Rutherford’s Nuclear Model of the Atom & the Nucleus | 原子核模型与核式结构

📚 Rutherford’s Nuclear Model of the Atom & the Nucleus | 原子核模型与核式结构

Few discoveries in physics have changed our picture of matter as dramatically as the Rutherford gold-foil experiment. Before 1911, scientists believed that the positive charge in an atom was spread out like a soft pudding. Rutherford’s team showed instead that almost all the mass and all the positive charge are squeezed into an incredibly small central core — the nucleus.

在物理学中，很少有发现能像卢瑟福金箔实验那样彻底改变我们对物质的认知。1911年之前，科学家们认为原子中的正电荷像软布丁一样均匀分布。而卢瑟福团队用实验证明：原子几乎全部的质量和全部的正电荷，都被压缩在一个极其微小的中心核——即原子核——之中。

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1. Historical Background: From Indivisible Atoms to Thomson’s Model | 历史背景：从不可分割原子到汤姆孙模型

In the 19th century, the atom was regarded as an indivisible, solid sphere. The discovery of the electron by J.J. Thomson in 1897 destroyed that simple picture, because atoms are electrically neutral yet they must somehow contain very light, negatively charged particles.

19世纪时，原子被视为不可分割的实心球体。1897年 J.J. 汤姆孙发现电子，打破了这一简单图景——因为原子是电中性的，却必然含有很轻的带负电粒子。

Thomson therefore proposed the “plum pudding” model: a sphere of positive charge with electrons embedded inside it, like raisins in a pudding. In this model, the positive charge was assumed to fill the entire atom uniformly.

汤姆孙因此提出了”葡萄干布丁”模型：一个带正电的球体，电子像布丁里的葡萄干一样镶嵌其中。该模型假设正电荷均匀地充满整个原子。

• Electron mass is about 1/1836 of a proton mass.

  电子质量约为质子质量的 1/1836。

• Any new atomic model must explain why atoms are neutral but contain electrons.

  任何新的原子模型都必须解释：原子呈电中性，却含有电子。

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2. The Rutherford Experiment: Setup | 卢瑟福实验：装置

In 1909, Geiger and Marsden, under Rutherford’s supervision, fired a beam of alpha particles into a very thin gold foil. The alpha particle is a helium nucleus, carrying a charge of +2e and a relatively large mass.

1909年，盖革和马斯登在卢瑟福指导下，将一束 α 粒子射向极薄的金箔。α 粒子是氦原子核，带 +2e 的正电荷，质量相对较大。

The experiment was conducted inside a vacuum chamber to prevent alpha particles from losing energy by collisions with air molecules. A movable zinc-sulfide screen was placed around the foil to detect the particles. When an alpha particle struck the screen, a tiny flash of light appeared.

实验在真空容器内进行，以避免 α 粒子与空气分子碰撞而损失能量。金箔周围安装了可移动的硫化锌荧光屏。每当 α 粒子击中屏时，就会出现一次微小的闪光。

• A tiny radioactive source emitted alpha particles | 微型放射源发射 α 粒子

• A sheet of gold foil about 10⁻⁶ m thick | 厚度约 10⁻⁶ m 的金箔

• A movable detector screen that could be rotated around the foil | 可绕金箔旋转的探测屏

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3. Predictions Based on Thomson’s Model | 基于汤姆孙模型的预期

If the plum-pudding model were correct, the positive charge in the gold atom would be spread over a large volume, producing only a very weak electric force on the fast-moving alpha particle.

若葡萄干布丁模型正确，则金原子中的正电荷应弥散于很大体积中，对高速运动的 α 粒子只能产生非常微弱的电场力。

Thomson’s model predicted that the vast majority of alpha particles would pass straight through the foil with almost no deflection. A very small number might be scattered by small angles, but the probability of a deflection greater than a few degrees was essentially zero.

汤姆孙模型预言：绝大多数 α 粒子将几乎不偏转地直穿金箔。极少数粒子可能发生小角度偏转，但偏转角大于几度的概率基本为零。

• Small-angle scattering expected: typically less than 0.1° | 预期以小角度散射为主：通常小于 0.1°

• No particle should ever bounce back towards the source | 不该有任何粒子反弹回放射源方向

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4. Actual Observations | 实际观测结果

The results stunned the physics community. Most alpha particles did pass straight through the foil, but a small fraction — about 1 in 8000 — were deflected through angles greater than 90°. A few were even scattered back through nearly 180°.

