Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus E4.1 to E4.6考纲 E4.1 至 E4.6PHYSICS HL
A mass of $0.20\ \mathrm{u}$ is converted entirely into energy. (a) state $E = \Delta m c^2$ and find the energy in $\mathrm{MeV}$; (b) express it in joules.$0.20\ \mathrm{u}$ 质量被完全转化为能量。(a) 写出 $E = \Delta m c^2$ 并求以 $\mathrm{MeV}$ 计的能量；(b) 以焦耳表示。
Mass and energy are linked by $E = \Delta m\, c^2$. With the mass in $\mathrm{u}$, multiply by the data-booklet equivalent $1\ \mathrm{u} = 931.5\ \mathrm{MeV}\,c^{-2}$: (M1)
$$ E = 0.20 \times 931.5 = 186.3 \approx 186\ \mathrm{MeV}. $$(A1)
Use $E = \Delta m\, c^2$ in SI units, with $\Delta m = 0.20 \times 1.66\times 10^{-27}\ \mathrm{kg}$: (M1)
$$ E = (0.20)(1.66\times 10^{-27})(3.00\times 10^{8})^2 \approx 2.99\times 10^{-11}\ \mathrm{J}. $$(A1)
质量与能量由 $E = \Delta m\, c^2$ 联系。质量以 $\mathrm{u}$ 计时，乘以数据手册当量 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}\,c^{-2}$：(M1)
$$ E = 0.20 \times 931.5 = 186.3 \approx 186\ \mathrm{MeV}. $$(A1)
用国际单位制的 $E = \Delta m\, c^2$，其中 $\Delta m = 0.20 \times 1.66\times 10^{-27}\ \mathrm{kg}$：(M1)
$$ E = (0.20)(1.66\times 10^{-27})(3.00\times 10^{8})^2 \approx 2.99\times 10^{-11}\ \mathrm{J}. $$(A1)
${}^{7}_{3}\mathrm{Li}$ has nuclear mass $7.01436\ \mathrm{u}$; $m_p = 1.00728$, $m_n = 1.00867\ \mathrm{u}$. (a) define mass defect; (b) calculate it; (c) total binding energy and BE per nucleon.${}^{7}_{3}\mathrm{Li}$ 核质量 $7.01436\ \mathrm{u}$；$m_p = 1.00728$、$m_n = 1.00867\ \mathrm{u}$。(a) 定义质量亏损；(b) 计算之；(c) 总结合能与比结合能。
The mass defect is the difference between the total mass of the separated, free nucleons and the actual mass of the bound nucleus: $\Delta m = (Z m_p + N m_n) - M_{\text{nucleus}}$. (B1)
Lithium-7 has $Z = 3$ protons and $N = 4$ neutrons: (M1)
$$ \Delta m = [3(1.00728) + 4(1.00867)] - 7.01436 = 7.05652 - 7.01436. $$ $$ \Delta m = 0.04216\ \mathrm{u}. $$(A1)
Convert the defect to energy with $1\ \mathrm{u} = 931.5\ \mathrm{MeV}$: (M1)
$$ E_b = 0.04216 \times 931.5 \approx 39.3\ \mathrm{MeV}. $$(A1)
Divide by the nucleon number $A = 7$:
$$ \frac{E_b}{A} = \frac{39.3}{7} \approx 5.61\ \mathrm{MeV}. $$(A1)
质量亏损是分离的自由核子总质量与结合核实际质量之差：$\Delta m = (Z m_p + N m_n) - M_{\text{nucleus}}$。(B1)
锂-7 有 $Z = 3$ 个质子与 $N = 4$ 个中子：(M1)
$$ \Delta m = [3(1.00728) + 4(1.00867)] - 7.01436 = 7.05652 - 7.01436. $$ $$ \Delta m = 0.04216\ \mathrm{u}. $$(A1)
用 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}$ 把亏损转成能量：(M1)
$$ E_b = 0.04216 \times 931.5 \approx 39.3\ \mathrm{MeV}. $$(A1)
除以核子数 $A = 7$：
$$ \frac{E_b}{A} = \frac{39.3}{7} \approx 5.61\ \mathrm{MeV}. $$(A1)
The BE-per-nucleon curve peaks near iron-56 at $\approx 8.8\ \mathrm{MeV}$. (a) define BE per nucleon and write it in terms of $\Delta m$; (b) why iron-56 is most stable; (c) the curve condition for energy release.比结合能曲线在铁-56 附近达峰，约 $8.8\ \mathrm{MeV}$。(a) 定义比结合能并用 $\Delta m$ 表示；(b) 铁-56 为何最稳定；(c) 释放能量的曲线条件。
