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Modern and Nuclear Physics · Solutions近代物理与核物理 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 24 marksAP 选择题 + 安/卑省考短答 · 共 24 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Photon energy光子能量 · HS-PS4-3 [3 marks][3 分]

A photon has frequency $5.0 \times 10^{14}\ \text{Hz}$. What is its energy?一个光子的频率为 $5.0 \times 10^{14}\ \text{Hz}$。它的能量是多少？

Answer:答案：  (A)  $3.3 \times 10^{-19}\ \text{J}$

(a) Apply the photon-energy relation $E = hf$套用光子能量关系 $E = hf$ M1·A1·A1

A photon's energy is its frequency times Planck's constant:光子能量等于频率乘以普朗克常数： $$ E \;=\; hf \;=\; (6.63 \times 10^{-34})(5.0 \times 10^{14}) \;=\; 3.315 \times 10^{-19} \;\approx\; 3.3 \times 10^{-19} \;\text{J}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $6.63 \times 10^{-34}\ \text{J}$: reports Planck's constant alone, forgetting to multiply by $f$.只写出普朗克常数，忘记乘以 $f$。
(C) $5.0 \times 10^{14}\ \text{J}$: mistakes the frequency (in Hz) for an energy.把频率（单位 Hz）误当成能量。
(D) $1.3 \times 10^{-48}\ \text{J}$: divides $h$ by $f$ instead of multiplying.把 $h$ 除以 $f$ 而非相乘。
Photon energy is set by frequency alone, not by brightness.光子能量仅由频率决定，与亮度无关。 $E = hf$ is the single most-used equation in modern physics. Each photon is an indivisible packet of energy; doubling the light's intensity adds more photons but leaves each photon's energy unchanged. Equivalently, $E = hc/\lambda$, so higher frequency (shorter wavelength) means a more energetic photon: a UV photon carries far more energy than an infrared one. Keep $h$ and $f$ in SI units (J·s and Hz) and the answer comes out in joules; divide by $1.60 \times 10^{-19}$ to convert to eV when a problem prefers that unit.$E = hf$ 是近代物理中使用最频繁的方程。每个光子是不可分割的能量包；将光强加倍只是增加光子数，而不改变每个光子的能量。等价地 $E = hc/\lambda$，所以频率越高（波长越短）光子能量越大：一个紫外光子携带的能量远大于红外光子。将 $h$ 与 $f$ 保持为国际单位（J·s 与 Hz），答案即为焦耳；当题目偏好 eV 时，除以 $1.60 \times 10^{-19}$ 即可换算。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Nuclear notation核符号 · HS-PS1-8 [3 marks][3 分]

How many neutrons are in the nucleus of ${}^{16}_{8}\mathrm{O}$?${}^{16}_{8}\mathrm{O}$ 的原子核中有多少个中子？

Answer:答案：  (A)  $8$

(a) Read off $A$ and $Z$, then subtract读出 $A$ 与 $Z$，再相减 M1·A1·A1

In notation ${}^A_Z X$, the superscript $A$ is the mass number (protons $+$ neutrons) and the subscript $Z$ is the atomic number (protons). Number of neutrons:在符号 ${}^A_Z X$ 中，上标 $A$ 是质量数（质子 $+$ 中子），下标 $Z$ 是原子序数（质子数）。中子数： $$ N \;=\; A - Z \;=\; 16 - 8 \;=\; 8. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $16$: reports the mass number $A$ (all nucleons), not the neutron count.报出质量数 $A$（全部核子），而非中子数。
(C) $24$: adds $A + Z = 16 + 8$ instead of subtracting.把 $A + Z = 16 + 8$ 相加而非相减。
(D) $4$: halves the atomic number for no physical reason.无物理依据地将原子序数取一半。
Mass number counts nucleons; atomic number counts protons; their difference is the neutron count.质量数数核子；原子序数数质子；两者之差就是中子数。 The atomic number $Z$ fixes the element's identity (8 protons is always oxygen). Different neutron counts give isotopes of the same element: ${}^{16}\mathrm{O}$, ${}^{17}\mathrm{O}$ and ${}^{18}\mathrm{O}$ all have $Z = 8$ but $8$, $9$ and $10$ neutrons respectively. The key relation $N = A - Z$ recurs throughout nuclear physics, especially when balancing decay equations, where the totals of $A$ (top) and $Z$ (bottom) must each be conserved across the arrow.原子序数 $Z$ 确定元素身份（8 个质子永远是氧）。不同中子数给出同一元素的同位素：${}^{16}\mathrm{O}$、${}^{17}\mathrm{O}$、${}^{18}\mathrm{O}$ 的 $Z$ 都是 $8$，但中子数分别为 $8$、$9$、$10$。关系式 $N = A - Z$ 贯穿整个核物理，尤其在配平衰变方程时，箭头两侧的 $A$（上标）总和与 $Z$（下标）总和必须各自守恒。
Q3MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Half-life半衰期 · SPH3U D3.11 [6 marks][6 分]

$800$ nuclei, half-life $T = 12$ h. (a) Nuclei after 3 half-lives. (b) Hours elapsed. (c) Why never exactly zero. (d) Does a freezer change $T$?$800$ 个核，半衰期 $T = 12$ h。(a) 3 个半衰期后的核数。(b) 经过的小时数。(c) 为何永不为零。(d) 冰箱是否改变 $T$？