实验结果震惊了物理学界。大多数 α 粒子确实直线穿过了金箔，但有一小部分——约 1/8000——偏转角度超过 90°。极少数甚至被反弹回来，偏转接近 180°。

The fact that any particle bounced backwards was considered impossible under Thomson’s model. To turn such a fast, massive particle around, the electric field inside the gold atom would have to be enormously stronger than the field produced by a uniformly distributed positive charge.

在汤姆孙模型中，任何粒子发生反弹都被认为是完全不可能的。要把如此高速且质量不小的 α 粒子掉头转向，金原子内部的电场必须远强于均匀分布正电荷所产生的场。

• Most particles: zero or very small deflection | 大多数粒子：零偏转或极小偏转

• About 1 in 8000: deflection greater than 90° | 约 1/8000 的粒子偏转超过 90°

• Very rare: deflection of up to 180° (backward scattering) | 极少数粒子：偏转接近 180°（背散射）

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5. Rutherford’s Nuclear Conclusion | 卢瑟福的核式结构结论

Rutherford concluded that in order to cause such large deflections, the positive charge of an atom must be concentrated in a tiny region called the nucleus. He estimated the nuclear radius to be about 10⁻¹⁵ m, while the atomic radius is about 10⁻¹⁰ m. The nucleus is therefore about 100,000 times smaller than the atom in diameter.

卢瑟福由此断定：要造成如此大的偏转，原子的正电荷必然集中在被称为原子核的极小区域内。他估算核半径约为 10⁻¹⁵ m，而原子半径约 10⁻¹⁰ m。因此，原子核的直径比原子本身小约十万倍。

Let d be the closest approach of the alpha particle to the nucleus. At this point, all initial kinetic energy has been converted into electric potential energy, so:

设 d 为 α 粒子到达离原子核最近的间距。在该点，初始动能全部转化为电势能，因此：

½mv² = (2e)(Ze) / (4πε₀d)

Solving for d gives an upper limit of the nuclear radius. This analysis uses conservation of energy and Coulomb’s law — two ideas you must always connect in nucleus questions.

解出 d 即可得到原子核半径的上限。这里运用了能量守恒与库仑定律——在核物理题目中你务必始终将两者联系起来。

• Positive charge is confined to a nucleus | 正电荷全部集中在原子核内

• Electrons orbit the nucleus at relatively huge distances | 电子在比核大得多的距离上绕核运动

• Most of the atom is empty space | 原子内部绝大部分是空的空间

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6. Size and Properties of the Nucleus | 原子核的大小与性质

Rutherford’s experiment did not directly measure the mass or the internal structure of the nucleus, but it gave a scale. The nuclear radius is typically of the order of a few femtometres, where 1 fm = 10⁻¹⁵ m.

卢瑟福实验并未直接测得原子核的质量或其内部结构，但给出了一个尺度。原子核半径通常在几飞米量级，1 fm = 10⁻¹⁵ m。

The nuclear radius R can be estimated using

核半径 R 可通过下式估算：

R ≈ r₀A¹ᐟ³

where A is the nucleon number and r₀ ≈ 1.2 fm. This formula is often used in data-analysis questions, and one of the most common methods is to plot R against A¹ᐟ³ to obtain a straight line through the origin.

其中 A 是核子数，r₀ ≈ 1.2 fm。该公式常用于数据分析题，最常见的方法之一是画出 R 对 A¹ᐟ³ 的图线，得到一条过原点的直线。

• A nucleus contains protons (charge +e) and neutrons (no charge)

  原子核包含质子（电荷 +e）和中子（不带电）

• The nucleus carries almost all the atom’s mass | 原子核承载了原子几乎全部的质量

• The electron cloud determines the atomic radius | 电子云决定了原子半径的大小

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7. Nuclear Model vs Plum-Pudding Model | 核式结构模型与葡萄干布丁模型对比

The table below summarises the key differences between the two models. You should be ready to reproduce and explain these points in any exam.

下表总结了两模型之间的关键区别。在考试中，你应当能够复述并解释这些要点。

Feature | 特征	Plum Pudding | 布丁模型	Nuclear Model | 核式结构
Positive charge distribution | 正电荷分布	Uniform over whole atom | 整个原子均匀分布	Concentrated in tiny nucleus | 集中在一个微小核内
Mass distribution | 质量分布	Spread out with positive charge | 与正电荷一同散布	Nucleus occupies almost all mass | 核占据几乎全部质量
Electron placement | 电子位置	Embedded | 镶嵌在球内	Orbiting the nucleus | 绕核运动
Explains large-angle scattering? | 能否解释大角度散射?	No | 不能	Yes | 能

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8. The Coulomb Scattering Formula | 库仑散射公式与考点公式

For a beam of alpha particles of charge q₁ scattering from a nucleus of charge q₂, the differential cross-section is described by the Rutherford scattering formula. You will not be asked to derive it, but you must understand the proportionalities.