It is the total binding energy of a nucleus shared equally over its nucleons, the fair measure for comparing the stability of nuclei of different sizes. (B1)
$$ \frac{E_b}{A} = \frac{\Delta m\, c^2}{A}, \qquad A = Z + N. $$(A1)
The peak means iron-56 has the greatest binding energy per nucleon of any nuclide, so its nucleons are the most tightly held. (R1)
Any reaction starting from iron-56, whether splitting or joining, would move to a lower point on the curve, which requires an input of energy rather than releasing it; hence iron-56 releases energy by neither fission nor fusion. (R1)
Energy is released only when the products lie higher on the curve than the reactants, that is, when the binding energy per nucleon increases. (A1)
The increase in binding energy means a more tightly bound, lower-mass final state, and the released binding energy appears as kinetic energy of the products. (R1)
它是核的总结合能平均分摊到每个核子上，是比较不同大小核稳定性的公平量度。(B1)
$$ \frac{E_b}{A} = \frac{\Delta m\, c^2}{A}, \qquad A = Z + N. $$(A1)
峰值意味着铁-56 在所有核素中具有最大的比结合能，故其核子结合得最紧。(R1)
从铁-56 出发的任何反应，无论分裂还是结合，都会移动到曲线上更低的点，这需要输入能量而非释放能量；故铁-56 既不裂变也不聚变释放能量。(R1)
只有当产物在曲线上高于反应物，即比结合能增大时，才释放能量。(A1)
比结合能增大意味着结合更紧、质量更低的末态，释放的结合能表现为产物的动能。(R1)
${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{140}_{\;54}\mathrm{Xe} + {}^{A}_{Z}\mathrm{Sr} + x\,{}^{1}_{0}\mathrm{n}$. (a) define induced fission and why a slow neutron; (b) find $A$, $Z$, $x$; (c) energy per fission.${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{140}_{\;54}\mathrm{Xe} + {}^{A}_{Z}\mathrm{Sr} + x\,{}^{1}_{0}\mathrm{n}$。(a) 定义诱发裂变及为何用慢中子；(b) 求 $A$、$Z$、$x$；(c) 每次裂变能量。
Induced fission is the splitting of a heavy nucleus triggered by its absorption of a neutron, which leaves it in a highly excited state that breaks apart almost immediately. (B1)
A slow (thermal) neutron is used because $^{235}\mathrm{U}$ captures neutrons far more efficiently at low (thermal) speeds, so the slow neutron is much more likely to be absorbed and cause a fission. (R1)
Conserve nucleon number (top) and proton number (bottom) separately. (M1)
Nucleon number: $1 + 235 = 140 + A + x$, so $A + x = 96$. Proton number: $0 + 92 = 54 + Z + 0$, so $Z = 38$. (A1)
Strontium has $Z = 38$, fixing the element; the standard fragment here is ${}^{94}_{38}\mathrm{Sr}$, giving $A = 94$ and hence $x = 96 - 94 = 2$ neutrons. (A1)
A single fission of $^{235}\mathrm{U}$ releases approximately $200\ \mathrm{MeV}$. (B1)
诱发裂变是重核因吸收一个中子而被触发的分裂，吸收使其处于高激发态，几乎立即裂开。(B1)
使用慢（热）中子，是因为 $^{235}\mathrm{U}$ 在低（热）速度下俘获中子的效率高得多，故慢中子更可能被吸收并引发裂变。(R1)
分别守恒核子数（上标）与质子数（下标）。(M1)
核子数：$1 + 235 = 140 + A + x$，故 $A + x = 96$。质子数：$0 + 92 = 54 + Z + 0$，故 $Z = 38$。(A1)