Answer:答案：  (a) $100$ nuclei个核  ·  (b) $36$ hours小时  ·  (c) each step only halves the count每步只减半  ·  (d) no, $T$ is independent of temperature不会，$T$ 与温度无关

(a) Halving sequence over 3 half-lives3 个半衰期的逐次减半 M1·A1

Each half-life halves the count: $800 \to 400 \to 200 \to 100$. Equivalently:每个半衰期使数量减半：$800 \to 400 \to 200 \to 100$。等价地： $$ N \;=\; N_0\!\left(\tfrac{1}{2}\right)^3 \;=\; 800 \times \tfrac{1}{8} \;=\; 100 \;\text{nuclei.} $$

(b) Elapsed time经过的时间 A1

$$ t \;=\; 3T \;=\; 3 \times 12 \;=\; 36 \;\text{hours.} $$

(c) Why the count never reaches exactly zero为何核数永远不会精确为零 A1

Each half-life removes only half of whatever remains, never the whole amount. So the count follows $800, 400, 200, 100, 50, \ldots$, halving forever and approaching but never reaching zero in any finite number of half-lives.每个半衰期只移除当前剩余量的一半，永远不会移除全部。因此核数依 $800, 400, 200, 100, 50, \ldots$ 不断减半，在任何有限个半衰期内都只能趋近而永远到不了零。

(d) Effect of a freezer on half-life冰箱对半衰期的影响 M1·A1

No change. Radioactive decay is a purely nuclear process; the half-life does not depend on temperature, pressure, or chemical state. Cooling the sample in a freezer leaves $T = 12$ h unchanged.不会改变。放射性衰变是纯核过程；半衰期不依赖温度、压力或化学状态。将样品放入冰箱冷却，$T = 12$ h 保持不变。
Half-life is an intrinsic nuclear constant, immune to ordinary lab conditions.半衰期是内禀的核常数，不受普通实验室条件影响。 Chemical reaction rates depend strongly on temperature (the Arrhenius law), so students often expect cooling or heating to slow or speed up decay. It does not: the decay of a nucleus is governed by the strong and weak nuclear forces, energies millions of times larger than the thermal energies of a freezer or furnace. This is exactly why radioisotope dating is reliable: a sample's half-life is the same whether it sat in Arctic ice or a desert. The exponential law $N = N_0 (1/2)^{t/T}$ has $T$ as a fixed property of the isotope alone.化学反应速率强烈依赖温度（阿伦尼乌斯定律），所以学生常以为冷却或加热能减慢或加快衰变。其实不然：原子核的衰变由强核力和弱核力支配，其能量比冰箱或熔炉的热能大数百万倍。这正是放射性同位素定年法可靠的原因：无论样品埋在北极冰层还是沙漠，其半衰期都相同。指数律 $N = N_0 (1/2)^{t/T}$ 中的 $T$ 仅是同位素本身的固定属性。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Photoelectric effect光电效应 · HS-PS4-3 [3 marks][3 分]

Light above threshold ejects electrons. Intensity is doubled, frequency unchanged. What happens?高于阈值的光射出电子。强度加倍，频率不变。会发生什么？

Answer:答案：  (A) number of electrons per second doubles; each electron's $E_k$ is unchanged每秒电子数加倍；每个电子的 $E_k$ 不变

(a) Apply the photon model运用光子模型 M1·A1·A1

In the photon picture, intensity is the number of photons arriving per second. Each photon still carries the same energy $E = hf$, so each ejected electron still receives the same maximum kinetic energy $E_k = hf - \phi$. Doubling the intensity doubles the number of photons, hence the number of electrons ejected per second, but leaves each electron's kinetic energy unchanged. Option (A).在光子图像中，强度即每秒到达的光子数。每个光子仍携带相同能量 $E = hf$，故每个被射出电子仍获得相同的最大动能 $E_k = hf - \phi$。强度加倍使光子数加倍，因而每秒射出的电子数加倍，但每个电子的动能不变。选 (A)。
Why the distractors fail.干扰项分析。
(B): kinetic energy depends on frequency, not intensity; $E_k = hf - \phi$ is unchanged.动能取决于频率而非强度；$E_k = hf - \phi$ 不变。
(C): the light is already above threshold (electrons are ejected), and threshold is set by frequency, not intensity.光已高于阈值（电子已射出），且阈值由频率而非强度决定。
(D): nothing in the photon model lowers individual energies when more photons arrive.光子模型中，光子增多并不会降低单个电子的能量。
Frequency sets each electron's energy; intensity sets how many electrons.频率决定每个电子的能量；强度决定电子的数量。 This is the single most important conceptual result of the photoelectric effect, and the one the classical wave model gets wrong. A wave model predicts that brighter light (more energy per second) should give each electron more energy, but experiment shows that brighter light just ejects more electrons at the same energy. Only raising the frequency raises $E_k$. Einstein's photon hypothesis ($E = hf$, one photon hits one electron) explains this perfectly and earned him the 1921 Nobel Prize. Remember the slogan: "colour controls energy, brightness controls count."这是光电效应最重要的概念性结论，也是经典波动模型出错之处。波动模型预测更亮的光（每秒更多能量）会让每个电子获得更多能量，但实验表明更亮的光只是以相同能量射出更多电子。只有提高频率才能提高 $E_k$。爱因斯坦的光子假设（$E = hf$，一个光子击中一个电子）对此作出完美解释，并使他赢得 1921 年诺贝尔奖。记住口诀："颜色控制能量，亮度控制数量。"
Q5MEDIUM中 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §5 Radioactive decay放射性衰变 · 30-D3.2k [9 marks][9 分]

