IB Physics HL · 鼎睿学苑

Unit E.4: Fission单元 E.4：核裂变

Part of Theme E "Nuclear and quantum physics". Mass-energy equivalence ties mass directly to energy through $E = mc^2$; the unified atomic mass unit and its energy equivalent ($931.5\ \mathrm{MeV/c^2}$) let you convert between the two in one step. Mass defect and binding energy explain why nuclei hold together, the binding-energy-per-nucleon curve explains which reactions release energy, and induced fission of $^{235}\mathrm{U}$ turns that release into a chain reaction. The unit closes with critical mass and the engineering of a controlled reactor: fuel, moderator, control rods, coolant, and containment.主题 E"核与量子物理"的一部分。质能等价通过 $E = mc^2$ 把质量与能量直接联系起来；统一原子质量单位及其能量当量（$931.5\ \mathrm{MeV/c^2}$）让你一步完成两者之间的换算。质量亏损与结合能解释了核为何能聚合在一起，比结合能曲线解释了哪些反应会释放能量，而 $^{235}\mathrm{U}$ 的诱发裂变把这种释放变成链式反应。本单元以临界质量及受控反应堆的工程收尾：燃料、慢化剂、控制棒、冷却剂与安全壳。

IB Physics · Theme E.4 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL6 个核心专题 · SL + HL
Read me first先读这里

How to use this guide本指南使用说明

E.4 is a "one equation, many disguises" unit. Almost every numerical question reduces to $E = \Delta m\, c^2$ applied to a mass defect, with the conversion $1\ \mathrm{u} \to 931.5\ \mathrm{MeV/c^2}$ doing the unit work. The marks split between clean arithmetic (track every $\mathrm{u}$, every $\mathrm{MeV}$) and clear conceptual answers (why iron-56 is the most stable, why a moderator is needed, what "critical" means). Learn the curve picture and the reactor parts cold; they carry most of the descriptive marks.E.4 是"一个公式、多种伪装"的单元。几乎每道计算题都归结为对质量亏损应用 $E = \Delta m\, c^2$，由换算 $1\ \mathrm{u} \to 931.5\ \mathrm{MeV/c^2}$ 完成单位转换。分数分布在干净的算术（盯紧每个 $\mathrm{u}$、每个 $\mathrm{MeV}$）与清晰的概念作答（为何铁-56 最稳定、为何需要慢化剂、"临界"是什么意思）。把曲线图像与反应堆部件背熟；它们承载了大部分描述性分数。

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If you are cramming如果你在临阵磨枪

Memorise $E = \Delta m\, c^2$ and $1\ \mathrm{u} = 931.5\ \mathrm{MeV/c^2}$. Energy released $=$ (BE of products) $-$ (BE of reactants), or equivalently $\Delta m\, c^2$ with $\Delta m =$ (mass before) $-$ (mass after). Iron-56 sits at the peak of the binding-energy-per-nucleon curve: fission of heavy nuclei and fusion of light nuclei both move toward it and release energy.

背熟 $E = \Delta m\, c^2$ 与 $1\ \mathrm{u} = 931.5\ \mathrm{MeV/c^2}$。释放能量 $=$（产物结合能）$-$（反应物结合能），等价地为 $\Delta m\, c^2$，其中 $\Delta m =$（反应前质量）$-$（反应后质量）。铁-56 位于比结合能曲线顶峰：重核裂变与轻核聚变都朝它移动并释放能量。

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If you are going for a 7如果你目标是 7 分

Be able to explain, not just compute. State why energy is released only when the products are more tightly bound (higher BE per nucleon). Conserve nucleon number and proton number in every fission equation. Explain the role of each reactor part in one sentence each, and connect "critical mass" to the neutron multiplication factor $k = 1$. Watch sign conventions: a positive mass defect means a bound, stable system that released energy on forming.

要能解释，而不仅是计算。说明只有当产物结合得更紧（比结合能更高）时才释放能量。每个裂变方程都要守恒核子数与质子数。用一句话说清每个反应堆部件的作用，并把"临界质量"与中子倍增因子 $k = 1$ 联系起来。注意符号约定：正的质量亏损意味着一个结合、稳定的系统，其形成时释放了能量。

HL flagHL 标记说明 E.4 Fission is common SL + HL content; there is no HL-only section in this super-topic. The deeper "going deeper" derivations below (e.g. the energetics of the binding-energy curve) are stretch material useful to all students aiming high, not formal HL-extension.E.4 核裂变是 SL + HL 共同内容；本超级专题没有 HL 专属小节。下文较深的"深入"推导（如比结合能曲线的能量学）是面向所有冲高分学生的拓展，而非正式的 HL 扩展内容。
Topic E4.1 · Mass-Energy EquivalenceTopic E4.1 · 质能等价

Mass-Energy Equivalence and the Unified Mass Unit质能等价与统一质量单位 E.4 SL+HL

The core relation. Mass and energy are two forms of the same thing. From the data booklet, E = mc^2: $$ E = m c^2, \qquad \Delta E = \Delta m\, c^2. $$ The unified atomic mass unit. $1\ \mathrm{u} = \tfrac{1}{12}$ of the mass of a carbon-12 atom $= 1.661 \times 10^{-27}\ \mathrm{kg}$. Its energy equivalent. Converting $1\ \mathrm{u}$ via $E = m c^2$ gives the data-booklet figure $$ 1\ \mathrm{u} = 931.5\ \mathrm{MeV}\,c^{-2}. $$ The one-step trick. Multiply a mass in $\mathrm{u}$ by $931.5$ to get its energy in $\mathrm{MeV}$. No need to go through $\mathrm{kg}$ and joules.
核心关系。质量与能量是同一事物的两种形式。数据手册公式 E = mc^2： $$ E = m c^2, \qquad \Delta E = \Delta m\, c^2. $$ 统一原子质量单位。$1\ \mathrm{u} = \tfrac{1}{12}$ 个碳-12 原子的质量 $= 1.661 \times 10^{-27}\ \mathrm{kg}$。 其能量当量。用 $E = m c^2$ 换算 $1\ \mathrm{u}$ 得数据手册数值 $$ 1\ \mathrm{u} = 931.5\ \mathrm{MeV}\,c^{-2}. $$ 一步换算技巧。把以 $\mathrm{u}$ 为单位的质量乘以 $931.5$ 即得以 $\mathrm{MeV}$ 为单位的能量。无需先转成 $\mathrm{kg}$ 与焦耳。
Worked Example E4.1 (energy of one unified mass unit)E4.1 例题（一个统一质量单位的能量）

Show that $1\ \mathrm{u} = 1.661 \times 10^{-27}\ \mathrm{kg}$ corresponds to an energy of about $931.5\ \mathrm{MeV}$. Take $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$ and $1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$.证明 $1\ \mathrm{u} = 1.661 \times 10^{-27}\ \mathrm{kg}$ 对应约 $931.5\ \mathrm{MeV}$ 的能量。取 $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$、$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$。

Identify. Convert mass to energy in joules with E = mc^2, then joules to $\mathrm{MeV}$.

