OCR物理 Medical Physics X射线 CT PET 超声 考点
医学物理是 OCR A-Level Physics（H556）Module 6「粒子与医学物理」的最后一部分，考试形式偏向情境应用：题目给出某位病人接受某项检查，要求你用物理原理解释成像过程、判断图像对比的来源、比较不同技术的图像质量与辐射剂量。这些知识点看似零散，其实可以串成一条主线：辐射或声波如何产生，如何与人体组织相互作用，如何被探测，最后如何变成可判读的图像。本文沿这条主线梳理 X 射线成像、CT、PET、超声成像与辐射防护五个板块，并给出 OCR 的答题模板与易错点。
Medical physics is examined at the end of OCR A-Level Physics (H556) Module 6, Particles and Medical Physics, and the questions are usually set in context: a patient has a particular scan, and you must use physics to explain how the image is formed, where the contrast comes from, and how the technique compares with alternatives in image quality and radiation dose. The apparently separate topics in fact form one chain of reasoning: how the radiation is produced, how it interacts with tissue, how it is detected, and how the signal becomes an interpretable image. This article works along that chain through X-ray imaging, CT, PET, ultrasound and radiation protection, and finishes with OCR-style answer templates and the mistakes that cost the most marks.
一、X 射线成像：产生、衰减与对比 / X-ray imaging: production, attenuation and contrast
X 射线由 X 射线管产生：钨丝阴极通电加热后发生热电子发射，电子在 50 至 150 kV 的高电压下被加速，高速撞击高原子序数的钨靶（阳极）后急剧减速，其动能中不足 1% 转化为 X 射线光子（轫致辐射），其余全部变成热，所以阳极必须旋转并用油或水冷却。产生的 X 射线束先经过铝或铜滤片，吸收那些只能增加剂量却无助于成像的低能光子，再用铅制准直器限制照射野，减少病人不必要的受照面积。X 射线穿过人体时按 I = I0 e^(-μx) 衰减，其中 μ 是衰减系数，强烈依赖于物质的原子序数 Z 和光子能量：骨中钙的原子序数高，光电吸收显著，μ 大，透过的强度小；软组织 μ 小，透过的强度大，于是在探测器（胶片或荧光屏）上形成明暗对比的投影图像。为了显示胃肠道、血管等对比度很低的结构，需要口服硫酸钡或注射含碘造影剂，用高 Z 物质人为提高局部的 μ。
An X-ray tube produces X-rays by thermionic emission: electrons boiled off a heated tungsten filament are accelerated through a potential difference of roughly 50 to 150 kV and strike a high atomic number tungsten target. Their sudden deceleration converts a small fraction, under 1%, of the kinetic energy into X-ray photons (bremsstrahlung), while nearly all the rest becomes heat, which is why the anode rotates and is cooled by oil or water. The beam is then filtered to remove low-energy photons that would add dose without improving the image, and collimated with lead to limit the exposed area. As the beam passes through the body its intensity falls according to I = I0 e^(-μx), where the attenuation coefficient μ depends strongly on atomic number and photon energy. Calcium-rich bone has a high μ, so little radiation is transmitted, whereas soft tissue has a low μ and transmits far more; the transmitted pattern is recorded on film or a detector as a shadow image. Barium sulfate swallowed for a barium meal or iodine compounds injected for angiography raise μ locally so that the gut or the blood vessels become visible.
二、X 射线的局限与 CT / Limitations of radiography and CT scanning
普通 X 光片有一个根本缺陷：它把三维结构压成一个二维投影，前后重叠的结构无法分开，软组织之间的对比也很差。CT（计算机断层扫描）正是为解决这个问题而设计的。CT 机把细准直的 X 射线束和一圈探测器装在可旋转的机架上，围绕病人旋转一周或数周，从上百个不同角度采集透射强度数据；计算机再求解这些数据构成的方程组，反推出体内每一个小体素（voxel）的衰减系数 μ，重建出一幅横断面图像。由于每一层都经过多角度、多束的扫描，CT 的辐射剂量明显高于单张 X 光片：一次胸部 X 光片约 0.02 mSv，而一次胸部或腹部 CT 约 7 至 10 mSv。所以 CT 换来的是高对比度的软组织成像、精确的解剖定位以及重叠结构的分离，代价是更高的剂量，因此必须由临床需要来证明其合理性。
A plain radiograph projects a three-dimensional body onto a two-dimensional image, so overlapping structures cannot be separated and soft-tissue contrast is poor. CT, computerised axial tomography, was designed to remove exactly this limitation. A narrow collimated X-ray beam and a ring of detectors rotate around the patient, and transmission data are collected from hundreds of different angles. A computer then solves the resulting set of simultaneous equations to deduce the attenuation coefficient of each small volume element, or voxel, and reconstructs a cross-sectional image. Because every slice is scanned with many beams at many angles, the dose is substantially higher than for a single radiograph: a chest radiograph delivers about 0.02 mSv, whereas a chest or abdominal CT delivers roughly 7 to 10 mSv. CT therefore buys high soft-tissue contrast, accurate anatomical localisation and the separation of overlapping structures at the cost of a higher dose, so each examination must be clinically justified.