(a) Alpha decay of ${}^{226}_{88}\mathrm{Ra}$. (b) Beta-minus decay of ${}^{14}_{6}\mathrm{C}$. (c) Rank alpha/beta/gamma by penetration.(a) ${}^{226}_{88}\mathrm{Ra}$ 的 alpha 衰变。(b) ${}^{14}_{6}\mathrm{C}$ 的 beta 负衰变。(c) alpha/beta/gamma 按穿透力排序。

Answer:答案：  (a) ${}^{226}_{88}\mathrm{Ra} \to {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\mathrm{He}$  ·  (b) ${}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}e + \bar\nu_e$  ·  (c) $\alpha < \beta < \gamma$

(a) Balanced alpha decay of Ra-226Ra-226 的配平 alpha 衰变 M1·A1·A1

An alpha particle is ${}^{4}_{2}\mathrm{He}$, so the daughter has $A = 226 - 4 = 222$ and $Z = 88 - 2 = 86$ (radon):alpha 粒子是 ${}^{4}_{2}\mathrm{He}$，故子核 $A = 226 - 4 = 222$，$Z = 88 - 2 = 86$（氡）： $$ {}^{226}_{88}\mathrm{Ra} \;\longrightarrow\; {}^{222}_{86}\mathrm{Rn} \;+\; {}^{4}_{2}\mathrm{He} $$ Check: top $226 = 222 + 4$; bottom $88 = 86 + 2$.核验：上标 $226 = 222 + 4$；下标 $88 = 86 + 2$。

(b) Balanced beta-minus decay of C-14C-14 的配平 beta 负衰变 M1·A1·A1

A neutron converts to a proton, emitting an electron ${}^{0}_{-1}e$ and an electron antineutrino $\bar\nu_e$: $A$ is unchanged, $Z$ rises by 1 (to nitrogen):一个中子转变为质子，发射一个电子 ${}^{0}_{-1}e$ 和一个电子反中微子 $\bar\nu_e$：$A$ 不变，$Z$ 增加 1（变为氮）： $$ {}^{14}_{6}\mathrm{C} \;\longrightarrow\; {}^{14}_{7}\mathrm{N} \;+\; {}^{0}_{-1}e \;+\; \bar\nu_e $$ Check: top $14 = 14 + 0$; bottom $6 = 7 + (-1)$.核验：上标 $14 = 14 + 0$；下标 $6 = 7 + (-1)$。

(c) Penetrating power, least to most穿透力从弱到强 A1·A1·A1

$\alpha$ (stopped by paper or a few cm of air) $<$ $\beta$ (stopped by a few mm of aluminium) $<$ $\gamma$ (needs centimetres of lead or metres of concrete). Heavier, more charged radiation ionises strongly and so loses energy fastest, giving it the shortest range.$\alpha$（被纸或几厘米空气阻挡）$<$ $\beta$（被几毫米铝阻挡）$<$ $\gamma$（需厘米级铅或米级混凝土）。质量越大、电荷越多的辐射电离能力越强，因而最快损失能量，射程最短。
Balance an equation by conserving the top ($A$) and bottom ($Z$) totals separately.配平方程要让上标（$A$）和下标（$Z$）的总和分别守恒。 Every nuclear equation must conserve nucleon number $A$ (the superscripts) and charge $Z$ (the subscripts) across the arrow. Alpha decay subtracts $(4, 2)$; beta-minus keeps $A$ fixed and adds $+1$ to $Z$ because a neutron ($Z=0$) becomes a proton ($Z=1$) plus an electron ($Z=-1$). The antineutrino carries no charge and (essentially) no mass, so it does not affect the balance, but it must be included to conserve energy, momentum, and lepton number, which is why Alberta 30-D3.2k requires it. Penetration ranks inversely with ionising power: alpha ionises most and travels least.每个核方程都必须让箭头两侧的核子数 $A$（上标）与电荷 $Z$（下标）守恒。alpha 衰变减去 $(4, 2)$；beta 负衰变保持 $A$ 不变并使 $Z$ 增 $1$，因为一个中子（$Z=0$）变为一个质子（$Z=1$）加一个电子（$Z=-1$）。反中微子不带电且（基本上）无质量，故不影响配平，但必须写出以守恒能量、动量和轻子数，这正是阿尔伯塔 30-D3.2k 要求它的原因。穿透力与电离能力成反比：alpha 电离最强、射程最短。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 31 marksAP 衔接简答题 + 荣誉级 · 共 31 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §1 Photoelectric effect光电效应 · HS-PS4-3 [8 marks][8 分]

Metal: $\phi = 2.30$ eV, light at $9.0 \times 10^{14}$ Hz. (a) Threshold $f_0$. (b) Are electrons ejected? (c) Max $E_k$ in J. (d) Max $E_k$ in eV.金属：$\phi = 2.30$ eV，光频率 $9.0 \times 10^{14}$ Hz。(a) 阈频 $f_0$。(b) 是否射出电子？(c) 最大 $E_k$（J）。(d) 最大 $E_k$（eV）。

Answer:答案：  (a) $f_0 \approx 5.55 \times 10^{14}\ \text{Hz}$  ·  (b) yes是  ·  (c) $E_k \approx 2.29 \times 10^{-19}\ \text{J}$  ·  (d) $E_k \approx 1.43\ \text{eV}$