识别。用 E = mc^2 把质量转成焦耳能量，再把焦耳转成 $\mathrm{MeV}$。

Set up. Energy in joules:

列式。焦耳能量：

$$ E = m c^2 = (1.661 \times 10^{-27})(3.00 \times 10^{8})^2 = 1.495 \times 10^{-10}\ \mathrm{J}. $$

Execute. Convert to electronvolts:

计算。转成电子伏：

$$ E = \frac{1.495 \times 10^{-10}}{1.60 \times 10^{-19}} \approx 9.34 \times 10^{8}\ \mathrm{eV} = 934\ \mathrm{MeV}. $$

Evaluate. The small discrepancy from $931.5\ \mathrm{MeV}$ is rounding in $c$ and the eV conversion; the data-booklet value uses more precise constants. Memorise $931.5$ for exam use.

评估。与 $931.5\ \mathrm{MeV}$ 的微小差异来自 $c$ 与电子伏换算的取整；数据手册数值用更精确的常数。考试时记住 $931.5$ 即可。

Going deeper: why mass and energy are interchangeable深入：为何质量与能量可互换

Special relativity treats mass as a form of "frozen" energy: a stationary object of mass $m$ has rest energy $E_0 = m c^2$. When a system loses internal energy (for example, nucleons binding together and emitting a photon), its mass decreases by exactly $\Delta E / c^2$. The factor $c^2 \approx 9 \times 10^{16}\ \mathrm{m^2\,s^{-2}}$ is enormous, so even a tiny mass change corresponds to a large energy. This is why nuclear reactions, with mass defects of order $0.1\%$ of the nucleus, release millions of times more energy per kilogram than chemical reactions.

狭义相对论把质量视为"凝固"的能量：静止质量为 $m$ 的物体具有静能 $E_0 = m c^2$。当系统失去内能（例如核子结合并发射一个光子）时，其质量恰好减少 $\Delta E / c^2$。因子 $c^2 \approx 9 \times 10^{16}\ \mathrm{m^2\,s^{-2}}$ 极大，故即使极小的质量变化也对应巨大的能量。这就是为何质量亏损约为核质量 $0.1\%$ 的核反应，每千克释放的能量是化学反应的数百万倍。

A nuclear process converts a mass of $3.0\ \mathrm{u}$ entirely into energy. The energy released is closest to:某核过程把 $3.0\ \mathrm{u}$ 的质量完全转化为能量。释放的能量最接近：
E4.1 · Q1
$310\ \mathrm{MeV}$
$931.5\ \mathrm{MeV}$
$2.79 \times 10^{3}\ \mathrm{MeV}$
$2.79 \times 10^{6}\ \mathrm{MeV}$
Multiply by the conversion: $3.0 \times 931.5 = 2.79 \times 10^{3}\ \mathrm{MeV}$. The $931.5\ \mathrm{MeV}$ figure is the energy of exactly one $\mathrm{u}$.乘以换算因子：$3.0 \times 931.5 = 2.79 \times 10^{3}\ \mathrm{MeV}$。$931.5\ \mathrm{MeV}$ 是恰好一个 $\mathrm{u}$ 的能量。
Each $\mathrm{u}$ of mass equals $931.5\ \mathrm{MeV}$. For a mass in $\mathrm{u}$, multiply by $931.5$ to get the energy in $\mathrm{MeV}$.每 $\mathrm{u}$ 质量等于 $931.5\ \mathrm{MeV}$。质量以 $\mathrm{u}$ 计时，乘以 $931.5$ 即得 $\mathrm{MeV}$ 能量。
Which statement about $E = mc^2$ is correct?关于 $E = mc^2$，哪项陈述正确？
E4.1 · Q2
A decrease in a system's mass corresponds to a release of energy.系统质量减小对应能量释放。
It applies only to objects moving close to the speed of light.它仅适用于接近光速运动的物体。
Mass can only be converted to energy, never the reverse.质量只能转化为能量，永远不能反向。
It applies to chemical reactions but not nuclear ones.它适用于化学反应而不适用于核反应。
$E = mc^2$ is universal: any energy change is accompanied by a mass change $\Delta m = \Delta E / c^2$. A mass decrease means energy was released; the conversion runs both ways.$E = mc^2$ 是普适的：任何能量变化都伴随质量变化 $\Delta m = \Delta E / c^2$。质量减小意味着释放能量；转换是双向的。
The rest-energy relation holds for all systems, fast or slow, chemical or nuclear, and works in both directions.静能关系对所有系统都成立，无论快慢、化学或核反应，且双向有效。
Topic E4.2 · Mass Defect and Binding EnergyTopic E4.2 · 质量亏损与结合能

Mass Defect and Nuclear Binding Energy质量亏损与核结合能 E.4 SL+HL

Mass defect. A bound nucleus has less mass than its separated constituents. The shortfall is the mass defect $\Delta m$: $$ \Delta m = \left[ Z m_p + N m_n \right] - M_{\text{nucleus}}. $$ Here $Z$ = protons, $N$ = neutrons, $m_p, m_n$ are the proton and neutron masses, and $M_{\text{nucleus}}$ is the actual nuclear mass. Binding energy. The energy released when the nucleus forms (equivalently, the energy needed to pull it fully apart). From the data booklet, E_b = Δm c^2: $$ E_b = \Delta m\, c^2. $$ Shortcut. With $\Delta m$ in $\mathrm{u}$: $E_b\,[\mathrm{MeV}] = \Delta m\,[\mathrm{u}] \times 931.5$.
质量亏损。一个结合的核质量小于其分离组分的质量之和。这个差额即质量亏损 $\Delta m$： $$ \Delta m = \left[ Z m_p + N m_n \right] - M_{\text{nucleus}}. $$ 其中 $Z$ 为质子数，$N$ 为中子数，$m_p, m_n$ 为质子与中子质量，$M_{\text{nucleus}}$ 为实际核质量。 结合能。核形成时释放的能量（等价地，把核完全拆散所需的能量）。数据手册公式 E_b = Δm c^2： $$ E_b = \Delta m\, c^2. $$ 捷径。$\Delta m$ 以 $\mathrm{u}$ 计：$E_b\,[\mathrm{MeV}] = \Delta m\,[\mathrm{u}] \times 931.5$。
Worked Example E4.2 (binding energy of helium-4)E4.2 例题（氦-4 的结合能）

The helium-4 nucleus has mass $4.00150\ \mathrm{u}$. Given $m_p = 1.00728\ \mathrm{u}$ and $m_n = 1.00867\ \mathrm{u}$, find the mass defect and the total binding energy.氦-4 核质量为 $4.00150\ \mathrm{u}$。已知 $m_p = 1.00728\ \mathrm{u}$、$m_n = 1.00867\ \mathrm{u}$，求质量亏损与总结合能。

Identify. Helium-4 has $Z = 2$ protons and $N = 2$ neutrons.

识别。氦-4 有 $Z = 2$ 个质子与 $N = 2$ 个中子。

Set up. Total mass of separated nucleons:

列式。分离核子的总质量：

$$ 2(1.00728) + 2(1.00867) = 4.03190\ \mathrm{u}. $$

Execute. Mass defect:

计算。质量亏损：

$$ \Delta m = 4.03190 - 4.00150 = 0.03040\ \mathrm{u}. $$ $$ E_b = \Delta m\, c^2 = 0.03040 \times 931.5 \approx 28.3\ \mathrm{MeV}. $$

Evaluate. A binding energy of $28.3\ \mathrm{MeV}$ for $4$ nucleons gives about $7.1\ \mathrm{MeV}$ per nucleon, consistent with helium-4's well-known stability (it is unusually tightly bound for a light nucleus).

