Take moments about the elbow. Sum of clockwise moments = sum of anticlockwise moments. The unknown Fₘ produces an anticlockwise moment (if arm is horizontal and biceps pulls upward). The weight and load produce clockwise moments. Solve for Fₘ, then use vertical force equilibrium to find joint reaction force.

分析：对肘关节取矩。顺时针力矩之和 = 逆时针力矩之和。未知力 Fₘ 产生逆时针力矩（假设手臂水平，肱二头肌向上拉）。前臂重量和负载产生顺时针力矩。解出 Fₘ，然后用竖向力平衡求出关节反力。

Solution: Taking anticlockwise positive: Fₘ × 0.040 m – (15 N × 0.15 m) – (50 N × 0.35 m) = 0 → Fₘ = (2.25 + 17.5) / 0.040 = 493.75 N ≈ 490 N. Vertical forces: F_elbow (up) + Fₘ (up) = 15 N + 50 N (down), so F_elbow = 65 – 493.75 = -428.75 N, i.e., direction is downward. The lever is a Class 3 lever (effort between fulcrum and load), which favours speed and range of motion over force.

解答：设逆时针为正：Fₘ × 0.040 m – (15 N × 0.15 m) – (50 N × 0.35 m) = 0 → Fₘ = (2.25 + 17.5) / 0.040 = 493.75 N ≈ 490 N。竖向力平衡：F_肘 (向上) + Fₘ (向上) = 15 N + 50 N (向下)，所以 F_肘 = 65 – 493.75 = -428.75 N，即方向向下。这是一个第三类杠杆（力点在支点和阻力点之间），以力量换取速度和运动幅度。

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7. Electricity and Chemistry: Electrolysis and Faraday’s Laws | 电学与化学：电解与法拉第定律

Question: A steady current of 2.0 A is passed through a copper(II) sulfate solution for 30 minutes. Copper ions Cu²⁺ each carry a charge of +2e. Calculate the mass of copper deposited on the cathode. (Elementary charge e = 1.60 × 10⁻¹⁹ C, Avogadro constant N_A = 6.02 × 10²³ mol⁻¹, molar mass of Cu = 63.5 g mol⁻¹.)

问题：一恒定电流 2.0 A 通过硫酸铜(II) 溶液，历时 30 分钟。铜离子 Cu²⁺ 各带 +2e 电荷。计算沉积在阴极上的铜的质量。（基本电荷 e = 1.60 × 10⁻¹⁹ C，阿伏伽德罗常数 N_A = 6.02 × 10²³ mol⁻¹，Cu 摩尔质量 = 63.5 g mol⁻¹。）

Analysis: Total charge Q = I t. Number of Cu²⁺ ions discharged = Q / (2e). Number of moles = N_ions / N_A. Mass = moles × molar mass.

分析：总电荷量 Q = I t。放电的 Cu²⁺ 离子数 = Q / (2e)。摩尔数 = 离子数 / N_A。质量 = 摩尔数 × 摩尔质量。

Solution: Q = 2.0 A × (30 × 60 s) = 3600 C. Number of ions = 3600 / (2 × 1.60×10⁻¹⁹) = 3600 / (3.20×10⁻¹⁹) = 1.125×10²². Moles = 1.125×10²² / 6.02×10²³ = 0.0187 mol. Mass = 0.0187 × 63.5 ≈ 1.19 g. This demonstrates the direct link between electric current and chemical change, a key interdisciplinary concept.

解答：Q = 2.0 A × (30 × 60 s) = 3600 C。离子数 = 3600 / (2 × 1.60×10⁻¹⁹) = 3600 / (3.20×10⁻¹⁹) = 1.125×10²²。摩尔数 = 1.125×10²² / 6.02×10²³ = 0.0187 mol。质量 = 0.0187 × 63.5 ≈ 1.19 g。这直接证明了电流与化学变化之间的联系，是一个关键的跨学科概念。

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8. Waves and Music: Standing Waves in String Instruments | 波与音乐：弦乐器中的驻波

Question: A violin string has length 0.33 m and linear density 0.65 g m⁻¹. It is tuned to a fundamental frequency of 440 Hz (A4). Calculate the tension required. When the violinist presses the string at a point 0.11 m from one end, what new fundamental frequency is produced? Relate this to an interval in music (e.g., a perfect fifth has a frequency ratio 3:2).