(a) Threshold frequency: set $E_k = 0$, so $hf_0 = \phi$阈频：令 $E_k = 0$，故 $hf_0 = \phi$ M1·A1

Convert $\phi$ to joules: $\phi = 2.30 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}$ J.将 $\phi$ 换算为焦耳：$\phi = 2.30 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}$ J。 $$ f_0 \;=\; \frac{\phi}{h} \;=\; \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}} \;\approx\; 5.55 \times 10^{14} \;\text{Hz.} $$

(b) Are electrons ejected?是否射出电子？ A1

Yes. The incident frequency $9.0 \times 10^{14}$ Hz exceeds $f_0 = 5.55 \times 10^{14}$ Hz, so each photon has more than enough energy to free an electron.是。入射频率 $9.0 \times 10^{14}$ Hz 超过 $f_0 = 5.55 \times 10^{14}$ Hz，故每个光子能量足以释放电子。

(c) Maximum kinetic energy using $E_k = hf - \phi$用 $E_k = hf - \phi$ 求最大动能 M1·A1·A1

$$ E_k \;=\; hf - \phi \;=\; (6.63 \times 10^{-34})(9.0 \times 10^{14}) - 3.68 \times 10^{-19} $$ $$ \;=\; 5.967 \times 10^{-19} - 3.68 \times 10^{-19} \;=\; 2.287 \times 10^{-19} \;\approx\; 2.29 \times 10^{-19}\ \text{J.} $$

(d) Convert to electronvolts换算为电子伏特 M1·A1

$$ E_k \;=\; \frac{2.287 \times 10^{-19}}{1.60 \times 10^{-19}} \;\approx\; 1.43\ \text{eV.} $$
The photoelectric equation $E_k = hf - \phi$ is a straight-line energy balance: photon energy in, escape cost out, kinetic energy left over.光电方程 $E_k = hf - \phi$ 是一条直线式能量守恒：光子能量进，逸出代价出，剩余即动能。 Plot $E_k$ against $f$ and you get a straight line of slope $h$ and frequency-intercept $f_0 = \phi/h$; this is exactly the graph Millikan measured in 1916 to confirm Einstein's theory. The single most common error here is unit mixing: $\phi$ is quoted in eV but $h$ and $f$ give joules, so $\phi$ must be converted to joules before subtracting (or work entirely in eV using $h = 4.14 \times 10^{-15}$ eV·s). Equivalently, $E_k$ in eV equals the stopping voltage in volts, here about $1.43$ V, which is what you would measure to bring the fastest electrons to rest.以 $E_k$ 对 $f$ 作图得到一条直线，斜率为 $h$，频率截距为 $f_0 = \phi/h$；这正是密立根 1916 年测得、用以验证爱因斯坦理论的图像。此处最常见的错误是单位混用：$\phi$ 以 eV 给出，而 $h$ 与 $f$ 给出焦耳，所以相减前必须把 $\phi$ 换算为焦耳（或全程用 eV，取 $h = 4.14 \times 10^{-15}$ eV·s）。等价地，以 eV 表示的 $E_k$ 等于以伏特表示的遏止电压，此处约 $1.43$ V，即让最快电子停下来所需测得的电压。
Q7MEDIUM中 🇨🇦 ON安 🇨🇦 AB阿 ON Provincial-style安大略省考风格 §3 Atomic spectra原子光谱 · 30-D2.5k [7 marks][7 分]

Hydrogen: electron drops $n=2$ ($-3.40$ eV) to $n=1$ ($-13.60$ eV). (a) Photon energy in eV. (b) In joules. (c) Frequency. (d) Wavelength and spectral region.氢：电子从 $n=2$（$-3.40$ eV）跃迁到 $n=1$（$-13.60$ eV）。(a) 光子能量（eV）。(b) 焦耳。(c) 频率。(d) 波长与光谱区域。

Answer:答案：  (a) $10.2\ \text{eV}$  ·  (b) $1.63 \times 10^{-18}\ \text{J}$  ·  (c) $f \approx 2.46 \times 10^{15}\ \text{Hz}$  ·  (d) $\lambda \approx 122\ \text{nm}$ (ultraviolet)（紫外）

(a) Photon energy is the level difference光子能量即能级差 M1·A1

$$ \Delta E \;=\; E_{\text{high}} - E_{\text{low}} \;=\; (-3.40) - (-13.60) \;=\; 10.2\ \text{eV.} $$

(b) Convert to joules换算为焦耳 A1

$$ \Delta E \;=\; 10.2 \times 1.60 \times 10^{-19} \;=\; 1.632 \times 10^{-18}\ \text{J.} $$

(c) Frequency from $\Delta E = hf$由 $\Delta E = hf$ 求频率 M1·A1

$$ f \;=\; \frac{\Delta E}{h} \;=\; \frac{1.632 \times 10^{-18}}{6.63 \times 10^{-34}} \;\approx\; 2.46 \times 10^{15}\ \text{Hz.} $$