三、PET：正电子湮灭与符合探测 / PET: positron annihilation and coincidence detection
PET（正电子发射断层扫描）不依赖外部 X 射线源，而是把放射性示踪剂注入体内。最常用的示踪剂是氟-18 标记的脱氧葡萄糖（FDG），代谢旺盛的细胞，例如肿瘤细胞和脑细胞，摄取得更多。氟-18 发生 β+ 衰变，放出一个正电子；正电子在组织中只前进几毫米就与一个电子相遇并湮灭，两个粒子的静质量完全转化为能量，按 E = mc² 产生两个能量各为 511 keV 的 γ 光子，而且为了守恒动量，它们沿几乎完全相反的方向（180 度）飞出。探测器环记录两个几乎同时到达的符合事件，把这两个探测器连成一条直线，湮灭点必定位于这条直线上，于是示踪剂的分布位置被确定。这种电子准直不需要铅准直器，灵敏度远高于 SPECT。氟-18 的半衰期约 110 分钟，因此必须在配备回旋加速器的医院现制现用；短半衰期既保证成像期间有足够活度，又限制了病人接受的剂量。
PET, positron emission tomography, uses no external X-ray source; instead a radioactive tracer is injected into the patient. The commonest tracer is fluorine-18 labelled fluorodeoxyglucose (FDG), which is taken up preferentially by metabolically active cells such as tumour cells and brain tissue. Fluorine-18 undergoes beta-plus decay and emits a positron, which travels only a few millimetres in tissue before meeting an electron. The two particles annihilate, converting their rest mass entirely into energy: by E = mc², two gamma photons of 511 keV each are produced, and momentum conservation requires them to travel in almost exactly opposite directions, 180 degrees apart. A ring of detectors records the coincidence of two near-simultaneous arrivals; the line joining the two detectors must pass through the point of annihilation, so the position of the tracer is located without any physical collimation. This electronic collimation makes PET far more sensitive than SPECT. Fluorine-18 has a half-life of about 110 minutes, so it must be made on site with a cyclotron; the short half-life gives enough activity for the scan while limiting the dose to the patient.
四、超声成像：压电效应、声阻抗与多普勒 / Ultrasound: piezoelectric effect, acoustic impedance and Doppler
超声成像使用 1 至 20 MHz 的机械纵波，医学诊断常用 2 至 18 MHz，由压电换能器产生：在压电晶体（石英或 PZT 陶瓷）两极加上交变电压，晶体的厚度随之周期性变化，从而在介质中激发超声波；同一个晶体也能把回到晶面的压力变化重新变成电压，因为压电效应是可逆的，所以一个探头既发射又接收。探头先发射一个短脉冲，随即切换为接收，记录回波的时间差 t，用 d = c t / 2 求出反射界面的深度，其中 c 是软组织中声速，约 1540 m/s，骨中约 4080 m/s，空气中只有 330 m/s；除以 2 是因为脉冲要走一个来回。声阻抗 Z = ρc 决定界面的反射比例，强度反射系数为 ((Z2 - Z1) / (Z1 + Z2))²。空气的声阻抗只有约 400 kg m⁻²s⁻¹，而皮肤约 1.6×10⁶ kg m⁻²s⁻¹，两者相差极大，几乎全部能量都会被反射回来，所以必须涂耦合凝胶排除空气层，让声波真正进入体内。频率越高分辨率越好，但衰减越快、穿透越浅：腹部深层结构常用 2 至 5 MHz，眼睛和浅表血管用 5 至 10 MHz。A 超把回波幅度对时间作图，用来测量眼轴长度或肿瘤深度；B 超让探头扫动，把回波幅度转换为屏幕上各点的亮度，得到二维断层图像，是产科和腹部检查的标准方式；多普勒超声利用运动血液产生的频移来测定血流速度与方向，可以查出血管狭窄。
Ultrasound imaging uses mechanical longitudinal waves of 1 to 20 MHz, with 2 to 18 MHz typical in diagnosis, generated by a piezoelectric transducer. An alternating potential difference applied across a piezoelectric crystal such as quartz or PZT makes the crystal change thickness periodically, launching ultrasound into the surrounding medium; the same crystal converts returning pressure variations back into a voltage, because the piezoelectric effect is reversible, so a single probe both transmits and receives. The probe emits a short pulse and then switches to receive, timing the echo delay t and giving the depth of the reflecting boundary as d = c t / 2, where c is the speed of sound in soft tissue, about 1540 m/s (about 4080 m/s in bone and only 330 m/s in air); the factor