问题：一根小提琴弦长 0.33 m，线密度为 0.65 g m⁻¹。将其调至基频 440 Hz（A4）。计算所需的张力。当演奏者按住弦上距一端 0.11 m 处时，产生的新的基频是多少？并说明该音程在音乐中的关系（如纯五度的频率比为 3:2）。

Analysis: For a fixed string, f₁ = (1/(2L)) √(T/µ). Solve for T with L = 0.33 m. The shortened length becomes L’ = 0.33 m – 0.11 m = 0.22 m (assuming pressing from one end, the vibrating length is the shorter segment). Then new f₁’ = (1/(2L’)) √(T/µ) = f₁ × (L/L’).

分析：对于两端固定的弦，f₁ = (1/(2L)) √(T/µ)。代入 L = 0.33 m 解出 T。缩短后弦长 L’ = 0.33 m – 0.11 m = 0.22 m（假设按住一端，振动段是较短部分）。新的基频 f₁’ = (1/(2L’)) √(T/µ) = f₁ × (L/L’)。

Solution: µ = 0.65×10⁻³ kg m⁻¹. 440 = (1/(2×0.33)) √(T / 0.65×10⁻³). Rearranging: √(T / 6.5×10⁻⁴) = 440 × 0.66 = 290.4. So T / 6.5×10⁻⁴ = 290.4² ≈ 84332, T ≈ 54.8 N. New length L’ = 0.22 m, f₁’ = 440 × (0.33/0.22) = 440 × 1.5 = 660 Hz. A frequency ratio 660:440 = 3:2 corresponds to a perfect fifth, the interval between A and E. This illustrates the physics behind musical harmony.

解答：µ = 0.65×10⁻³ kg m⁻¹。440 = (1/(2×0.33)) √(T / 0.65×10⁻³)。整理得：√(T / 6.5×10⁻⁴) = 440 × 0.66 = 290.4。所以 T / 6.5×10⁻⁴ = 290.4² ≈ 84332，T ≈ 54.8 N。新弦长 L’ = 0.22 m，f₁’ = 440 × (0.33/0.22) = 440 × 1.5 = 660 Hz。频率比 660:440 = 3:2，是纯五度，即 A 到 E 的音程。这揭示了音乐和声背后的物理原理。

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9. Optics and Biology: The Human Eye and Corrective Lenses | 光学与生物学：人眼与矫正镜片

Question: A short-sighted person cannot see clearly objects beyond 2.0 m. The distance from the eye lens to the retina is 2.5 cm. Calculate the power of the eye lens when viewing a distant object (at infinity) for a normal eye, and the power required for this myopic eye. Determine the power of the spectacle lens needed to correct this defect, assuming it is placed 1.5 cm in front of the eye.

问题：一近视者无法看清 2.0 m 以外的物体。眼睛晶状体到视网膜的距离为 2.5 cm。计算正常眼注视无穷远物体时晶状体的屈光力，以及该近视眼所需的屈光力。并确定矫正此缺陷所需的眼镜片屈光力，设镜片位于眼前 1.5 cm 处。

Analysis: For a normal eye, the eye lens focuses parallel rays onto the retina, so object distance u = ∞, image distance v = 2.5 cm. Lens power P = 1/f = 1/u + 1/v. For the myopic eye, the far point is at 2.0 m; when looking at this far point, the eye lens must produce an image on the retina. Use u = 2.0 m, v = 2.5 cm to find the maximum power of the myopic eye. For correction, spectacles must take an object at ∞ and produce a virtual image at the far point. The spectacle lens power P_s = 1/f_s. Consider the distance from spectacle to eye.

