AQA物理 卢瑟福散射 核半径 半衰期
核物理是 AQA 7408 大纲第 3.8 节的内容，只出现在 A-Level 第二年，卷面落在 Paper 2。它的题目看着碎，主线其实只有几条：散射实验给出的核模型、核半径与恒定密度、三种衰变各自的守恒量、指数衰变、质量亏损与结合能。这几条串起来，计算题和解释题都不会脱轨。
Nuclear physics is section 3.8 of the AQA 7408 specification, taught in the second year of the A-level and examined in Paper 2. The questions look fragmented, but they follow a small number of threads: the scattering evidence for the nuclear model, the radius law and constant density, the conservation rules behind each decay mode, exponential decay, and mass defect with binding energy. Tie those threads together and both the calculations and the six-mark explanations stay on track.
卢瑟福散射的三条观察与结论
卢瑟福用 α 粒子轰击金箔，看到的现象只有三条。绝大多数 α 粒子几乎沿直线穿过，说明原子内部大部分是空的。少数偏转角超过 90 度，说明存在一个体积很小、集中了正电荷的中心。极少数被弹回，说明这个中心几乎集中了原子的全部质量。枣糕模型无法解释大角度散射，核式模型才站得住。考纲 3.8.1.1 只要求定性描述，不要求推导，把观察和结论一一对应就够，多写公式不加分。
Rutherford fired alpha particles at thin gold foil and recorded three features. Almost all of them passed through with little deflection, so most of the atom must be empty space. A small fraction scattered through angles greater than 90 degrees, which needs a tiny region holding concentrated positive charge. A very few came back, which means nearly all the mass sits in that same region. The plum pudding model cannot produce large-angle scattering; the nuclear model can. AQA asks only for a qualitative treatment here, so describe each observation and attach its conclusion.
最近接近距离为什么只是半径上限
把 α 粒子的初动能全部当成转化为静电势能，就能估出它离核最近的距离。E_k = (1/4πε0)(Qq/r)，其中 Q = Ze，q = 2e，所以 r = (1/4πε0)(2Ze^2/E_k)。这个 r 只是核半径的上限，因为 α 粒子还没接触到核就被弹开了。算例：E_k = 7.7 MeV、Z = 79 时，r 约为 3 × 10^-14 m，比真实核半径大一个数量级。单位统一是这题的主要坑：MeV 要乘 1.6 × 10^-13 换成 J，fm 要换成 m，漏一步结果就差好几个数量级。
Treating the alpha particle's initial kinetic energy as fully converted into electrostatic potential energy gives an estimate of how close it gets. Setting Ek equal to (1/4 pi epsilon0)(Qq/r) with Q = Ze and q = 2e gives r = (1/4 pi epsilon0)(2Ze squared / Ek). That value is only an upper limit on the nuclear radius, because the alpha particle turns back before it touches the nucleus. For Ek = 7.7 MeV and Z = 79 the result is about 3 × 10^-14 m, roughly ten times larger than the real radius.
电子衍射与 R = R0 A^(1/3)
电子衍射是目前测核半径最准的办法。电子几乎不受强相互作用力影响，衍射图样第一个极小值对应的角度随核变大而变小，由 de Broglie 波长和衍射条件可以反推半径。大量数据拟合出 R = R0 A^(1/3)，R0 通常取 1.2 fm，AQA 部分教材与真题用 1.05 fm，做题时以卷面给的数值为准。A = 56 时，56^(1/3) ≈ 3.83，R ≈ 4.0 fm。
Electron diffraction gives the most reliable measurements of nuclear radius. Electrons feel almost nothing from the strong force, and the angle of the first minimum in the diffraction pattern shrinks as the nucleus grows, so the radius follows from the de Broglie wavelength and the diffraction condition. Fitting many nuclei produces R = R0 A to the power one third, with R0 near 1.2 fm; some AQA texts and papers use 1.05 fm, so take the value printed on the paper. For A = 56 the cube root is about 3.83 and R comes out near 4.0 fm.