of two appears because the pulse must travel there and back. The acoustic impedance Z = ρc controls how much is reflected, the intensity reflection coefficient being ((Z2 - Z1) / (Z1 + Z2))². Air has an acoustic impedance of only about 400 kg m⁻²s⁻¹ compared with about 1.6×10⁶ for skin, so almost all the energy would be reflected at the skin-air boundary; coupling gel therefore replaces that air layer. Higher frequencies give better resolution but are attenuated faster, so penetration is poorer: 2 to 5 MHz suits deep abdominal structures, while 5 to 10 MHz suits the eye and superficial blood vessels. An A-scan plots echo amplitude against time and measures the axial length of the eye or the depth of a tumour; a B-scan sweeps the beam and maps echo amplitude onto the brightness of the display, producing the two-dimensional cross-sections used in obstetric and abdominal scanning; Doppler ultrasound measures the frequency shift caused by moving blood to determine flow speed and direction, which reveals narrowed arteries.
五、剂量、示踪剂选择与防护 / Dose, tracer choice and protection
电离辐射的剂量用希沃特（Sv）衡量，因为这个单位同时考虑了辐射的类型和组织的敏感程度。作为参照：一次牙科 X 光约 0.005 mSv，一次胸片约 0.02 mSv，一次 CT 约 2 至 10 mSv，一次 PET 约 7 mSv，而天然本底辐射每年约 2.2 至 3 mSv。选择示踪剂有几条物理原则。第一，辐射必须能穿透人体：纯 α 或纯 β 发射体如锶-90 会被组织吸收，只贡献剂量而不成像，所以常用纯 γ 发射体或正电子发射体。第二，半衰期要恰到好处，既长到足以完成制备与成像，又短到不致让病人在体内长期累积剂量，这正是锝-99m（约 6 小时，140 keV）和氟-18（约 110 分钟）成为主流的原因。第三，示踪剂必须化学无毒，并能选择性地聚集在目标器官。防护遵循 ALARA 原则，即在合理可行的范围内尽可能低：只在确有诊断收益时才使用电离辐射，能用超声或 MRI 取得同等信息时优先替代，缩短曝光时间、增大与源的距离、使用铅屏蔽，而剂量按距离的平方反比规律迅速下降。超声属于非电离辐射，在诊断强度下没有已知危害，只是会使组织轻微升温。
Radiation dose is measured in sieverts, a unit that takes account of both the type of radiation and the sensitivity of the irradiated tissue. For reference, a dental radiograph gives about 0.005 mSv, a chest radiograph about 0.02 mSv, a CT scan roughly 2 to 10 mSv and a PET scan about 7 mSv, while natural background radiation is about 2.2 to 3 mSv per year. Several physical principles govern the choice of tracer. First, the radiation must be able to escape the body: a pure alpha or beta emitter such as strontium-90 is absorbed in tissue, contributing dose without forming an image, so pure gamma emitters or positron emitters are preferred. Second, the half-life must be neither too long nor too short: long enough to prepare the tracer and complete the scan, yet short enough to limit the dose that accumulates in the patient, which is why technetium-99m (about 6 hours, 140 keV) and fluorine-18 (about 110 minutes) dominate practice. Third, the tracer must be chemically non-toxic and must concentrate selectively in the organ of interest. Protection follows the ALARA principle, as low as reasonably achievable: use ionising radiation only where there is a real diagnostic benefit, prefer ultrasound or MRI when they give the same information, keep exposure times short, increase the distance from the source and use lead shielding, since dose falls off with the inverse square of distance. Ultrasound is non-ionising and has no known risk at diagnostic intensities, although it does heat tissue very slightly.