锶的 $Z = 38$，从而确定元素；此处标准碎片为 ${}^{94}_{38}\mathrm{Sr}$，给出 $A = 94$，进而 $x = 96 - 94 = 2$ 个中子。(A1)
$^{235}\mathrm{U}$ 单次裂变释放约 $200\ \mathrm{MeV}$。(B1)
Reactor parts: fuel, moderator, control rods, coolant, containment. (a) function of moderator, control rods, coolant; (b) enrichment and why natural uranium needs it; (c) why fission fragments are radioactive waste.反应堆部件：燃料、慢化剂、控制棒、冷却剂、安全壳。(a) 慢化剂、控制棒、冷却剂的作用；(b) 浓缩及天然铀为何需要；(c) 裂变碎片为何是放射性废料。
Moderator: slows the fast fission neutrons to thermal speeds (by elastic collisions) so they are absorbed efficiently by $^{235}\mathrm{U}$. (A1)
Control rods: absorb neutrons (boron or cadmium) and are inserted or withdrawn to keep the multiplication factor at $k = 1$. (A1)
Coolant: carries heat away from the core to a heat exchanger, where it raises steam to drive the turbine. (A1)
Enrichment is increasing the proportion of the fissile isotope $^{235}\mathrm{U}$ above its natural abundance. (B1)
Natural uranium is over $99\%$ $^{238}\mathrm{U}$, which is not fissile by thermal neutrons and tends to absorb them without fissioning. (R1)
Raising the $^{235}\mathrm{U}$ fraction to a few percent ensures enough fissile nuclei are present for the chain reaction to reach $k = 1$ once the neutrons are moderated. (R1)
The two fission fragments are neutron-rich compared with stable nuclei of the same proton number. (B1)
They are therefore unstable and undergo further radioactive decay (mostly beta and gamma), so the spent fuel remains radioactive long after it leaves the reactor. (R1)
慢化剂：把快裂变中子（通过弹性碰撞）减速到热速度，使其被 $^{235}\mathrm{U}$ 高效吸收。(A1)
控制棒：吸收中子（硼或镉），通过插入或抽出维持倍增因子 $k = 1$。(A1)
冷却剂：把热量从堆芯带到热交换器，在那里产生蒸汽驱动汽轮机。(A1)
浓缩是把可裂变同位素 $^{235}\mathrm{U}$ 的比例提高到其天然丰度之上。(B1)
天然铀中超过 $99\%$ 是 $^{238}\mathrm{U}$，它不能被热中子裂变，反而倾向于吸收中子却不裂变。(R1)
把 $^{235}\mathrm{U}$ 比例提高到百分之几，可保证有足够可裂变核，使链式反应在中子被慢化后达到 $k = 1$。(R1)
两块裂变碎片相比同质子数的稳定核而言是富中子的。(B1)
因而它们不稳定，会继续放射性衰变（主要为贝塔与伽马），故乏燃料在离开反应堆后长期仍具放射性。(R1)
${}^{236}\mathrm{U}$ ($E_b/A = 7.6$) fissions to ${}^{141}\mathrm{Ba}$ ($8.3$) and ${}^{92}\mathrm{Kr}$ ($8.5$), values in $\mathrm{MeV}$. (a) why products lie higher; (b) total BE of reactant and products; (c) energy released; (d) % uncertainty in the $^{236}\mathrm{U}$ reading ($\pm 0.1$); (e) main form of the released energy.${}^{236}\mathrm{U}$（$E_b/A = 7.6$）裂变为 ${}^{141}\mathrm{Ba}$（$8.3$）与 ${}^{92}\mathrm{Kr}$（$8.5$），单位 $\mathrm{MeV}$。(a) 产物为何更高；(b) 反应物与产物总结合能；(c) 释放能量；(d) $^{236}\mathrm{U}$ 读数的百分比不确定度（$\pm 0.1$）；(e) 释放能量的主要形式。
The fragments ${}^{141}\mathrm{Ba}$ and ${}^{92}\mathrm{Kr}$ have larger binding energy per nucleon ($8.3$, $8.5\ \mathrm{MeV}$) than ${}^{236}\mathrm{U}$ ($7.6\ \mathrm{MeV}$). (R1)
A heavy nucleus splitting moves its nucleons up the curve toward the iron-56 peak, into more tightly bound, lower-mass states; this is why the reaction releases energy. (R1)