评估。$4$ 个核子的结合能为 $28.3\ \mathrm{MeV}$，约 $7.1\ \mathrm{MeV}$ 每核子，与氦-4 著名的稳定性一致（对轻核而言它结合得异常紧密）。

Sign sense符号含义 A larger mass defect means a larger binding energy means a more tightly bound, more stable nucleus. Binding energy is positive by convention: it is the energy you would have to supply to dismantle the nucleus.质量亏损越大，结合能越大，核结合得越紧、越稳定。按惯例结合能为正值：它是把核拆散所需供给的能量。
Going deeper: atomic mass vs nuclear mass and electron masses深入：原子质量与核质量以及电子质量

Tables usually quote atomic masses (nucleus plus electrons), not bare nuclear masses. If you use atomic masses for the nucleus, you should also use the hydrogen-atom mass ($1.00783\ \mathrm{u}$) in place of the bare proton, so the $Z$ electron masses cancel. The electron binding energies (a few $\mathrm{eV}$) are negligible compared with nuclear binding energies (millions of $\mathrm{eV}$), so this bookkeeping rarely changes a $3$-significant-figure answer. For IB problems, use whichever masses the question supplies and stay consistent.

数据表通常给出原子质量（核加电子），而非裸核质量。若用原子质量代表核，则应同时用氢原子质量（$1.00783\ \mathrm{u}$）代替裸质子，使 $Z$ 个电子质量相消。电子结合能（几 $\mathrm{eV}$）相比核结合能（数百万 $\mathrm{eV}$）可忽略，故这种记账很少改变 $3$ 位有效数字的答案。IB 题目中用题目所给的质量并保持一致即可。

The mass of a nucleus is always:一个核的质量总是：
E4.2 · Q1
Greater than the sum of its separated nucleon masses.大于其分离核子质量之和。
Less than the sum of its separated nucleon masses.小于其分离核子质量之和。
Equal to the sum of its separated nucleon masses.等于其分离核子质量之和。
Greater for stable nuclei, less for unstable ones.稳定核更大，不稳定核更小。
Binding releases energy, so a bound nucleus has less mass than its free constituents. The shortfall, the mass defect, equals $E_b / c^2$.结合释放能量，故结合的核质量小于其自由组分。这个差额即质量亏损，等于 $E_b / c^2$。
Every bound nucleus has a positive mass defect: its mass is less than the summed masses of the free protons and neutrons.每个结合的核都有正的质量亏损：其质量小于自由质子与中子的质量之和。
A nucleus has a mass defect of $0.50\ \mathrm{u}$. Its total binding energy is closest to:某核的质量亏损为 $0.50\ \mathrm{u}$。其总结合能最接近：
E4.2 · Q2
$235\ \mathrm{MeV}$
$932\ \mathrm{MeV}$
$466\ \mathrm{MeV}$
$1863\ \mathrm{MeV}$
$E_b = \Delta m\, c^2 = 0.50 \times 931.5 \approx 466\ \mathrm{MeV}$. Just multiply the defect in $\mathrm{u}$ by $931.5$.$E_b = \Delta m\, c^2 = 0.50 \times 931.5 \approx 466\ \mathrm{MeV}$。把以 $\mathrm{u}$ 计的亏损乘以 $931.5$ 即可。
Use $E_b = \Delta m \times 931.5\ \mathrm{MeV}$ with $\Delta m$ in $\mathrm{u}$. Here $0.50 \times 931.5 \approx 466\ \mathrm{MeV}$.用 $E_b = \Delta m \times 931.5\ \mathrm{MeV}$（$\Delta m$ 以 $\mathrm{u}$ 计）。此处 $0.50 \times 931.5 \approx 466\ \mathrm{MeV}$。
Topic E4.3 · Binding Energy per NucleonTopic E4.3 · 比结合能

The Binding-Energy-per-Nucleon Curve比结合能曲线 E.4 SL+HL

Binding energy per nucleon. $$ \frac{E_b}{A} = \frac{\Delta m\, c^2}{A}, $$ where $A = Z + N$ is the nucleon (mass) number. This is the fair way to compare the stability of different nuclei. The curve. Plotting $E_b / A$ against $A$ rises steeply for light nuclei, peaks near $A \approx 56$ (iron-56, $\approx 8.8\ \mathrm{MeV}$ per nucleon), then falls gently for heavy nuclei. The rule that runs the unit. Reactions that increase $E_b / A$ release energy. Heavy nuclei move up the curve by splitting (fission); light nuclei move up by joining (fusion). Iron-56 is the turning point: it cannot release energy by either.
比结合能。 $$ \frac{E_b}{A} = \frac{\Delta m\, c^2}{A}, $$ 其中 $A = Z + N$ 为核子数（质量数）。这是比较不同核稳定性的公平方式。 曲线。把 $E_b / A$ 对 $A$ 作图：轻核段陡升，在 $A \approx 56$（铁-56，约 $8.8\ \mathrm{MeV}$ 每核子）附近达峰，随后对重核缓降。 主导本单元的规则。使 $E_b / A$ 增大的反应释放能量。重核通过裂变（分裂）沿曲线上移；轻核通过聚变（结合）上移。铁-56 是转折点：两种方式都无法使其释放能量。
Worked Example E4.3 (reading energy release off the curve)E4.3 例题（从曲线读出能量释放）

A heavy nucleus with $A = 236$ and $E_b / A = 7.6\ \mathrm{MeV}$ splits into two fragments, each with $A = 118$ and $E_b / A = 8.5\ \mathrm{MeV}$. Estimate the energy released.质量数 $A = 236$、比结合能 $E_b / A = 7.6\ \mathrm{MeV}$ 的重核裂成两块，每块 $A = 118$、$E_b / A = 8.5\ \mathrm{MeV}$。估算释放的能量。

Identify. Energy released $=$ (total BE of products) $-$ (BE of reactant), because products are more tightly bound.

识别。释放能量 $=$（产物总结合能）$-$（反应物结合能），因为产物结合得更紧。

Set up. Binding energy before:

列式。反应前结合能：

$$ E_{b,\text{before}} = 236 \times 7.6 = 1793.6\ \mathrm{MeV}. $$

Execute. Binding energy after (two fragments):

计算。反应后结合能（两块碎片）：

$$ E_{b,\text{after}} = 2 \times (118 \times 8.5) = 2006\ \mathrm{MeV}. $$ $$ E_{\text{released}} = 2006 - 1793.6 \approx 212\ \mathrm{MeV}. $$

Evaluate. About $200\ \mathrm{MeV}$ per fission, the standard figure for uranium. The products sit higher on the curve, so the extra binding is released as kinetic energy of the fragments and neutrons.