六、OCR 答题要点与常见错误 / OCR answer technique and common mistakes
OCR H556 的医学物理题通常先给一个情境，例如某位病人接受了某种扫描，然后要求你描述物理过程、解释图像对比的来源、或比较两种成像方式的优劣。写描述题时按「产生、与组织相互作用、探测、成像」的顺序把整条链条写全：以 PET 为例，必须写到正电子发射、正电子与电子湮灭、产生两个 511 keV 的 γ 光子、两者方向相差 180 度、以及符合探测。数据表会给出 I = I0 e^(-μx)，但不会解释 μ 的物理含义，所以一定要补一句「μ 与物质的原子序数和光子能量有关」。常见失分点包括：把 CT 说成发射扫描，其实 CT 使用外部 X 射线源，属于透射成像；把 PET 与 SPECT 混为一谈；把超声说成电离辐射；计算深度时忘记 d = c t / 2 中的 2；以及混淆电子准直与铅准直。比较题则要点明剂量与图像质量的取舍：CT 剂量高但解剖细节好，超声无电离辐射但受气体和骨骼遮挡，PET 反映代谢功能但分辨率较低。
Medical physics questions in OCR H556 usually present a scenario, such as a patient undergoing a particular scan, and then ask you to describe the physics, explain where the image contrast comes from, or compare two imaging methods. When describing a technique, write the whole chain in order: production, interaction with tissue, detection, image formation. For PET, for example, you must mention positron emission, annihilation with an electron, the two 511 keV gamma photons, their 180 degree separation and coincidence detection. The data sheet gives I = I0 e^(-μx) but does not explain the meaning of μ, so always add that μ depends on atomic number and photon energy. Common mistakes include calling CT an emission scan when it uses an external X-ray source and is therefore transmission imaging, confusing PET with SPECT, describing ultrasound as ionising, forgetting the factor of two in d = c t / 2, and mixing up electronic collimation with lead collimation. Comparison questions should spell out the trade-off between dose and image quality: CT gives the best anatomical detail at a high dose, ultrasound avoids ionising radiation but is blocked by gas and bone, and PET shows metabolic function at comparatively low spatial resolution.
学习建议 / Study tips
第一，把五种成像方式整理成一张对比清单：辐射源在外部还是内部、成像机制是透射、发射还是反射、典型能量或频率、典型剂量、主要优势与局限，考前默写一遍。第二，所有计算题先写公式再代入，注意单位：μ 的单位是 m⁻¹，Z 的单位是 kg m⁻²s⁻¹，c 用 m/s，深度题不要漏掉除以 2。第三，只背结论不够，explain 类题目需要因果链，例如「骨比软组织吸收更多 X 射线，因为钙的原子序数高，光电吸收截面大，透过的光子因此更少，在探测器上形成更强的对比」。第四，记住三个「为什么」：为什么必须涂耦合凝胶，因为要排除空气、实现声阻抗匹配；为什么用高 Z 的钨靶，因为产生 X 射线的效率更高；为什么示踪剂半衰期要短，因为要限制病人剂量。第五，剂量数字不必全背，但要记住数量级顺序：牙片小于胸片，胸片远小于 CT 与 PET，并牢记天然本底每年约 2 至 3 mSv 这个参照点。第六，把 Module 6 的公式单独抄一页：I = I0 e^(-μx)、E = mc²、d = c t / 2、Z = ρc 以及强度反射系数公式，考前只看这一页。
First, build one comparison list of the five techniques: external or internal source, transmission, emission or reflection mechanism, typical energy or frequency, typical dose, and the main advantages and limitations, and write it out from memory before the exam. Second, always write the equation before substituting numbers, and watch the units: μ is measured in m⁻¹, Z in kg m⁻²s⁻¹ and c in m/s, and never drop the factor of two in a depth calculation. Third, learning conclusions is not enough: explain questions need causal chains, for example that bone absorbs more X-rays than soft tissue because calcium has a higher atomic number and a larger photoelectric absorption cross-section, so fewer photons are transmitted and the contrast on the detector is stronger. Fourth, remember three whys: why coupling gel is essential, because it removes the air layer and matches acoustic impedance; why a high-Z tungsten target is used, because X-ray production is more efficient; and why tracer half-lives are short, because this limits the dose to the patient. Fifth, you do not need to memorise every dose figure, but do know the order of magnitude: a dental radiograph is less than a chest radiograph, which is far less than a CT or PET scan, with natural background at about 2 to 3 mSv per year as the reference point. Sixth, copy the Module 6 equations onto a single page, namely I = I0 e^(-μx), E = mc², d = c t / 2, Z = ρc and the intensity reflection coefficient, and revise from that page only.
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