Total binding energy $= (E_b / A) \times A$ for each nuclide. (M1)
$$ E_{b,\text{reactant}} = 236 \times 7.6 = 1793.6\ \mathrm{MeV}. $$(A1)
$$ E_{b,\text{products}} = (141 \times 8.3) + (92 \times 8.5) = 1170.3 + 782.0 = 1952.3\ \mathrm{MeV}. $$(A1)
Energy released is the gain in total binding energy (products minus reactant): (M1)
$$ E = 1952.3 - 1793.6 = 158.7 \approx 159\ \mathrm{MeV}. $$(A1)
For the ${}^{236}\mathrm{U}$ reading of $7.6\ \mathrm{MeV}$ with $\pm 0.1\ \mathrm{MeV}$: (M1)
$$ \frac{0.1}{7.6}\times 100\% \approx 1.3\% \approx 1\%. $$(A1)
The released energy appears mainly as kinetic energy of the two fission fragments. (A1)
The two fragments are both positively charged and are created very close together, so they fly apart under intense electrostatic (Coulomb) repulsion. (R1)
This carries off roughly $165\ \mathrm{MeV}$ of the total; the prompt neutrons and gamma rays take much smaller shares. (R1)
碎片 ${}^{141}\mathrm{Ba}$ 与 ${}^{92}\mathrm{Kr}$ 的比结合能（$8.3$、$8.5\ \mathrm{MeV}$）大于 ${}^{236}\mathrm{U}$（$7.6\ \mathrm{MeV}$）。(R1)
重核分裂使其核子沿曲线向铁-56 峰上移，进入结合更紧、质量更低的状态；这就是反应释放能量的原因。(R1)
每个核素的总结合能 $= (E_b / A) \times A$。(M1)
$$ E_{b,\text{反应物}} = 236 \times 7.6 = 1793.6\ \mathrm{MeV}. $$(A1)
$$ E_{b,\text{产物}} = (141 \times 8.3) + (92 \times 8.5) = 1170.3 + 782.0 = 1952.3\ \mathrm{MeV}. $$(A1)
释放能量是总结合能的增量（产物减反应物）：(M1)
$$ E = 1952.3 - 1793.6 = 158.7 \approx 159\ \mathrm{MeV}. $$(A1)
对 ${}^{236}\mathrm{U}$ 读数 $7.6\ \mathrm{MeV}$，$\pm 0.1\ \mathrm{MeV}$：(M1)
$$ \frac{0.1}{7.6}\times 100\% \approx 1.3\% \approx 1\%. $$(A1)
释放的能量主要表现为两块裂变碎片的动能。(A1)
两块碎片都带正电且生成时彼此非常靠近，故在强烈的静电（库仑）斥力下飞散。(R1)
这带走总能量中约 $165\ \mathrm{MeV}$；瞬发中子与伽马射线所占份额小得多。(R1)
Cumulative fissions logged: $7.5\times 10^{19}$ each second (linear), each releasing $200\ \mathrm{MeV}$. (a) graph shape and meaning of gradient; (b) fission rate; (c) convert $200\ \mathrm{MeV}$ to joules and find thermal power; (d) why the line is straight, in terms of $k$.记录的累计裂变：每秒 $7.5\times 10^{19}$（线性），每次释放 $200\ \mathrm{MeV}$。(a) 图形与斜率含义；(b) 裂变率；(c) 把 $200\ \mathrm{MeV}$ 转成焦耳并求热功率；(d) 用 $k$ 说明直线原因。
The cumulative number of fissions increases by equal amounts in equal times, so the graph is a straight line through the origin. (A1)
Its gradient is the change in fission count per unit time, that is, the fission rate (fissions per second). (A1)
Read the gradient from two well-separated points, e.g. $(0,\,0)$ and $(4.0,\,30.0\times 10^{19})$: (M1)
$$ \text{rate} = \frac{30.0\times 10^{19} - 0}{4.0 - 0} = 7.5\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
Energy per fission in joules: (M1)
$$ 200\ \mathrm{MeV} = 200 \times 1.60\times 10^{-13} = 3.20\times 10^{-11}\ \mathrm{J}. $$Power is energy per fission times fissions per second: (M1)