评估。约每次裂变 $200\ \mathrm{MeV}$，是铀裂变的标准数值。产物在曲线上更高，多出的结合能以碎片与中子的动能形式释放。

Going deeper: why the curve has a peak at all深入：曲线为何会出现峰值

Two competing effects shape the curve. The strong nuclear force is short-ranged, so each nucleon only binds to its near neighbours; as nuclei grow, surface nucleons (with fewer neighbours) become proportionally less important, which raises $E_b / A$ at first. But the electrostatic (Coulomb) repulsion between protons is long-ranged and grows with $Z^2$; in heavy nuclei it pushes $E_b / A$ back down. The balance of a saturating strong force against accumulating Coulomb repulsion produces the broad maximum near iron-56. Beyond that, adding nucleons costs more Coulomb energy than the strong force can recoup, so heavy nuclei are primed to release energy by fission.

两种竞争效应塑造了曲线。强核力短程，每个核子只与近邻结合；核增大时表面核子（近邻较少）所占比例下降，这在初期抬高 $E_b / A$。但质子间的静电（库仑）斥力是长程的，并随 $Z^2$ 增长；在重核中它把 $E_b / A$ 压回。饱和的强核力与累积的库仑斥力之间的平衡，在铁-56 附近形成宽阔的极大值。越过此点，增加核子带来的库仑能多于强核力所能补偿，故重核倾向于通过裂变释放能量。

Iron-56 sits at the peak of the binding-energy-per-nucleon curve. This means iron-56:铁-56 位于比结合能曲线顶峰。这意味着铁-56：
E4.3 · Q1
Is the most stable nucleus and cannot release energy by fission or fusion.是最稳定的核，无法通过裂变或聚变释放能量。
Has the largest total binding energy of all nuclei.在所有核中具有最大的总结合能。
Releases the most energy when it undergoes fission.裂变时释放的能量最多。
Has the largest mass defect per nucleon of any unstable nucleus.在所有不稳定核中具有最大的每核子质量亏损。
Peak $E_b / A$ means maximum binding per nucleon, so iron-56 is the most stable. Any reaction moving away from it lowers $E_b / A$ and would require energy input, so it releases none by fission or fusion.$E_b / A$ 达峰意味着每核子结合最紧，故铁-56 最稳定。任何远离它的反应都降低 $E_b / A$ 且需要输入能量，故它无法通过裂变或聚变释放能量。
"Peak" refers to $E_b / A$ (per nucleon), not total $E_b$. The peak marks maximum stability, so iron-56 releases energy by neither fission nor fusion."峰"指 $E_b / A$（每核子），不是总 $E_b$。峰标志最大稳定性，故铁-56 既不裂变也不聚变释放能量。
A reaction releases energy when:什么情况下反应释放能量：
E4.3 · Q2
The total nucleon number increases.总核子数增加。
The products have lower binding energy per nucleon than the reactants.产物比结合能低于反应物。
The products have larger total mass than the reactants.产物总质量大于反应物。
The products have higher binding energy per nucleon than the reactants.产物比结合能高于反应物。
Higher $E_b / A$ in the products means a more tightly bound, lower-mass final state. The released binding energy appears as kinetic energy. This is why both heavy-nucleus fission and light-nucleus fusion can release energy.产物 $E_b / A$ 更高意味着结合更紧、质量更低的末态。释放的结合能表现为动能。这就是为何重核裂变与轻核聚变都能释放能量。
Energy is released when products are more tightly bound (higher $E_b / A$) and therefore lower in mass than the reactants.当产物结合更紧（$E_b / A$ 更高）因而质量低于反应物时，释放能量。
Topic E4.4 · Nuclear FissionTopic E4.4 · 核裂变

Nuclear Fission and Induced Fission of Uranium-235核裂变与铀-235 的诱发裂变 E.4 SL+HL

Fission. A heavy nucleus splits into two lighter fragments (plus neutrons), releasing energy because the fragments lie higher on the $E_b / A$ curve. Induced fission. A slow ("thermal") neutron is absorbed by $^{235}_{\;92}\mathrm{U}$, forming an excited $^{236}\mathrm{U}$ that splits almost immediately. A typical equation. $$ {}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \;\to\; {}^{141}_{\;56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}. $$ Conservation always holds. Top numbers (nucleon $A$): $1 + 235 = 141 + 92 + 3$. Bottom numbers (proton $Z$): $0 + 92 = 56 + 36 + 0$. Energy per fission. $\approx 200\ \mathrm{MeV}$, mostly as kinetic energy of the two fragments.
裂变。重核分裂为两块较轻碎片（外加中子），因碎片在 $E_b / A$ 曲线上更高而释放能量。 诱发裂变。慢（"热"）中子被 $^{235}_{\;92}\mathrm{U}$ 吸收，形成激发态 $^{236}\mathrm{U}$，几乎立即分裂。 典型方程。 $$ {}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \;\to\; {}^{141}_{\;56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}. $$ 守恒始终成立。上标（核子数 $A$）：$1 + 235 = 141 + 92 + 3$。下标（质子数 $Z$）：$0 + 92 = 56 + 36 + 0$。 每次裂变能量。约 $200\ \mathrm{MeV}$，主要为两块碎片的动能。
Worked Example E4.4 (balancing a fission equation)E4.4 例题（配平裂变方程）

A uranium-235 nucleus absorbs a neutron and fissions into a barium-144 nucleus, a krypton-89 nucleus, and some neutrons: ${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{144}_{\;56}\mathrm{Ba} + {}^{89}_{36}\mathrm{Kr} + x\,{}^{1}_{0}\mathrm{n}$. Find $x$, and confirm the equation is balanced.铀-235 核吸收一个中子并裂变成钡-144 核、氪-89 核及若干中子：${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{144}_{\;56}\mathrm{Ba} + {}^{89}_{36}\mathrm{Kr} + x\,{}^{1}_{0}\mathrm{n}$。求 $x$，并验证方程配平。

Identify. Conserve nucleon number $A$ (top) and proton number $Z$ (bottom) separately.

识别。分别守恒核子数 $A$（上标）与质子数 $Z$（下标）。

Set up (nucleon number). Left: $1 + 235 = 236$. Right: $144 + 89 + x$.

列式（核子数）。左：$1 + 235 = 236$。右：$144 + 89 + x$。

$$ 236 = 233 + x \;\Rightarrow\; x = 3. $$

Execute (proton number check). Left: $0 + 92 = 92$. Right: $56 + 36 + 3(0) = 92$. Balanced.

计算（质子数核对）。左：$0 + 92 = 92$。右：$56 + 36 + 3(0) = 92$。配平。

Evaluate. Three neutrons are emitted. Each can induce a further fission, the basis of the chain reaction in the next section. Note that the fission of $^{235}\mathrm{U}$ does not have a single fixed equation; many fragment pairs occur, but every one conserves $A$ and $Z$.