(d) Wavelength from $\lambda = c/f$ and spectral region由 $\lambda = c/f$ 求波长并判断光谱区 M1·A1

$$ \lambda \;=\; \frac{c}{f} \;=\; \frac{3.00 \times 10^{8}}{2.46 \times 10^{15}} \;\approx\; 1.22 \times 10^{-7}\ \text{m} \;=\; 122\ \text{nm.} $$ At $122$ nm the photon lies in the ultraviolet (shorter than the $400$ nm violet edge of visible light). It is the first line of the hydrogen Lyman series.$122$ nm 处光子位于紫外区（短于可见光紫端的 $400$ nm）。它是氢莱曼系的第一条谱线。
Discrete energy levels produce a discrete line spectrum, the fingerprint of each element.离散能级产生离散线光谱，是每种元素的指纹。 Bohr's key idea: an electron can only occupy fixed energy levels, so a downward jump emits a photon of exactly $\Delta E = E_{\text{high}} - E_{\text{low}}$, giving sharp spectral lines rather than a continuous rainbow. Transitions ending at $n = 1$ (the Lyman series) are large energy drops and land in the ultraviolet; transitions ending at $n = 2$ (the Balmer series) are smaller and include the visible red $H_\alpha$ line at $656$ nm. Astronomers read these line patterns in starlight to identify which elements a distant star contains, without ever sampling it directly.玻尔的核心思想：电子只能占据固定能级，所以向下跃迁发射的光子能量恰为 $\Delta E = E_{\text{高}} - E_{\text{低}}$，产生锐利的谱线而非连续彩虹。终态为 $n = 1$ 的跃迁（莱曼系）能量降幅大，落在紫外区；终态为 $n = 2$ 的跃迁（巴尔末系）降幅较小，包含 $656$ nm 处可见的红色 $H_\alpha$ 线。天文学家通过分析星光中的谱线图样来识别遥远恒星所含元素，无需直接取样。
Q8HARD难 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 Binding energy结合能 · 30-D3.6k [8 marks][8 分]

${}^{7}_{3}\mathrm{Li}$, mass $7.01600$ u; $m_p = 1.00728$ u, $m_n = 1.00867$ u. (a) Protons & neutrons. (b) Free-nucleon mass. (c) Mass defect. (d) Binding energy (MeV). (e) Per nucleon.${}^{7}_{3}\mathrm{Li}$，质量 $7.01600$ u；$m_p = 1.00728$ u，$m_n = 1.00867$ u。(a) 质子与中子数。(b) 自由核子质量。(c) 质量亏损。(d) 结合能（MeV）。(e) 每核子。

Answer:答案：  (a) $3p,\ 4n$  ·  (b) $7.05652\ \text{u}$  ·  (c) $\Delta m = 0.04052\ \text{u}$  ·  (d) $E_B \approx 37.7\ \text{MeV}$  ·  (e) $\approx 5.39\ \text{MeV/nucleon}$

(a) Composition from ${}^{7}_{3}\mathrm{Li}$由 ${}^{7}_{3}\mathrm{Li}$ 读出组成 A1

$Z = 3$ protons; $N = A - Z = 7 - 3 = 4$ neutrons.$Z = 3$ 个质子；$N = A - Z = 7 - 3 = 4$ 个中子。

(b) Total mass of the free nucleons自由核子的总质量 M1·A1

$$ m_{\text{free}} \;=\; 3m_p + 4m_n \;=\; 3(1.00728) + 4(1.00867) \;=\; 3.02184 + 4.03468 \;=\; 7.05652\ \text{u.} $$

(c) Mass defect质量亏损 M1·A1

$$ \Delta m \;=\; m_{\text{free}} - m_{\text{nucleus}} \;=\; 7.05652 - 7.01600 \;=\; 0.04052\ \text{u.} $$

(d) Binding energy via $E_B = \Delta m \cdot c^2$由 $E_B = \Delta m \cdot c^2$ 求结合能 M1·A1

$$ E_B \;=\; 0.04052 \times 931.5\ \text{MeV/u} \;\approx\; 37.7\ \text{MeV.} $$

(e) Binding energy per nucleon每核子结合能 A1

$$ \frac{E_B}{A} \;=\; \frac{37.7}{7} \;\approx\; 5.39\ \text{MeV/nucleon.} $$
The mass defect is the "missing" mass converted to the energy that binds the nucleus; binding energy per nucleon measures stability.质量亏损是转化为结合能的"缺失"质量；每核子结合能衡量稳定性。 A bound nucleus is always lighter than the sum of its free nucleons because forming it released energy ($E_B = \Delta m c^2$); to pull it apart you must supply exactly that much energy. Dividing by the nucleon count $A$ gives binding energy per nucleon, the fair way to compare nuclei of different sizes. Lithium-7 at $\approx 5.4$ MeV/nucleon sits below the iron peak ($\approx 8.8$ MeV/nucleon near $A = 56$), so fusing light nuclei toward iron releases energy. The dominant exam error is mixing up which mass is larger: always compute $m_{\text{free}} - m_{\text{nucleus}}$ so $\Delta m$ comes out positive.束缚态核总比其自由核子之和更轻，因为形成它时释放了能量（$E_B = \Delta m c^2$）；要拆开它必须正好补回这么多能量。除以核子数 $A$ 得每核子结合能，是比较不同大小原子核稳定性的公平方式。锂-7 约 $5.4$ MeV/核子，低于铁峰（$A = 56$ 附近约 $8.8$ MeV/核子），故轻核向铁聚变会释放能量。考试中最常见的错误是搞反哪个质量更大：始终计算 $m_{\text{free}} - m_{\text{nucleus}}$，使 $\Delta m$ 为正。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Wave-particle duality波粒二象性 · HS-PS4-3 (above NGSS floor)（超出 NGSS 基准） [8 marks][8 分]