对于一束电荷为 q₁ 的 α 粒子被电荷为 q₂ 的核散射，微分截面由卢瑟福散射公式描述。通常不会要求你推导该式，但必须理解其中的正比关系。

Scattering rate ∝ (Z₁Z₂ / Eₖ)² × 1/sin⁴(θ/2)

散射率 ∝ (Z₁Z₂ / Eₖ)² × 1/sin⁴(θ/2)

Three conclusions follow directly. First, particles that pass far from the nucleus are hardly deflected. Second, a head-on collision is the only way a particle can be turned back completely. Third, increasing the alpha-particle energy reduces the chance of large-angle scattering, because faster particles spend less time near the nucleus and can get closer to it.

由此可直接得到三个结论。第一，从离核较远区域穿过的粒子几乎不发生偏转。第二，只有正碰才有可能使粒子完全反弹。第三，增大 α 粒子能量会降低大角度散射概率，因为快粒子在核附近停留时间更短，而且能够更接近核。

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9. Classic Exam Question Patterns | 经典考题思路

Examiners love asking why alpha particles are used in this experiment instead of electrons or X-rays. The usual answer involves two reasons. Alpha particles are relatively heavy and highly charged, so they can be scattered detectably, and they do not penetrate materials as easily as electrons do.

考官非常喜欢问：为什么实验中用 α 粒子而不用电子或 X 射线。常见答案包含两个要点。α 粒子相对较重且电荷量较大，能够产生可探测的散射；同时，α 粒子的穿透能力远弱于电子。

Another common question gives you the kinetic energy of the alpha particle and the atomic number of the target. You are asked to estimate the maximum nuclear radius using energy conservation.

另一类常见考题会给出 α 粒子的动能和靶的原子序数，要求你用能量守恒估算核半径的最大值。

• Step 1: Convert energy from eV to J when necessary

  第一步：必要时将能量从 eV 换算为 J

• Step 2: Equate ½mv² with q₁q₂/(4πε₀d)

  第二步：令 ½mv² 等于 q₁q₂/(4πε₀d)

• Step 3: State that the true nuclear radius is smaller than d

  第三步：说明真实核半径一定小于 d

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10. Common Misconceptions | 常见误区与易错点

Many students think that Rutherford’s gold-foil experiment proved the nucleus contains protons and neutrons. That is incorrect. The experiment only proved that the positive charge and most of the mass are localised in a tiny central region.

许多学生以为卢瑟福金箔实验证明了原子核内含有质子和中子。这是不对的。该实验只证明了正电荷和大部分质量集中在一个很小的中心区域。

A second misconception is the claim that electrons orbit the nucleus like planets around the Sun. In fact, the Rutherford model left the electron dynamics unexplained. The modern picture involves quantum orbitals, not classical circular orbits.

第二个常见误区是：电子像行星绕太阳一样绕核运行。实际上，卢瑟福模型并未解释电子动力学，现代图像使用量子轨道，而非经典圆周轨道。

• Alpha particles are not electrons; they have charge +2e

  α 粒子不是电子；其电荷为 +2e

• The detector measures the number of particles per unit time, not their velocity

  探测器测量单位时间内的粒子数，而非粒子的速度

• Gold was chosen because it can be hammered into extremely thin foil

  选择金是因为金可被锤成极薄的箔片

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11. Exam Strategy and Revision Tips | 考场技巧与复习建议

In a six-mark question, you will often be asked to explain how the observations from the gold-foil experiment led to the nuclear model. Structure your answer in three parts: observation one, conclusion one; observation two, conclusion two.

在 6 分题中，常常要求你解释金箔实验的观测结果如何支持核式结构模型。作答时按三段式结构组织：第一个观测及其结论；第二个观测及其结论；最后总结。

Also be prepared to state that the nucleus is positively charged because alpha particles are repelled by it. Since both the alpha particle and the nucleus are positive, the electrostatic force is repulsive and produces the large-angle deflection.

同时要准备好表述：”由于 α 粒子与原子核均带正电，库仑斥力造成大角度偏转”。既然两者均为正电荷，静电力便是斥力。

• Draw a diagram of the apparatus at least once before the exam

  考试前至少亲手画一遍实验装置图

• Memorise the magnitude comparisons: R_nucleus ≈ 10⁻¹⁵ m, R_atom ≈ 10⁻¹⁰ m

  牢记数量级对比：核半径 ≈ 10⁻¹⁵ m，原子半径 ≈ 10⁻¹⁰ m

• Practise converting nuclear radius equations from log-linear form

  练习将核半径公式转化为线性对数形式作图

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