评估。放出三个中子。每个都能诱发下一次裂变，这是下一节链式反应的基础。注意 $^{235}\mathrm{U}$ 的裂变没有唯一固定方程；会出现多种碎片对，但每一种都守恒 $A$ 与 $Z$。

Common exam trap常见考试陷阱 The incoming neutron counts on the left-hand side. Forgetting it makes the nucleon numbers fail to balance. Also remember: fission fragments are typically unequal in size and are themselves neutron-rich, so they are radioactive and decay further (a source of nuclear waste).入射中子在左边要计入。漏掉它会使核子数配不平。还要记住：裂变碎片通常大小不等，且本身富中子，因而具放射性并继续衰变（核废料的来源）。
Going deeper: where the 200 MeV actually goes深入：那 200 MeV 究竟去了哪里

Of the roughly $200\ \mathrm{MeV}$ per fission, about $165\ \mathrm{MeV}$ appears as kinetic energy of the two charged fragments (they fly apart under intense Coulomb repulsion), about $5\ \mathrm{MeV}$ as kinetic energy of the prompt neutrons, and the rest as gamma rays and, over longer times, beta particles, antineutrinos, and decay energy from the radioactive fragments. In a reactor, the fragment kinetic energy is what heats the fuel; the neutrinos ($\sim 10\ \mathrm{MeV}$ worth) escape entirely and represent unrecoverable energy. IB questions usually quote the round figure $\approx 200\ \mathrm{MeV}$ and ask you to scale it up to per-kilogram or per-second power outputs.

每次裂变约 $200\ \mathrm{MeV}$ 中，约 $165\ \mathrm{MeV}$ 表现为两块带电碎片的动能（它们在强库仑斥力下飞散），约 $5\ \mathrm{MeV}$ 为瞬发中子的动能，其余为伽马射线，以及在更长时间内放射性碎片衰变产生的贝塔粒子、反中微子与衰变能。在反应堆中，碎片动能用于加热燃料；中微子（约 $10\ \mathrm{MeV}$）完全逃逸，是不可回收的能量。IB 题目通常引用整数 $\approx 200\ \mathrm{MeV}$，让你换算为每千克或每秒的功率输出。

In ${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{140}_{\;54}\mathrm{Xe} + {}^{Z}_{A}\mathrm{Sr} + 2\,{}^{1}_{0}\mathrm{n}$, the strontium fragment is:在 ${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{140}_{\;54}\mathrm{Xe} + {}^{Z}_{A}\mathrm{Sr} + 2\,{}^{1}_{0}\mathrm{n}$ 中，锶碎片为：
E4.4 · Q1
${}^{96}_{40}\mathrm{Sr}$
${}^{94}_{36}\mathrm{Sr}$
${}^{94}_{38}\mathrm{Sr}$
${}^{96}_{38}\mathrm{Sr}$
Nucleon: $1 + 235 = 140 + A + 2 \Rightarrow A = 94$. Proton: $0 + 92 = 54 + Z + 0 \Rightarrow Z = 38$. So ${}^{94}_{38}\mathrm{Sr}$.核子数：$1 + 235 = 140 + A + 2 \Rightarrow A = 94$。质子数：$0 + 92 = 54 + Z + 0 \Rightarrow Z = 38$。故 ${}^{94}_{38}\mathrm{Sr}$。
Balance $A$: $236 = 140 + A + 2$ gives $A = 94$. Balance $Z$: $92 = 54 + Z$ gives $Z = 38$.配平 $A$：$236 = 140 + A + 2$ 得 $A = 94$。配平 $Z$：$92 = 54 + Z$ 得 $Z = 38$。
Most of the energy released in a single fission of $^{235}\mathrm{U}$ appears as:$^{235}\mathrm{U}$ 单次裂变释放的大部分能量表现为：
E4.4 · Q2
Gamma-ray energy.伽马射线能量。
Kinetic energy of the two fission fragments.两块裂变碎片的动能。
Kinetic energy of the emitted neutrons.放出中子的动能。
Energy carried off by neutrinos.中微子带走的能量。
The two positively charged fragments repel strongly and fly apart, carrying roughly $165\ \mathrm{MeV}$ of the $\approx 200\ \mathrm{MeV}$ as kinetic energy. This is what heats a reactor's fuel.两块带正电的碎片强烈相斥并飞散，在约 $200\ \mathrm{MeV}$ 中带走约 $165\ \mathrm{MeV}$ 的动能。这正是加热反应堆燃料的来源。
The dominant share goes to the kinetic energy of the heavy fragments, driven apart by Coulomb repulsion; neutrons and gammas take much smaller shares.主要份额是重碎片的动能，由库仑斥力驱动飞散；中子与伽马所占份额小得多。
Topic E4.5 · Chain Reactions and Critical MassTopic E4.5 · 链式反应与临界质量

Chain Reactions, Critical Mass, Controlled vs Uncontrolled链式反应、临界质量、受控与不受控 E.4 SL+HL

Chain reaction. Each fission emits $2$–$3$ neutrons. If on average one of them goes on to cause a further fission, the process sustains itself. Multiplication factor $k$. The average number of further fissions caused by the neutrons from one fission.
  • $k < 1$ subcritical: reaction dies out.
  • $k = 1$ critical: steady, self-sustaining rate (a power reactor).
  • $k > 1$ supercritical: reaction grows exponentially (a bomb, if uncontrolled).
Critical mass. The minimum mass of fissile material for which $k \ge 1$. Below it, too many neutrons escape through the surface before causing a fission. Controlled vs uncontrolled. A reactor is held at $k = 1$ by absorbing surplus neutrons; a weapon is engineered to reach $k > 1$ deliberately.
链式反应。每次裂变放出 $2$–$3$ 个中子。若平均其中一个继续引发下一次裂变，过程便能自持。 倍增因子 $k$。一次裂变放出的中子平均引发的进一步裂变次数。
  • $k < 1$ 次临界：反应熄灭。
  • $k = 1$ 临界：稳定自持（功率反应堆）。
  • $k > 1$ 超临界：反应指数增长（不受控时即为炸弹）。
临界质量。使 $k \ge 1$ 的裂变材料最小质量。低于此值时，过多中子在引发裂变前从表面逃逸。 受控与不受控。反应堆通过吸收多余中子维持 $k = 1$；武器则被设计为蓄意达到 $k > 1$。
Worked Example E4.5 (the geometry of critical mass)E4.5 例题（临界质量的几何原理）

Explain, in terms of surface area and volume, why a sphere of fissile material below its critical mass cannot sustain a chain reaction, but the same material above the critical mass can.用表面积与体积解释：为何低于临界质量的裂变材料球不能维持链式反应，而同种材料高于临界质量时却能。

Identify. Neutrons are produced throughout the volume but lost only across the surface.

识别。中子在整个体积内产生，但只从表面损失。

Reason. Production rate scales with volume ($\propto r^3$); escape rate scales with surface area ($\propto r^2$). The ratio of escape to production scales as $r^2 / r^3 = 1/r$, so it falls as the sphere grows.

推理。产生率随体积变化（$\propto r^3$）；逃逸率随表面积变化（$\propto r^2$）。逃逸与产生之比为 $r^2 / r^3 = 1/r$，故随球增大而下降。

Execute. For a small mass (small $r$), the surface-to-volume ratio is large, so a high fraction of neutrons escape before causing fission: $k < 1$, subcritical. Increase the mass and $r$, and proportionally fewer neutrons escape; at the critical mass $k = 1$ and the reaction self-sustains.

计算。对于小质量（小 $r$），表面体积比大，故高比例中子在引发裂变前逃逸：$k < 1$，次临界。增大质量与 $r$，逃逸比例相应减小；在临界质量处 $k = 1$，反应自持。

Evaluate. This is why critical mass depends on shape (a sphere minimises surface area), density (compression raises density and lowers the critical mass), and any surrounding neutron reflector (which bounces escaping neutrons back).