Electron, $m = 9.11 \times 10^{-31}$ kg, $v = 2.0 \times 10^{6}$ m/s. (a) Momentum. (b) de Broglie wavelength. (c) Why it diffracts off a $0.1$ nm lattice. (d) Why no matter waves for a tennis ball.电子，$m = 9.11 \times 10^{-31}$ kg，$v = 2.0 \times 10^{6}$ m/s。(a) 动量。(b) 德布罗意波长。(c) 为何能在 $0.1$ nm 晶格上衍射。(d) 为何网球无物质波。

Answer:答案：  (a) $p \approx 1.82 \times 10^{-24}\ \text{kg}\cdot\text{m/s}$  ·  (b) $\lambda \approx 3.64 \times 10^{-10}\ \text{m}\ (0.364\ \text{nm})$  ·  (c) $\lambda \sim$ lattice spacing与晶格间距相当  ·  (d) $\lambda$ far too small to detect太小无法探测

(a) Momentum $p = mv$动量 $p = mv$ M1·A1

$$ p \;=\; mv \;=\; (9.11 \times 10^{-31})(2.0 \times 10^{6}) \;=\; 1.822 \times 10^{-24} \;\text{kg}\cdot\text{m/s.} $$

(b) de Broglie wavelength $\lambda = h/p$德布罗意波长 $\lambda = h/p$ M1·A1·A1

$$ \lambda \;=\; \frac{h}{p} \;=\; \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} \;\approx\; 3.64 \times 10^{-10}\ \text{m} \;=\; 0.364\ \text{nm.} $$

(c) Why diffraction occurs为何发生衍射 A1

Diffraction is significant only when the wavelength is comparable to the obstacle spacing. Here $\lambda \approx 0.36$ nm is the same order as the $\sim 0.1$ nm spacing between atoms in a crystal lattice, so the electron's matter wave diffracts off the lattice, just as X-rays do.只有当波长与障碍物间距相当时衍射才显著。此处 $\lambda \approx 0.36$ nm 与晶格中原子间距 $\sim 0.1$ nm 处于同一数量级，故电子的物质波能在晶格上衍射，正如 X 射线一样。

(d) Why everyday objects show no matter waves为何日常物体不显物质波 M1·A1

A tennis ball has an enormous mass compared with an electron, so its momentum $p = mv$ is huge and $\lambda = h/p \sim 10^{-34}$ m is fantastically small, far smaller than any object or slit it could interact with. With no obstacle of comparable size, no diffraction is ever observable, so matter waves are hidden at the macroscopic scale.网球的质量比电子大得多，故其动量 $p = mv$ 极大，$\lambda = h/p \sim 10^{-34}$ m 小得惊人，远小于它可能相互作用的任何物体或缝隙。由于不存在尺寸相当的障碍物，永远观察不到衍射，所以物质波在宏观尺度被隐藏。
de Broglie unified waves and particles: every moving object has a wavelength $\lambda = h/p$, but only tiny, light particles have a wavelength large enough to matter.德布罗意统一了波与粒子：每个运动物体都有波长 $\lambda = h/p$，但只有微小轻质粒子的波长才足够大到起作用。 The factor that decides whether wave behaviour shows up is the size of $\lambda$ relative to the apparatus. Because Planck's constant $h$ is so small ($\sim 10^{-34}$), $\lambda$ is only appreciable when momentum $p$ is also tiny, which means low mass and modest speed. Electrons fit perfectly: their $0.1$ to $1$ nm wavelengths matched crystal lattices, and the Davisson-Germer experiment (1927) confirmed de Broglie's 1924 prediction. This same wave nature is what lets electron microscopes resolve far finer detail than light microscopes: shorter wavelength means sharper resolution.决定是否显现波动行为的因素，是 $\lambda$ 相对于仪器尺度的大小。由于普朗克常数 $h$ 极小（$\sim 10^{-34}$），只有当动量 $p$ 也很小时 $\lambda$ 才可观，这意味着低质量与适中速度。电子恰好符合：其 $0.1$ 至 $1$ nm 的波长与晶格匹配，戴维孙-革末实验（1927 年）证实了德布罗意 1924 年的预言。正是这种波动性使电子显微镜能分辨远比光学显微镜更精细的细节：波长越短，分辨率越高。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Half-life (applied)半衰期（应用） · 30-D3.3k [9 marks][9 分]

I-131, $T = 8.02$ d, $N_0 = 1.60 \times 10^{12}$ atoms. (a) Atoms after $24.06$ d. (b) Fraction after $32.08$ d. (c) Why negligible after $\sim 5$ half-lives. (d) Why warming will not speed decay.I-131，$T = 8.02$ 天，$N_0 = 1.60 \times 10^{12}$ 个原子。(a) $24.06$ 天后原子数。(b) $32.08$ 天后分数。(c) 为何约 $5$ 个半衰期后可忽略。(d) 为何加热不能加快衰变。

Answer:答案：  (a) $2.00 \times 10^{11}$ atoms个原子  ·  (b) $1/16 = 6.25\%$  ·  (c) $\approx 3\%$ remains残留  ·  (d) decay rate is independent of temperature衰变速率与温度无关

(a) Atoms after $24.06$ days ($t/T = 3$)$24.06$ 天后的原子数（$t/T = 3$） M1·A1·A1

Number of half-lives: $t/T = 24.06 / 8.02 = 3$.半衰期数：$t/T = 24.06 / 8.02 = 3$。 $$ N \;=\; N_0\!\left(\tfrac{1}{2}\right)^3 \;=\; 1.60 \times 10^{12} \times \tfrac{1}{8} \;=\; 2.00 \times 10^{11}\ \text{atoms.} $$