$$ P = (3.20\times 10^{-11})(7.5\times 10^{19}) = 2.4\times 10^{9}\ \mathrm{W} = 2.4\ \mathrm{GW}. $$(A1)
A straight line means the fission rate is constant in time. (A1)
This is the critical condition $k = 1$: on average each fission triggers exactly one further fission, so the number of fissions per second neither grows nor decays. (R1)
If $k > 1$ the rate would rise (the graph would curve upward) and if $k < 1$ it would fall away; the operator holds $k = 1$ with the control rods to keep the power steady. (R1)
累计裂变次数在相等时间内增加相等的量，故图线为过原点的直线。(A1)
其斜率是单位时间内裂变计数的变化，即裂变率（每秒裂变次数）。(A1)
用相距较远的两点读斜率，如 $(0,\,0)$ 与 $(4.0,\,30.0\times 10^{19})$：(M1)
$$ \text{率} = \frac{30.0\times 10^{19} - 0}{4.0 - 0} = 7.5\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
每次裂变能量（焦耳）：(M1)
$$ 200\ \mathrm{MeV} = 200 \times 1.60\times 10^{-13} = 3.20\times 10^{-11}\ \mathrm{J}. $$功率为每次裂变能量乘以每秒裂变次数：(M1)
$$ P = (3.20\times 10^{-11})(7.5\times 10^{19}) = 2.4\times 10^{9}\ \mathrm{W} = 2.4\ \mathrm{GW}. $$(A1)
直线意味着裂变率随时间恒定。(A1)
这是临界条件 $k = 1$：平均每次裂变恰好引发一次后续裂变，故每秒裂变次数既不增长也不衰减。(R1)
若 $k > 1$ 速率会上升（图线上凸），若 $k < 1$ 会衰减；操作员用控制棒维持 $k = 1$ 以保持功率恒定。(R1)
${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{141}_{\;56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}$; $m(\mathrm{U}) = 235.04393$, $m(\mathrm{Ba}) = 140.91440$, $m(\mathrm{Kr}) = 91.92617$, $m_n = 1.00867\ \mathrm{u}$. (a) check conservation; (b) $\Delta m$; (c) energy in MeV and J; (d) fissions/s for $1.8\ \mathrm{GW}$; (e) why electrical power is less.${}^{1}_{0}\mathrm{n} + {}^{235}_{\;92}\mathrm{U} \to {}^{141}_{\;56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}$；$m(\mathrm{U}) = 235.04393$、$m(\mathrm{Ba}) = 140.91440$、$m(\mathrm{Kr}) = 91.92617$、$m_n = 1.00867\ \mathrm{u}$。(a) 验证守恒；(b) $\Delta m$；(c) 能量 MeV 与 J；(d) $1.8\ \mathrm{GW}$ 所需裂变/秒；(e) 电功率为何更小。
Nucleon number: $1 + 235 = 236$ on the left; $141 + 92 + 3(1) = 236$ on the right. Balanced. (A1)
Proton number: $0 + 92 = 92$ on the left; $56 + 36 + 3(0) = 92$ on the right. Balanced. (A1)
Total mass before (one neutron plus the uranium): (M1)
$$ m_{\text{before}} = 1.00867 + 235.04393 = 236.05260\ \mathrm{u}. $$Total mass after (the two fragments plus three neutrons): (M1)
$$ m_{\text{after}} = 140.91440 + 91.92617 + 3(1.00867) = 235.86658\ \mathrm{u}. $$ $$ \Delta m = 236.05260 - 235.86658 = 0.18602\ \mathrm{u}. $$(A1)
Convert the mass difference with $1\ \mathrm{u} = 931.5\ \mathrm{MeV}$: (M1)
$$ E = 0.18602 \times 931.5 \approx 173\ \mathrm{MeV}. $$(A1)
In joules, $E = 173 \times 1.60\times 10^{-13} \approx 2.77\times 10^{-11}\ \mathrm{J}$. (A1)
The fission rate is the power divided by the energy released per fission: (M1)