核密度与核的大小无关
R 正比于 A^(1/3)，意味着体积 V = (4/3)πR^3 正比于 A，而质量也正比于 A，所以核密度与核的大小无关。把核近似成半径 R0 的球，每个核子的质量取 1.66 × 10^-27 kg，得到密度约 3.4 × 10^17 kg m^-3；若 R0 取 1.2 fm，则约 2.3 × 10^17 kg m^-3。这个数字大约是水的密度的 3 × 10^14 倍，中子星物质的密度也在同一量级。
Because R scales as A to the one third, the volume (4/3)pi R cubed is proportional to A, and the mass is proportional to A as well, so the density does not depend on how big the nucleus is. Treating the nucleus as a sphere of radius R0 and each nucleon as 1.66 × 10^-27 kg gives about 3.4 × 10^17 kg per cubic metre, or 2.3 × 10^17 if R0 is 1.2 fm. That is roughly 3 × 10^14 times the density of water, and neutron star material sits in the same range.
强相互作用力的 0.5 fm 与 3 fm 两条界线
强相互作用力只在 10^-15 m 的尺度上起作用。两核子间距小于约 0.5 fm 时表现为斥力，0.5 到 3 fm 之间是引力，超过 3 fm 就迅速衰减到可以忽略。斥力那一段解释了核为什么不会被压塌，引力那一段解释了核子为什么能抱成团。它还有电荷无关性：质子与质子、中子与中子、质子与中子之间的强作用力强度相同，电荷间的库仑斥力只是叠加在上面。
The strong force acts over roughly 10^-15 m. Below about 0.5 fm the force between two nucleons is repulsive, between 0.5 and 3 fm it is attractive, and beyond 3 fm it falls away quickly. The repulsive core stops nuclei from collapsing, while the attractive region holds nucleons together. The force is also charge independent: proton-proton, neutron-neutron and proton-neutron attraction has the same strength, with Coulomb repulsion simply added on top for protons.
α 衰变与 N-Z 图上的稳定带
α 衰变放出氦-4 核，核子数减 4，质子数减 2，写成 U-238 -> Th-234 + He-4 就是完整方程。每个衰变方程都要先查两件事：核子数守恒，电荷数守恒。N-Z 图上的稳定核集中成一条带，轻核区中子数与质子数接近，重核区为了抵消质子间的库仑斥力需要更多中子，所以稳定带向右上方弯。α 衰变适合质量数大的核，因为一次带走两个质子和两个中子，能有效降低库仑能。
Alpha decay emits a helium-4 nucleus, so the nucleon number drops by four and the proton number by two: uranium-238 becomes thorium-234 plus an alpha particle. Every decay equation should be checked for conservation of nucleon number and of charge before anything else. On the N against Z plot the stable nuclei form a band that bends upwards at high mass, because extra neutrons are needed to offset Coulomb repulsion between protons. Alpha decay therefore dominates at the heavy end of the chart, where removing two protons at once cuts the Coulomb energy efficiently.
β- 衰变里的夸克变化
β- 衰变本质是核里的一个中子变成质子：n -> p + e- + 反中微子。核子数不变，质子数加一，例如 C-14 -> N-14 + e- + 反中微子。再往下一层是夸克：中子 udd，质子 uud，真正发生的是 d -> u + e- + 反中微子。AQA 大纲要求写出这种夸克变化，答题时把电荷和核子数一起核对，别只写电子。
Beta minus decay turns a neutron inside the nucleus into a proton, written as n goes to p plus an electron plus an antineutrino. The nucleon number is unchanged and the proton number rises by one, so carbon-14 becomes nitrogen-14 in the same step. At quark level the neutron is udd and the proton is uud, which means one down quark has turned into an up quark while emitting the electron and the antineutrino. AQA expects that quark change in your answer, and the antineutrino is not decoration: without it the beta spectrum would show a single energy instead of a continuous distribution.
β+ 衰变与电子俘获的区别
β+ 衰变方向相反：p -> n + e+ + 中微子，质子数减一，核子数不变。它主要在人工核素里出现，因为母核必须比子核多出足够能量才能造出正电子。电子俘获是同一个变化的另一种写法：核内一个质子抓走内层电子变成中子，p + e- -> n + 中微子，质子数同样减一，但不放正电子。判断该写哪一个，看题目里有没有轨道电子参与。
Beta plus decay runs the other way: a proton becomes a neutron while emitting a positron and a neutrino, so Z falls by one and A stays the same. It is mainly seen in artificial nuclides, because the parent must have enough extra energy to create the positron. Electron capture is the same change written differently, with a proton absorbing an inner-shell electron and turning into a neutron, so Z again falls by one but no positron appears. Read the question to see whether an orbital electron is involved before choosing between the two.