(b) Fraction after $32.08$ days ($t/T = 4$)$32.08$ 天后的分数（$t/T = 4$） M1·A1

$$ \frac{N}{N_0} \;=\; \left(\tfrac{1}{2}\right)^4 \;=\; \frac{1}{16} \;=\; 6.25\%. $$ Activity is proportional to $N$, so the activity also drops to $1/16$ of its original value.活度正比于 $N$，故活度也降至原来的 $1/16$。

(c) Why clinically negligible after $\sim 5$ half-lives为何约 $5$ 个半衰期后临床可忽略 A1·A1

After 5 half-lives the fraction remaining is $(1/2)^5 = 1/32 \approx 3\%$. With only about $3\%$ of the original activity left, the radiation dose to the patient is small enough to be treated as negligible for clinical purposes.5 个半衰期后剩余分数为 $(1/2)^5 = 1/32 \approx 3\%$。仅剩约 $3\%$ 的原始活度时，患者所受辐射剂量已小到在临床上可视为可忽略。

(d) Why warming will not help为何加热无效 M1·A1

Radioactive decay is a nuclear process whose rate is fixed by the half-life. Half-life does not depend on temperature, pressure, or chemical environment, so warming the sample leaves the decay rate (and $T = 8.02$ d) completely unchanged.放射性衰变是核过程，其速率由半衰期固定。半衰期不依赖温度、压力或化学环境，故加热样品对衰变速率（及 $T = 8.02$ 天）毫无影响。
Counting half-lives turns awkward exponentials into simple powers of one-half.数半衰期能把麻烦的指数变成简单的二分之一幂。 Whenever the elapsed time is a whole-number multiple of the half-life, skip logarithms entirely: just compute $n = t/T$ and apply $(1/2)^n$. Here $24.06 / 8.02 = 3$ and $32.08 / 8.02 = 4$ are clean integers by design. The rule of thumb that a radioisotope is "gone" after about 5 to 10 half-lives ($\le 3\%$ to $0.1\%$ remaining) is why iodine-131, with its convenient 8-day half-life, is ideal for thyroid therapy: it delivers a strong dose quickly, then fades within weeks. A longer-lived isotope would irradiate the patient for years; a shorter-lived one would decay before treatment.只要经过的时间是半衰期的整数倍，就完全不必用对数：直接算 $n = t/T$ 再套 $(1/2)^n$。此处 $24.06 / 8.02 = 3$、$32.08 / 8.02 = 4$ 都是特意设计的整数。"约 5 到 10 个半衰期后放射性同位素基本消失"（剩余 $\le 3\%$ 至 $0.1\%$）这一经验法则，正是碘-131（半衰期约 8 天）适合甲状腺治疗的原因：它能快速给出强剂量，再在数周内消退。半衰期更长的同位素会让患者受辐射数年；更短的则在治疗前就衰变殆尽。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §7 Fusion & mass-energy聚变与质能 · 30-D3.5k [9 marks][9 分]

D-T fusion ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$; masses $2.01410$, $3.01605$, $4.00260$, $1.00867$ u. (a) Reactant mass. (b) Product mass. (c) $\Delta m$. (d) Energy (MeV). (e) Why energy is released.D-T 聚变 ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$；质量 $2.01410$、$3.01605$、$4.00260$、$1.00867$ u。(a) 反应物质量。(b) 产物质量。(c) $\Delta m$。(d) 能量（MeV）。(e) 为何释放能量。

Answer:答案：  (a) $5.03015\ \text{u}$  ·  (b) $5.01127\ \text{u}$  ·  (c) $\Delta m = 0.01888\ \text{u}$  ·  (d) $E \approx 17.6\ \text{MeV}$  ·  (e) products more tightly bound产物结合更紧

(a) Mass on the reactant side反应物一侧的质量 A1

$$ m_{\text{before}} \;=\; 2.01410 + 3.01605 \;=\; 5.03015\ \text{u.} $$

(b) Mass on the product side产物一侧的质量 A1

$$ m_{\text{after}} \;=\; 4.00260 + 1.00867 \;=\; 5.01127\ \text{u.} $$

(c) Mass defect质量亏损 M1·A1

$$ \Delta m \;=\; m_{\text{before}} - m_{\text{after}} \;=\; 5.03015 - 5.01127 \;=\; 0.01888\ \text{u.} $$

(d) Energy released via $E = \Delta m \cdot c^2$由 $E = \Delta m \cdot c^2$ 求释放能量 M1·A1·A1

$$ E \;=\; \Delta m \times 931.5\ \text{MeV/u} \;=\; 0.01888 \times 931.5 \;\approx\; 17.6\ \text{MeV.} $$