$$ N = \frac{P}{E} = \frac{1.8\times 10^{9}}{2.77\times 10^{-11}} \approx 6.5\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
The fission energy first becomes heat, which a reactor turns into electricity through a steam turbine and generator. (B1)
That heat-engine stage is limited by thermodynamic efficiency, so only a fraction (roughly a third) of the thermal power is delivered as electrical power; the rest is rejected as waste heat. (R1)
核子数：左边 $1 + 235 = 236$；右边 $141 + 92 + 3(1) = 236$。配平。(A1)
质子数：左边 $0 + 92 = 92$；右边 $56 + 36 + 3(0) = 92$。配平。(A1)
反应前总质量（一个中子加铀）：(M1)
$$ m_{\text{前}} = 1.00867 + 235.04393 = 236.05260\ \mathrm{u}. $$反应后总质量（两块碎片加三个中子）：(M1)
$$ m_{\text{后}} = 140.91440 + 91.92617 + 3(1.00867) = 235.86658\ \mathrm{u}. $$ $$ \Delta m = 236.05260 - 235.86658 = 0.18602\ \mathrm{u}. $$(A1)
用 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}$ 换算质量差：(M1)
$$ E = 0.18602 \times 931.5 \approx 173\ \mathrm{MeV}. $$(A1)
以焦耳计，$E = 173 \times 1.60\times 10^{-13} \approx 2.77\times 10^{-11}\ \mathrm{J}$。(A1)
裂变率为功率除以每次裂变释放的能量：(M1)
$$ N = \frac{P}{E} = \frac{1.8\times 10^{9}}{2.77\times 10^{-11}} \approx 6.5\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
裂变能量首先变为热，反应堆通过蒸汽汽轮机与发电机把热转为电。(B1)
该热机环节受热力学效率限制，故只有一部分（约三分之一）热功率作为电功率输出；其余作为废热排放。(R1)
Each ${}^{235}\mathrm{U}$ fission gives $\approx 2.5$ neutrons. (a) define $k$ and its value at steady power; (b) define critical mass and explain it via surface-area-to-volume ratio; (c) distinguish controlled and uncontrolled reactions via $k$.每次 ${}^{235}\mathrm{U}$ 裂变给出约 $2.5$ 个中子。(a) 定义 $k$ 及稳定功率时的取值；(b) 定义临界质量并用表面积体积比解释；(c) 用 $k$ 区分受控与不受控反应。
The neutron multiplication factor $k$ is the average number of neutrons from one fission that go on to cause a further fission. (B1)
At steady power the rate is constant, so each fission produces exactly one further fission on average: $k = 1$. (A1)
The critical mass is the minimum mass of fissile material for which the chain reaction is just self-sustaining, that is, $k \ge 1$. (B1)
Neutrons are produced throughout the volume (rate $\propto r^3$) but are lost only across the surface (rate $\propto r^2$), so the fraction escaping scales as $r^2 / r^3 = 1 / r$. (M1)
For a small mass ($r$ small) this surface-to-volume ratio is large, so a high fraction of neutrons escape before causing a fission. (R1)
Too few neutrons then remain to sustain the chain, giving $k < 1$, and the reaction dies out; only at or above the critical mass do enough neutrons stay in to reach $k = 1$. (R1)
A power reactor is held at $k = 1$ by absorbing surplus neutrons in control rods, so the power stays steady (a controlled chain reaction). (A1)
A fission weapon is engineered to reach $k > 1$, so the fission rate grows exponentially and releases its energy in a single rapid, uncontrolled burst. (A1)