评估。这就是为何临界质量取决于形状（球形表面积最小）、密度（压缩提高密度、降低临界质量），以及周围的中子反射层（把逃逸中子反弹回来）。

Going deeper: prompt vs delayed neutrons and why reactors are controllable深入：瞬发中子与缓发中子，以及反应堆为何可控

If all fission neutrons were emitted instantly, a reactor at $k$ slightly above $1$ would run away in milliseconds, far too fast to control mechanically. The saving grace is that a small fraction (under $1\%$) of neutrons are delayed, released seconds later by the decay of certain fission fragments. Operators run the reactor so that it is subcritical on prompt neutrons alone and only reaches $k = 1$ when the delayed neutrons are included. This stretches the response time from milliseconds to seconds, which is what makes a power reactor mechanically controllable with moving control rods.

若所有裂变中子都瞬时放出，$k$ 略大于 $1$ 的反应堆会在毫秒内失控，远快于机械可控的速度。挽救之处在于：一小部分（不到 $1\%$）中子是缓发的，由某些裂变碎片在数秒后衰变释放。操作员让反应堆仅靠瞬发中子时处于次临界，只有计入缓发中子才达到 $k = 1$。这把响应时间从毫秒延长到秒，正是功率反应堆能用可移动控制棒机械控制的原因。

A reactor operates at a steady power level. The neutron multiplication factor $k$ is:某反应堆稳定运行在恒定功率。中子倍增因子 $k$ 为：
E4.5 · Q1
$k > 1$$k > 1$
$k = 1$$k = 1$
$k < 1$$k < 1$
$k = 0$$k = 0$
Steady power means each fission triggers exactly one further fission on average: $k = 1$, the critical condition. $k > 1$ would mean rising power; $k < 1$ would mean the reaction dying out.恒定功率意味着每次裂变平均恰好引发一次后续裂变：$k = 1$，即临界条件。$k > 1$ 表示功率上升；$k < 1$ 表示反应熄灭。
"Steady" power is the definition of critical operation, $k = 1$: one fission in, one fission out, on average."恒定"功率即临界运行的定义，$k = 1$：平均一进一出。
Why does a mass of fissile material below the critical mass fail to sustain a chain reaction?为何低于临界质量的裂变材料无法维持链式反应？
E4.5 · Q2
The fissile nuclei are too far apart to interact.裂变核相距太远无法相互作用。
The neutrons are moving too fast to be absorbed.中子运动太快无法被吸收。
There is not enough energy released per fission.每次裂变释放的能量不足。
Too high a fraction of neutrons escape through the surface before causing a fission.过高比例的中子在引发裂变前从表面逃逸。
A small mass has a large surface-area-to-volume ratio, so a large fraction of the fission neutrons leak out before they can be absorbed by another nucleus, driving $k$ below $1$.小质量的表面体积比大，故大比例裂变中子在被另一个核吸收前泄漏，使 $k$ 低于 $1$。
It is a geometry effect: below the critical mass, surface leakage of neutrons exceeds what production can replace, so $k < 1$.这是几何效应：低于临界质量时，中子表面泄漏超过产生所能补充，故 $k < 1$。
Topic E4.6 · Nuclear Reactor ComponentsTopic E4.6 · 核反应堆部件

Nuclear Reactor Components and Safety核反应堆部件与安全 E.4 SL+HL

The five parts and one job each.
PartFunction
FuelFissile material (enriched $^{235}\mathrm{U}$) where fission occurs.
ModeratorSlows fast neutrons to thermal speeds (water, graphite) so they are absorbed efficiently by $^{235}\mathrm{U}$.
Control rodsAbsorb neutrons (boron, cadmium); inserted/withdrawn to hold $k = 1$.
CoolantCarries heat from the core to a heat exchanger to raise steam.
Containment / shieldingThick concrete/steel that blocks radiation and contains material in an accident.
Energy balance. Fragment kinetic energy $\to$ heat in fuel $\to$ coolant $\to$ steam $\to$ turbine $\to$ generator.
五个部件各司一职。
部件作用
燃料发生裂变的裂变材料（浓缩 $^{235}\mathrm{U}$）。
慢化剂把快中子减速到热速度（水、石墨），使其被 $^{235}\mathrm{U}$ 高效吸收。
控制棒吸收中子（硼、镉）；通过插入/抽出维持 $k = 1$。
冷却剂把热量从堆芯带到热交换器以产生蒸汽。
安全壳 / 屏蔽层厚混凝土/钢，阻挡辐射并在事故中封存材料。
能量平衡。碎片动能 $\to$ 燃料中的热 $\to$ 冷却剂 $\to$ 蒸汽 $\to$ 汽轮机 $\to$ 发电机。
Worked Example E4.6 (reactor power from fission rate)E4.6 例题（由裂变率求反应堆功率）

A reactor sustains $6.0 \times 10^{19}$ fissions of $^{235}\mathrm{U}$ per second, each releasing $200\ \mathrm{MeV}$. Find the thermal power output. Take $1\ \mathrm{MeV} = 1.60 \times 10^{-13}\ \mathrm{J}$.某反应堆每秒维持 $6.0 \times 10^{19}$ 次 $^{235}\mathrm{U}$ 裂变，每次释放 $200\ \mathrm{MeV}$。求热功率输出。取 $1\ \mathrm{MeV} = 1.60 \times 10^{-13}\ \mathrm{J}$。

Identify. Power $=$ energy per fission $\times$ fissions per second.

识别。功率 $=$ 每次裂变能量 $\times$ 每秒裂变次数。

Set up. Energy per fission in joules:

列式。每次裂变能量（焦耳）：

$$ 200\ \mathrm{MeV} = 200 \times 1.60 \times 10^{-13} = 3.20 \times 10^{-11}\ \mathrm{J}. $$

Execute. Power:

计算。功率：

$$ P = (3.20 \times 10^{-11})(6.0 \times 10^{19}) \approx 1.9 \times 10^{9}\ \mathrm{W} = 1.9\ \mathrm{GW}. $$

Evaluate. About $1.9\ \mathrm{GW}$ thermal, a realistic large-reactor figure. The electrical output is lower (roughly a third) because the steam cycle is limited by thermodynamic efficiency.

评估。约 $1.9\ \mathrm{GW}$ 热功率，是大型反应堆的现实数值。电功率更低（约三分之一），因蒸汽循环受热力学效率限制。

Safety in one line each安全要点（各一句） Control rods drop in to shut the reaction down ($k \ll 1$) in an emergency; the coolant must keep flowing even after shutdown because radioactive fragments keep producing decay heat; the containment building is the last barrier against release of radioactive material.紧急时控制棒落入以关停反应（$k \ll 1$）；停堆后冷却剂也必须持续流动，因放射性碎片仍产生衰变热；安全壳建筑是防止放射性物质泄漏的最后屏障。
Going deeper: why a moderator is needed and why enrichment matters深入：为何需要慢化剂，以及浓缩为何重要

The neutrons emitted by fission are "fast" (high kinetic energy), but $^{235}\mathrm{U}$ captures neutrons most efficiently when they are "thermal" (slow, in thermal equilibrium with the surroundings). The moderator slows neutrons through elastic collisions with light nuclei without absorbing them, sharply raising the chance that each neutron triggers a further fission. Natural uranium is over $99\%$ non-fissile $^{238}\mathrm{U}$, which absorbs fast neutrons without fissioning; enriching the fuel to a few percent $^{235}\mathrm{U}$ ensures enough fissile nuclei are present for $k$ to reach $1$ once the neutrons are moderated. Weapons-grade material is enriched far higher and uses no moderator, relying on fast neutrons and supercritical geometry.