(e) Why the reaction releases energy为何该反应释放能量 A1·A1

The helium-4 product sits much higher on the binding-energy-per-nucleon curve than the light deuterium and tritium inputs. Because the product is more tightly bound, its total mass is less than the reactants' (the mass defect), and that missing mass is released as energy via $E = mc^2$.氦-4 产物在每核子结合能曲线上的位置远高于轻质的氘、氚原料。由于产物结合更紧，其总质量小于反应物（即质量亏损），这部分缺失质量通过 $E = mc^2$ 以能量形式释放。
Fusion and fission both release energy by climbing toward the iron peak of the binding-energy curve.聚变与裂变都通过向结合能曲线的铁峰靠拢来释放能量。 The binding-energy-per-nucleon curve rises steeply from hydrogen to a peak near iron ($A \approx 56$), then falls slowly toward uranium. Light nuclei release energy by fusing upward toward the peak; heavy nuclei release energy by fissioning downward toward it. Both increase binding energy per nucleon, both shed mass, both convert that mass to energy. One D-T fusion event releases $17.6$ MeV, about a million times more energy per reaction than burning one carbon atom ($\sim 4$ eV). This is why fusion powers the Sun and is pursued as a clean terrestrial power source, although confining a plasma hot enough to overcome proton-proton repulsion remains the engineering challenge.每核子结合能曲线从氢陡升至铁附近（$A \approx 56$）的顶峰，然后向铀缓降。轻核通过向上聚变靠近峰顶释放能量；重核通过向下裂变靠近峰顶释放能量。两者都提高每核子结合能、都损失质量、都把质量转化为能量。一次 D-T 聚变释放 $17.6$ MeV，每次反应约比燃烧一个碳原子（$\sim 4$ eV）多出百万倍能量。这就是聚变为太阳供能、并被作为清洁地面能源追求的原因，尽管约束足够高温的等离子体以克服质子间斥力仍是工程难题。
Q12HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Fission chain reaction裂变链式反应 · HS-PS1-8 (above NGSS floor)（超出 NGSS 基准） [7 marks][7 分]

${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + x\,{}^{1}_{0}\mathrm{n}$, $\Delta m = 0.1860$ u. (a) Find $x$. (b) Energy per fission (MeV). (c) In joules. (d) How neutrons sustain a chain reaction.${}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + x\,{}^{1}_{0}\mathrm{n}$，$\Delta m = 0.1860$ u。(a) 求 $x$。(b) 每次裂变能量（MeV）。(c) 焦耳。(d) 中子如何维持链式反应。

Answer:答案：  (a) $x = 3$  ·  (b) $E \approx 173\ \text{MeV}$  ·  (c) $E \approx 2.77 \times 10^{-11}\ \text{J}$  ·  (d) each fission frees neutrons that trigger further fissions每次裂变释放的中子引发更多裂变

(a) Balance the equation for $x$配平方程求 $x$ M1·A1

Conserve mass number $A$ (top): $1 + 235 = 141 + 92 + x(1)$, so $236 = 233 + x \Rightarrow x = 3$. Charge $Z$ (bottom) already balances: $0 + 92 = 56 + 36 + 0$.守恒质量数 $A$（上标）：$1 + 235 = 141 + 92 + x(1)$，故 $236 = 233 + x \Rightarrow x = 3$。电荷 $Z$（下标）已配平：$0 + 92 = 56 + 36 + 0$。

(b) Energy per fission via $E = \Delta m \cdot c^2$由 $E = \Delta m \cdot c^2$ 求每次裂变能量 M1·A1

$$ E \;=\; 0.1860 \times 931.5\ \text{MeV/u} \;\approx\; 173\ \text{MeV.} $$

(c) Convert to joules换算为焦耳 M1·A1

Using $1\ \text{MeV} = 1.60 \times 10^{-13}$ J:取 $1\ \text{MeV} = 1.60 \times 10^{-13}$ J： $$ E \;=\; 173.3 \times 1.60 \times 10^{-13} \;\approx\; 2.77 \times 10^{-11}\ \text{J.} $$

(d) How the released neutrons sustain a chain reaction释放的中子如何维持链式反应 A1

Each fission is triggered by one neutron but releases three. If on average at least one of those neutrons goes on to split another ${}^{235}\mathrm{U}$ nucleus, the reactions sustain themselves generation after generation, a self-sustaining chain reaction.每次裂变由一个中子引发，却释放三个中子。若平均至少有一个中子继续去劈裂另一个 ${}^{235}\mathrm{U}$ 核，反应便能代代自持，形成自持链式反应。
One neutron in, several out: this multiplication is what makes a chain reaction, and the multiplication factor decides reactor vs. bomb.一个中子进、多个中子出：这种倍增正是链式反应的关键，而增殖系数决定是反应堆还是炸弹。 Each ${}^{235}\mathrm{U}$ fission consumes one neutron and produces about three, so the reaction can multiply: $1 \to 3 \to 9 \to \cdots$ if every neutron triggers a new fission. The multiplication factor $k$ is the average number of those neutrons that cause a further fission. A reactor uses control rods (boron or cadmium) to absorb excess neutrons and hold $k = 1$ exactly, giving steady power; a weapon allows $k > 1$ and uncontrolled exponential growth. Although $173$ MeV per fission sounds tiny, summed over the $\sim 10^{24}$ nuclei in a kilogram of uranium it is enormous, which is why nuclear fuel is millions of times more energy-dense than chemical fuel.每次 ${}^{235}\mathrm{U}$ 裂变消耗一个中子、产生约三个，所以反应能够倍增：若每个中子都引发新裂变，则 $1 \to 3 \to 9 \to \cdots$。增殖系数 $k$ 是这些中子中平均引发进一步裂变的数目。反应堆用控制棒（硼或镉）吸收多余中子、使 $k = 1$ 精确保持，从而稳定输出功率；武器则允许 $k > 1$ 而不受控指数增长。虽然每次裂变 $173$ MeV 听起来微不足道，但对一千克铀中约 $10^{24}$ 个核求和后却极其巨大，这正是核燃料的能量密度比化学燃料高数百万倍的原因。