分析：正常眼中，眼晶状体将平行光线会聚到视网膜上，物距 u = ∞，像距 v = 2.5 cm。透镜屈光力 P = 1/f = 1/u + 1/v。近视眼的远点在 2.0 m；注视远点物体时，晶状体须将图像聚焦在视网膜上。用 u = 2.0 m，v = 2.5 cm 求近视眼的最大屈光力。矫正时，眼镜片须将无限远物体成虚像于远点处。眼镜片屈光力 P_s = 1/f_s，并考虑镜眼距。

Solution: Normal eye: u = ∞, v = 0.025 m → P = 1/0.025 = 40 D. Myopic eye: u = 2.0 m, v = 0.025 m → P_myopic = 1/2.0 + 1/0.025 = 0.5 + 40 = 40.5 D. So myopic eye is too powerful by 0.5 D. For spectacles: the lens must take an object at ∞ and form a virtual image at the far point (2.0 m from eye). If lens is 1.5 cm = 0.015 m from eye, the image distance from lens v’ = -(2.0 – 0.015) = -1.985 m (virtual upright image). Then 1/f_s = 1/u + 1/v’ = 0 + 1/(-1.985) = -0.504 D. The required spectacle power is -0.50 D (concave lens).

解答：正常眼：u = ∞，v = 0.025 m → P = 1/0.025 = 40 D。近视眼：u = 2.0 m，v = 0.025 m → P_近视 = 1/2.0 + 1/0.025 = 0.5 + 40 = 40.5 D。近视眼屈光力高出 0.5 D。眼镜片：须将无限远物体成虚像于远点（距眼 2.0 m）。镜片距眼 0.015 m，像距 v’ = -(2.0 – 0.015) = -1.985 m。1/f_s = 1/u + 1/v’ = 0 + 1/(-1.985) = -0.504 D。所需眼镜屈光力为 -0.50 D（凹透镜）。

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10. Nuclear Physics and Medicine: Radioactive Tracers and Half-Life | 核物理与医学：放射性示踪剂与半衰期

Question: Technetium-99m (⁹⁹ᵐTc) is a gamma emitter used in medical imaging. It has a half-life of 6.0 hours. A patient is injected with a sample having an initial activity of 800 MBq. Calculate the activity after 18 hours. The detector requires at least 50 MBq to produce a clear image. How long after injection can the scan be usefully performed? Also, explain why a pure gamma emitter with a short half-life is chosen for this purpose, linking to biological clearance and radiation dose.

问题：锝-99m (⁹⁹ᵐTc) 是一种用于医学成像的伽马射线源，半衰期为 6.0 小时。患者被注射初始活度为 800 MBq 的样品。计算 18 小时后的活度。探测器至少需要 50 MBq 才能生成清晰图像。注射后多长时间内进行扫描仍有效？此外，结合生物清除和辐射剂量，解释为何选择短半衰期的纯伽马发射体。

Analysis: Radioactive decay follows A = A₀ e^{-λt}, where λ = ln2 / T₁/₂. After 18 h, number of half-lives = 3, activity reduces by factor 2³ = 8. To find the time when A = 50 MBq, use A/A₀ = 1/16 = e^{-λt}, giving t = T₁/₂ × log₂(16) = 4 half-lives = 24 h. Short half-life ensures quick decay after procedure, minimizing patient dose; pure gamma emission avoids alpha/beta damage to tissues and allows easy detection outside the body.

分析：放射性衰变遵循 A = A₀ e^{-λt}，其中 λ = ln2 / T₁/₂。18 小时后，半衰期数 = 3，活度降低为原来的 1/8。求 A = 50 MBq 的时间：A/A₀ = 1/16 = e^{-λt}，得 t = T₁/₂ × log₂(16) = 4 个半衰期 = 24 小时。短半衰期确保检查后快速衰变，减少患者剂量；纯伽马发射避免 α/β 粒子对组织损伤，并易于体外探测。

Solution: After 18 h: A = 800 / 2³ = 100 MBq. Useful scanning up to 24 hours (when A = 50 MBq). Choice of isotope:

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