中子倍增因子 $k$ 是一次裂变放出的中子中平均引发下一次裂变的个数。(B1)
稳定功率时速率恒定，故平均每次裂变恰好产生一次后续裂变：$k = 1$。(A1)
临界质量是链式反应恰好能自持（即 $k \ge 1$）所需的裂变材料最小质量。(B1)
中子在整个体积内产生（率 $\propto r^3$），但只从表面损失（率 $\propto r^2$），故逃逸比例按 $r^2 / r^3 = 1 / r$ 变化。(M1)
对小质量（$r$ 小），该表面体积比大，故高比例中子在引发裂变前逃逸。(R1)
于是剩余中子太少无法维持链式，给出 $k < 1$，反应熄灭；只有达到或超过临界质量，才有足够中子留下以达到 $k = 1$。(R1)
功率反应堆通过控制棒吸收多余中子维持 $k = 1$，故功率保持恒定（受控链式反应）。(A1)
裂变武器被设计为达到 $k > 1$，故裂变率指数增长，在一次快速、不受控的爆发中释放能量。(A1)
Reactor: $3.0\ \mathrm{GW}$ thermal, $200\ \mathrm{MeV}$ per fission, $33\%$ efficiency, $N_A = 6.02\times 10^{23}$, $M = 235\ \mathrm{g\,mol^{-1}}$, $1\ \mathrm{yr} = 3.15\times 10^{7}\ \mathrm{s}$. (a) electrical power; (b) fissions/s; (c) mass of $^{235}\mathrm{U}$ fissioned per year; (d) why spent fuel stays hazardous.反应堆：$3.0\ \mathrm{GW}$ 热，每次裂变 $200\ \mathrm{MeV}$，效率 $33\%$，$N_A = 6.02\times 10^{23}$，$M = 235\ \mathrm{g\,mol^{-1}}$，$1\ \mathrm{yr} = 3.15\times 10^{7}\ \mathrm{s}$。(a) 电功率；(b) 裂变/秒；(c) 每年裂变的 $^{235}\mathrm{U}$ 质量；(d) 乏燃料为何长期危险。
Electrical power is the efficiency times the thermal power: $P_{\text{elec}} = 0.33 \times 3.0\times 10^{9} \approx 0.99\times 10^{9}\ \mathrm{W} = 0.99\ \mathrm{GW}$. (A1)
Each fission gives $200\ \mathrm{MeV} = 3.20\times 10^{-11}\ \mathrm{J}$. The thermal power requires: (M1)
$$ N = \frac{P_{\text{thermal}}}{E} = \frac{3.0\times 10^{9}}{3.20\times 10^{-11}} \approx 9.4\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
Number of fissions in one year: $N_{\text{yr}} = (9.375\times 10^{19})(3.15\times 10^{7}) \approx 2.95\times 10^{27}$. (M1)
Convert nuclei to moles and then to mass using $m = \dfrac{N_{\text{yr}}}{N_A}\,M$: (M1)
$$ m = \frac{2.95\times 10^{27}}{6.02\times 10^{23}} \times 235\ \mathrm{g} \approx 1.15\times 10^{6}\ \mathrm{g} = 1.15\times 10^{3}\ \mathrm{kg}. $$(A1)
The fission fragments are neutron-rich and unstable, and some have long half-lives. (B1)
They keep emitting ionising radiation (beta and gamma) for many years, and continue to release decay heat after shutdown, so the spent fuel must be cooled and shielded long after it leaves the core. (R1)
电功率为效率乘以热功率：$P_{\text{电}} = 0.33 \times 3.0\times 10^{9} \approx 0.99\times 10^{9}\ \mathrm{W} = 0.99\ \mathrm{GW}$。(A1)
每次裂变给出 $200\ \mathrm{MeV} = 3.20\times 10^{-11}\ \mathrm{J}$。该热功率需要：(M1)
$$ N = \frac{P_{\text{热}}}{E} = \frac{3.0\times 10^{9}}{3.20\times 10^{-11}} \approx 9.4\times 10^{19}\ \mathrm{s^{-1}}. $$(A1)
一年内裂变次数：$N_{\text{年}} = (9.375\times 10^{19})(3.15\times 10^{7}) \approx 2.95\times 10^{27}$。(M1)
用 $m = \dfrac{N_{\text{年}}}{N_A}\,M$ 把核数转成摩尔再转成质量：(M1)
$$ m = \frac{2.95\times 10^{27}}{6.02\times 10^{23}} \times 235\ \mathrm{g} \approx 1.15\times 10^{6}\ \mathrm{g} = 1.15\times 10^{3}\ \mathrm{kg}. $$(A1)
裂变碎片富中子且不稳定，其中一些半衰期很长。(B1)
它们多年持续放出电离辐射（贝塔与伽马），并在停堆后继续释放衰变热，故乏燃料在离开堆芯后仍需长期冷却与屏蔽。(R1)