γ 跃迁与锝-99m 的 140 keV
γ 射线来自激发态核退激，质子数和中子数都不变，所以方程只写光子。它常跟在 α 或 β 衰变后面出现，因为子核可能留在激发态。锝-99m 是大纲点名的例子：m 表示亚稳态，退激放出的 γ 光子能量约 140 keV，半衰期 6 小时，医院用它做骨扫描和心肌显像。选它是因为这个能量适合探测，半衰期短到不会在体内停留太久。
Gamma emission comes from a nucleus dropping out of an excited state, so Z and A are unchanged and the equation shows only a photon. It usually follows alpha or beta decay because the daughter can be left excited. Technetium-99m is the example named in the specification: the m marks a metastable state, the gamma photon carries about 140 keV, and the half-life is six hours, which is why hospitals use it for bone scans and heart imaging. The energy suits detection and the half-life is short without making the scan impractical.
γ 强度的逆平方定律与背景扣除
γ 射线的强度随距离按逆平方衰减，I = k/x^2，考纲把验证这条定律列为必做实验 12。做法是把源和探测器拉开不同距离，测计数率，先测本底再扣除，然后画 I 对 1/x^2 的图，过原点的直线才算验证成功。本底辐射来自宇宙线、氡气和建筑材料，不扣掉的话近距离数据会偏高，直线就不过原点。距离要量到探测器灵敏体积，量外壳会带进系统误差。
Gamma intensity falls with the inverse square of distance, I = k/x squared, and the specification lists verifying this as required practical 12. Take count rates at several source-to-detector distances, subtract the background first, then plot intensity against 1 over x squared: only a straight line through the origin counts as verification. Background from cosmic rays, radon and building materials inflates the near readings and pulls the line off the origin, so measure to the sensitive volume rather than the casing.
指数衰变定律与活度 A = λN
放射性衰变是随机的，每个核在单位时间内衰变的概率 λ 是常数，所以 dN/dt = -λN，积分得到 N = N0 e^(-λt)。活度 A = λN，单位贝可勒尔，1 Bq 表示每秒衰变一次。算例：样品含 5.0 × 10^18 个核，λ = 1.2 × 10^-4 s^-1，则 A = 6.0 × 10^14 Bq。注意 λ 的单位必须是 s^-1，题目给年或小时要先换算。
Radioactive decay is random, with a constant probability λ per nucleus per second, so dN by dt equals minus λN and integrating gives N equals N0 times e to the minus λt. Activity is A = λN in becquerel, where one becquerel is one decay per second. For 5.0 × 10^18 nuclei with λ equal to 1.2 × 10^-4 per second, the activity is 6.0 × 10^14 becquerel. Keep λ in per second, because a value left in per year shifts the answer by a factor of about thirty million.
半衰期 ln2/λ 与对数图斜率
半衰期是活度或核数降到一半所需的时间，T½ = ln2/λ ≈ 0.693/λ。算例：某样品活度 45 天内从 800 Bq 降到 100 Bq，降了 3 个半衰期，所以 T½ = 15 天。图形题里，普通衰变曲线只能读数，取对数后 lnN 对 t 是直线，斜率等于 -λ，用斜率求 λ 比读图更准。很多学生以为半衰期会随温度或压强改变，其实它只由核本身决定。读衰变曲线取半衰期，可以从任意读数出发找它的一半对应的时间，不必从初始值开始。
Half-life is the time for the activity or the number of nuclei to halve, and it equals ln2 divided by λ, about 0.693 over λ. An activity falling from 800 to 100 becquerel over 45 days has passed through three half-lives, so the half-life is 15 days. On a graph, a plain decay curve is hard to read precisely, while plotting lnN against time gives a straight line of gradient minus λ. Half-life is fixed by the nucleus itself; temperature and pressure do not change it.