裂变放出的中子是"快"的（高动能），但 $^{235}\mathrm{U}$ 在中子为"热"（慢、与环境热平衡）时俘获效率最高。慢化剂通过与轻核的弹性碰撞使中子减速而不吸收它们，大幅提高每个中子引发下一次裂变的概率。天然铀中超过 $99\%$ 是不可裂变的 $^{238}\mathrm{U}$，它吸收快中子却不裂变；将燃料浓缩到百分之几的 $^{235}\mathrm{U}$，可保证在中子被慢化后有足够的裂变核使 $k$ 达到 $1$。武器级材料浓缩得高得多且不用慢化剂，依赖快中子与超临界几何结构。

The purpose of the moderator in a thermal nuclear reactor is to:热核反应堆中慢化剂的作用是：
E4.6 · Q1
Slow fast neutrons so they are absorbed more readily by $^{235}\mathrm{U}$.使快中子减速，以便更易被 $^{235}\mathrm{U}$ 吸收。
Absorb surplus neutrons to keep $k = 1$.吸收多余中子以维持 $k = 1$。
Carry heat away from the core.把热量从堆芯带走。
Shield operators from gamma radiation.屏蔽伽马辐射以保护操作员。
The moderator slows fast fission neutrons to thermal speeds through elastic collisions, where the $^{235}\mathrm{U}$ absorption cross-section is much larger. Absorbing neutrons is the control rods' job; carrying heat is the coolant's; shielding is the containment's.慢化剂通过弹性碰撞把快裂变中子减速到热速度，此时 $^{235}\mathrm{U}$ 的吸收截面大得多。吸收中子是控制棒的工作；带走热量是冷却剂的；屏蔽是安全壳的。
Distinguish the parts: moderator slows neutrons, control rods absorb them, coolant removes heat, containment shields. The moderator's job is the first of these.区分各部件：慢化剂减速中子、控制棒吸收中子、冷却剂带走热量、安全壳屏蔽。慢化剂负责其中第一项。
To reduce the reaction rate in a reactor, the operators should:要降低反应堆的反应速率，操作员应：
E4.6 · Q2
Remove the moderator entirely.完全移除慢化剂。
Increase the coolant flow rate.增大冷却剂流量。
Insert the control rods further into the core.把控制棒进一步插入堆芯。
Add more fuel rods to the core.向堆芯加入更多燃料棒。
Control rods (boron or cadmium) absorb neutrons. Pushing them further in removes more neutrons from the cycle, lowering $k$ below $1$ and slowing the reaction. Coolant flow controls temperature, not reaction rate; adding fuel would raise it.控制棒（硼或镉）吸收中子。把它们更深插入会从循环中移走更多中子，使 $k$ 低于 $1$ 并减缓反应。冷却剂流量控制温度而非反应速率；加燃料反而会提高速率。
Reaction rate is set by the neutron balance. Inserting neutron-absorbing control rods lowers $k$; coolant flow and fuel quantity do not directly control $k$ this way.反应速率由中子平衡决定。插入吸收中子的控制棒降低 $k$；冷却剂流量与燃料量不以这种方式直接控制 $k$。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Unit discipline (every paper)单位纪律（每张试卷）
  • Keep masses in $\mathrm{u}$ until the last step, then multiply by $931.5$ for energy in $\mathrm{MeV}$. Converting to $\mathrm{kg}$ early invites arithmetic slips.
  • 把质量保持为 $\mathrm{u}$ 直到最后一步，再乘以 $931.5$ 得 $\mathrm{MeV}$ 能量。过早转成 $\mathrm{kg}$ 容易算错。
  • Carry enough significant figures in the mass defect. $\Delta m$ is a small difference of large numbers; rounding the input masses too soon destroys the answer.
  • 质量亏损要保留足够有效数字。$\Delta m$ 是大数之间的小差；过早对输入质量取整会毁掉答案。
Balancing nuclear equations配平核方程
  • Conserve top numbers (nucleon $A$) and bottom numbers (proton $Z$) separately. Two independent sums; both must balance.
  • 分别守恒上标（核子数 $A$）与下标（质子数 $Z$）。两个独立等式，都要配平。
  • Count the incoming neutron on the left. The most common error is leaving the trigger neutron out of the nucleon balance.
  • 把入射中子计入左边。最常见的错误是把触发中子漏在核子数平衡之外。
Energy-release reasoning (Paper 2 explain)能量释放推理（Paper 2 简答）
  • Anchor every "why does this release energy?" answer in the $E_b / A$ curve. Products higher on the curve $\Rightarrow$ more tightly bound $\Rightarrow$ energy released.
  • 每个"为何释放能量？"都要锚定在 $E_b / A$ 曲线上。产物在曲线上更高 $\Rightarrow$ 结合更紧 $\Rightarrow$ 释放能量。
  • Use "energy released $=$ BE(products) $-$ BE(reactants)" for curve data, or $\Delta m\, c^2$ for mass data. Pick whichever the question hands you.
  • 曲线数据用"释放能量 $=$ 结合能(产物) $-$ 结合能(反应物)"，质量数据用 $\Delta m\, c^2$。题目给哪个就用哪个。
Reactor descriptive marks反应堆描述题得分
  • Name the part and state its job in one clause. "Moderator: slows neutrons to thermal speeds." Vague answers like "controls the reaction" score nothing.
  • 说出部件名并用一个分句说明其作用。"慢化剂：把中子减速到热速度。"像"控制反应"这样的含糊作答得不到分。
  • Tie "critical" to $k = 1$ explicitly. Markschemes credit the neutron-multiplication language, not just "self-sustaining".
  • 把"临界"明确联系到 $k = 1$。评分认可中子倍增的表述，而不仅是"自持"。
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Mass-energy equivalence?质能等价公式？
$$E = m c^2$$
Energy equivalent of $1\ \mathrm{u}$?$1\ \mathrm{u}$ 的能量当量？
$$1\ \mathrm{u} = 931.5\ \mathrm{MeV}\,c^{-2}$$
Mass defect $\Delta m$?质量亏损 $\Delta m$？
$$\Delta m = (Z m_p + N m_n) - M_{\text{nuc}}$$
Binding energy?结合能？
$$E_b = \Delta m\, c^2$$
Binding energy per nucleon?比结合能？
$$\frac{E_b}{A} = \frac{\Delta m\, c^2}{A}$$
Most stable nucleus?最稳定的核？
Iron-56, peak of the $E_b / A$ curve.铁-56，$E_b / A$ 曲线顶峰。
Condition for a reaction to release energy?反应释放能量的条件？
Products have higher $E_b / A$ than reactants.产物的 $E_b / A$ 高于反应物。
Induced fission of $^{235}\mathrm{U}$?$^{235}\mathrm{U}$ 的诱发裂变？
$${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}$$
Energy per fission of $^{235}\mathrm{U}$?$^{235}\mathrm{U}$ 每次裂变能量？
$\approx 200\ \mathrm{MeV}$, mostly fragment kinetic energy.$\approx 200\ \mathrm{MeV}$，主要为碎片动能。
Conservation rules in a fission equation?裂变方程的守恒规则？
Nucleon number $A$ and proton number $Z$ each conserved.核子数 $A$ 与质子数 $Z$ 各自守恒。
Critical condition for a reactor?反应堆的临界条件？
$$k = 1$$Steady, self-sustaining rate.稳定自持的速率。
Critical mass?临界质量？
Minimum fissile mass for which $k \ge 1$.使 $k \ge 1$ 的最小裂变材料质量。
Job of the moderator?慢化剂的作用？
Slows neutrons to thermal speeds for efficient absorption.把中子减速到热速度以便高效吸收。
Job of the control rods?控制棒的作用？
Absorb neutrons (boron/cadmium) to hold $k = 1$.吸收中子（硼/镉）以维持 $k = 1$。
Mixed Practice综合练习

Unit E.4 Practice Quiz单元 E.4 练习测验

A fission reaction has a total mass defect of $0.21\ \mathrm{u}$. The energy released is closest to:某裂变反应总质量亏损为 $0.21\ \mathrm{u}$。释放的能量最接近：
Q1
$93\ \mathrm{MeV}$
$140\ \mathrm{MeV}$
$196\ \mathrm{MeV}$
$931\ \mathrm{MeV}$
$E = \Delta m\, c^2 = 0.21 \times 931.5 \approx 196\ \mathrm{MeV}$, the familiar $\approx 200\ \mathrm{MeV}$ per fission.$E = \Delta m\, c^2 = 0.21 \times 931.5 \approx 196\ \mathrm{MeV}$，即熟悉的每次裂变 $\approx 200\ \mathrm{MeV}$。
Multiply the mass defect in $\mathrm{u}$ by $931.5$: $0.21 \times 931.5 \approx 196\ \mathrm{MeV}$.把以 $\mathrm{u}$ 计的质量亏损乘以 $931.5$：$0.21 \times 931.5 \approx 196\ \mathrm{MeV}$。
In ${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{137}_{\;52}\mathrm{Te} + {}^{A}_{Z}\mathrm{Zr} + 2\,{}^{1}_{0}\mathrm{n}$, the zirconium fragment is:在 ${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{137}_{\;52}\mathrm{Te} + {}^{A}_{Z}\mathrm{Zr} + 2\,{}^{1}_{0}\mathrm{n}$ 中，锆碎片为：
Q2
${}^{99}_{40}\mathrm{Zr}$
${}^{97}_{40}\mathrm{Zr}$
${}^{97}_{38}\mathrm{Zr}$
${}^{95}_{42}\mathrm{Zr}$
Nucleon: $236 = 137 + A + 2 \Rightarrow A = 97$. Proton: $92 = 52 + Z \Rightarrow Z = 40$, which is zirconium. So ${}^{97}_{40}\mathrm{Zr}$.核子数：$236 = 137 + A + 2 \Rightarrow A = 97$。质子数：$92 = 52 + Z \Rightarrow Z = 40$，即锆。故 ${}^{97}_{40}\mathrm{Zr}$。
Balance $A$: $236 = 137 + A + 2 \Rightarrow A = 97$. Balance $Z$: $92 = 52 + Z \Rightarrow Z = 40$.配平 $A$：$236 = 137 + A + 2 \Rightarrow A = 97$。配平 $Z$：$92 = 52 + Z \Rightarrow Z = 40$。
A nucleus of mass number $A = 240$ with $E_b / A = 7.5\ \mathrm{MeV}$ fissions into two equal fragments each with $E_b / A = 8.4\ \mathrm{MeV}$. Energy released:质量数 $A = 240$、$E_b / A = 7.5\ \mathrm{MeV}$ 的核裂成两块相等碎片，每块 $E_b / A = 8.4\ \mathrm{MeV}$。释放能量：
Q3
$90\ \mathrm{MeV}$
$108\ \mathrm{MeV}$
$180\ \mathrm{MeV}$
$216\ \mathrm{MeV}$
Before: $240 \times 7.5 = 1800\ \mathrm{MeV}$. After: $240 \times 8.4 = 2016\ \mathrm{MeV}$ (total nucleons unchanged). Released $= 2016 - 1800 = 216\ \mathrm{MeV}$.前：$240 \times 7.5 = 1800\ \mathrm{MeV}$。后：$240 \times 8.4 = 2016\ \mathrm{MeV}$（总核子数不变）。释放 $= 2016 - 1800 = 216\ \mathrm{MeV}$。
Total nucleons stay at $240$, so energy released $= 240 \times (8.4 - 7.5) = 240 \times 0.9 = 216\ \mathrm{MeV}$.总核子数保持 $240$，故释放能量 $= 240 \times (8.4 - 7.5) = 216\ \mathrm{MeV}$。
A reactor runs with neutron multiplication factor $k = 1.02$. Without intervention, its power output will:某反应堆以中子倍增因子 $k = 1.02$ 运行。若不干预，其功率输出将：
Q4
Increase over time (supercritical).随时间增大（超临界）。
Stay constant (critical).保持不变（临界）。
Decrease over time (subcritical).随时间减小（次临界）。
Drop instantly to zero.瞬间降为零。
$k > 1$ means each generation of fissions is larger than the last, so power rises. Operators would insert control rods to bring $k$ back to $1$.$k > 1$ 意味着每一代裂变都比上一代多，故功率上升。操作员会插入控制棒把 $k$ 拉回 $1$。
$k = 1.02 > 1$ is supercritical: the fission rate grows generation over generation, so power increases until controlled.$k = 1.02 > 1$ 为超临界：裂变率逐代增长，故在受控前功率上升。
Why can both fission of heavy nuclei and fusion of light nuclei release energy?为何重核裂变与轻核聚变都能释放能量？
Q5
Both increase the total number of nucleons.两者都增加总核子数。
Both move the products toward the iron-56 peak, raising the binding energy per nucleon.两者都使产物趋向铁-56 峰，提高比结合能。
Both convert protons directly into neutrons.两者都把质子直接转化为中子。
Both reduce the binding energy per nucleon of the products.两者都降低产物的比结合能。
Heavy nuclei sit to the right of the iron-56 peak, light nuclei to the left. Splitting heavy nuclei and joining light nuclei both shift the products closer to the peak, raising $E_b / A$ and releasing the difference as energy.重核在铁-56 峰右侧，轻核在左侧。分裂重核与结合轻核都使产物更靠近峰，提高 $E_b / A$ 并以能量形式释放差额。
The unifying idea is the $E_b / A$ curve: any reaction whose products are closer to the iron-56 peak releases energy, whether by fission or fusion.统一思路是 $E_b / A$ 曲线：凡产物更靠近铁-56 峰的反应都释放能量，无论裂变还是聚变。
Self-check自查

Readiness Checklist备考清单

Tick each item when you can do it cold, without notes, on your first attempt.

每一条都要"裸做"做对（不看笔记、一次过）才打勾。

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IB Physics HL · Unit E.4 Fission · v1 · First Assessment 2025IB 物理 HL · 单元 E.4 核裂变 · v1 · 首次考核 2025