质量亏损 1 u = 931.5 MeV 与结合能
质量亏损等于把核拆成单个核子后的总质量减去核本身的质量，乘 c^2 就是结合能。1 u = 931.5 MeV，这个换算值由 AQA 大纲 3.8.1.6 明确给出，考卷也会印。算例：氦-4 核由 2 个质子和 2 个中子组成，分开时质量合计 2 × 1.007276 + 2 × 1.008665 = 4.031882 u，核质量 4.001506 u，亏损 0.030376 u，结合能 0.030376 × 931.5 ≈ 28.3 MeV，每核子约 7.07 MeV。题目若给的是原子质量，要先减掉 Z 个电子（每个 0.000549 u），漏掉这一步答案会差 1 MeV 上下。
Mass defect is the total mass of the separated nucleons minus the mass of the nucleus, and multiplying by c squared converts it into binding energy. The specification states that 1 u = 931.5 MeV, and that conversion is printed on the paper. For helium-4, two protons and two neutrons add up to 4.031882 u while the nucleus itself is 4.001506 u, so 0.030376 u has gone missing. That is about 28.3 MeV of binding energy, or 7.07 MeV per nucleon.
结合能曲线的铁峰与裂变聚变
平均结合能对核子数的曲线在 A ≈ 56 附近最高，约 8.8 MeV，铁-56 就在这个位置。曲线左端上升快，说明轻核聚变能放出能量；右端缓慢下降，说明重核裂变也能放出能量。判断一个反应是否放能，比较反应前后的总结合能，或用质量亏损乘 931.5 MeV，两条路必须给出一致结果。中等质量的核之所以占优，是因为那里的每个核子都被邻居充分包围，表面效应和库仑斥力都还不算大。
The graph of average binding energy per nucleon against nucleon number peaks near A = 56 at about 8.8 MeV, where iron-56 sits. The curve rises steeply on the light side, so fusing light nuclei releases energy, and it falls slowly on the heavy side, so splitting heavy nuclei does too. To decide whether a reaction releases energy, compare total binding energy before and after, or multiply the mass loss by 931.5 MeV; both routes must agree. Mid-mass nuclei win because every nucleon there is fully surrounded by neighbours while surface effects and Coulomb repulsion stay modest.
慢化剂 控制棒与临界质量的分工
热中子诱发裂变的条件是中子先被慢化到热能附近，慢化剂用水或石墨，靠弹性碰撞把中子能量降下来。控制棒吸收中子，插入越深链式反应越弱；冷却剂把热量带走，同时防止燃料棒过热。临界质量指的是能维持链式反应的最小质量，低于它中子逃逸多于产生，反应就停下来。这三样东西在题里经常一起问，答之前先看清问的是慢化、吸收还是传热。
Thermal fission needs neutrons slowed to near thermal energies, and the moderator, water or graphite, does that through elastic collisions. Control rods absorb neutrons, so pushing them deeper weakens the chain reaction, while the coolant carries heat away and keeps the fuel rods from overheating. Critical mass is the smallest mass that sustains a chain reaction; below it more neutrons escape than are produced. Questions often mix these three roles, so check whether you are being asked about moderation, absorption or heat transfer before writing.
阅卷时错得最多的四类失分
我批改这类题时，错得最多的是四类。一类是把核子数守恒和电荷守恒当成同一件事，α 衰变方程里质子数没减 2。一类是 β 衰变只写电子、不写反中微子，或者把 β+ 放出的正电子写成电子。一类是算活度时 λ 用了年的单位，结果差三千万倍。还有一类是核半径公式里的 A 写成原子序数 Z，题目换个元素就全错。写完方程逐项核对，这四种都能挡住。
Marking these questions, the same slips turn up repeatedly. Mass number and charge conservation get treated as one condition, so an alpha equation is left with the wrong proton count. A beta equation appears with only an electron and no antineutrino, or a positron is written as an electron. Activities come out wrong by a factor of about thirty million because λ was left in per year. And A gets swapped for Z in the radius formula, which ruins the answer the moment the element changes. Writing the equation first and checking each quantity line by line catches all four.
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