Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus E1.1 to E1.6考纲 E1.1 至 E1.6PHYSICS HL
Geiger-Marsden: (i) most alphas pass straight through; (ii) a very few deflect by more than $90^{\circ}$. (a) what each shows; (b) ratio of atomic to nuclear radius and why few large deflections.盖革-马斯登：(i) 多数 α 直穿；(ii) 极少数偏转超过 $90^{\circ}$。(a) 各说明什么；(b) 原子与原子核半径之比及为何少有大角度偏转。
(i) Most alpha particles pass straight through almost undeflected, so most of the atom is empty space. (A1)
(ii) A very few are deflected through large angles, so the positive charge and nearly all the mass of the atom are concentrated in a tiny, dense central nucleus that can repel and reverse a fast positive alpha. (A1)
Ratio of radii: (M1)
$$ \frac{r_{\text{atom}}}{r_{\text{nucleus}}} = \frac{10^{-10}}{10^{-15}} = 10^{5}. $$The nucleus is about $10^{5}$ times smaller than the atom, so its cross-sectional target area is a tiny fraction of the atom's. Only an alpha that happens to head almost straight at a nucleus comes close enough to feel a strong Coulomb repulsion, so large-angle deflections are extremely rare. (A1)
(i) 多数 α 粒子几乎不偏转地直穿，故原子大部分是空的。(A1)
(ii) 极少数被大角度偏转，故原子的正电荷与几乎全部质量集中在一个微小、致密的中心原子核中，它能排斥并使快速带正电的 α 反向。(A1)
半径之比：(M1)
$$ \frac{r_{\text{atom}}}{r_{\text{nucleus}}} = \frac{10^{-10}}{10^{-15}} = 10^{5}. $$原子核比原子约小 $10^{5}$ 倍，故其截面靶面积只占原子的极小一部分。只有恰好几乎正对原子核飞行的 α 才能靠近到感受强库仑斥力，因此大角度偏转极其罕见。(A1)
Nuclide $^{35}_{17}\mathrm{Cl}$. (a) protons, neutrons, nucleons; (b) name an isotope and justify with $Z$, $N$; (c) estimate the nuclear mass in kg with $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$.核素 $^{35}_{17}\mathrm{Cl}$。(a) 质子、中子、核子数；(b) 写出一个同位素并用 $Z$、$N$ 说明；(c) 取 $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$ 估算核质量（kg）。
$Z = 17$ gives 17 protons; nucleons $= A = 35$. (A1)
Neutrons $= A - Z = 35 - 17 = 18$. (A1)
For example $^{37}_{17}\mathrm{Cl}$. (A1)
It has the same proton number $Z = 17$ but a different neutron number ($N = 37 - 17 = 20$ instead of 18). Same $Z$ with different $N$ is exactly the definition of an isotope. (A1)
Take each of the 35 nucleons as $1\ \mathrm{u}$: (M1)
$$ m \approx 35 \times 1.66\times10^{-27} = 5.81\times10^{-26}\ \mathrm{kg} \approx 5.8\times10^{-26}\ \mathrm{kg}. $$(A1)
$Z = 17$ 给出 17 个质子；核子数 $= A = 35$。(A1)
中子数 $= A - Z = 35 - 17 = 18$。(A1)
例如 $^{37}_{17}\mathrm{Cl}$。(A1)
它的质子数 $Z = 17$ 相同，但中子数不同（$N = 37 - 17 = 20$，而非 18）。$Z$ 相同而 $N$ 不同正是同位素的定义。(A1)
取 35 个核子各为 $1\ \mathrm{u}$：(M1)
$$ m \approx 35 \times 1.66\times10^{-27} = 5.81\times10^{-26}\ \mathrm{kg} \approx 5.8\times10^{-26}\ \mathrm{kg}. $$(A1)
A transition releases $4.00\ \mathrm{eV}$ as one photon. (a) energy in J; (b) wavelength from $E = hc/\lambda$; (c) which EM region and why.某跃迁以一个光子释放 $4.00\ \mathrm{eV}$。(a) 能量（J）；(b) 由 $E = hc/\lambda$ 求波长；(c) 属哪个电磁波段及理由。
Multiply the energy in eV by $1.60\times10^{-19}\ \mathrm{J\,eV^{-1}}$: (M1)
$$ E = 4.00 \times 1.60\times10^{-19} = 6.40\times10^{-19}\ \mathrm{J}. $$(A1)
Rearrange $E = hc/\lambda$ to $\lambda = hc/E$: (M1)
$$ \lambda = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{6.40\times10^{-19}} = \frac{1.989\times10^{-25}}{6.40\times10^{-19}} \approx 3.11\times10^{-7}\ \mathrm{m}. $$That is $\lambda \approx 311\ \mathrm{nm}$. (A1)
The photon is ultraviolet. (A1)
Its wavelength $311\ \mathrm{nm}$ is shorter than the violet edge of the visible band (about $400\ \mathrm{nm}$), so it lies in the ultraviolet. (R1)
把以 eV 计的能量乘以 $1.60\times10^{-19}\ \mathrm{J\,eV^{-1}}$：(M1)
$$ E = 4.00 \times 1.60\times10^{-19} = 6.40\times10^{-19}\ \mathrm{J}. $$(A1)
把 $E = hc/\lambda$ 变形为 $\lambda = hc/E$：(M1)
$$ \lambda = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{6.40\times10^{-19}} = \frac{1.989\times10^{-25}}{6.40\times10^{-19}} \approx 3.11\times10^{-7}\ \mathrm{m}. $$即 $\lambda \approx 311\ \mathrm{nm}$。(A1)
该光子属紫外。(A1)
其波长 $311\ \mathrm{nm}$ 短于可见光紫端（约 $400\ \mathrm{nm}$），故处于紫外波段。(R1)
Cool H gas: (1) white light passed through it; (2) the gas excited and viewed directly. (a) appearance of each spectrum; (b) why the dark lines coincide with the bright lines.低温氢气：(1) 白光穿过；(2) 气体被激发后直接观测。(a) 两种光谱外观；(b) 为何暗线与亮线波长重合。
Case (1) is an absorption spectrum: a continuous (rainbow) background crossed by a set of dark lines where photons matching upward transitions have been removed. (A1)
Case (2) is an emission spectrum: a dark background showing only a set of bright coloured lines, the photons released as electrons fall between levels. (A1)
Both processes involve the same atom, so the same set of discrete energy levels and therefore the same set of energy differences $\Delta E = E_{\text{high}} - E_{\text{low}}$. (R1)
Each photon has $E = hf = hc/\lambda$ fixed by its $\Delta E$, so absorption removes exactly the wavelengths that emission produces. The dark lines therefore fall at the same wavelengths as the bright lines. (R1)
情形 (1) 为吸收光谱：连续（彩虹）背景上叠加一组暗线，对应匹配向上跃迁、被移除的光子。(A1)
情形 (2) 为发射光谱：暗背景上仅显示一组亮彩色线，对应电子在能级间下落所释放的光子。(A1)
两过程涉及同一原子，故有同一套分立能级，因而有同一套能级差 $\Delta E = E_{\text{high}} - E_{\text{low}}$。(R1)
每个光子 $E = hf = hc/\lambda$ 由其 $\Delta E$ 决定，故吸收所移除的恰是发射所产生的波长。因此暗线与亮线落在相同波长处。(R1)
Protons repel each other but nuclei hold together. (a) three properties of the strong force and how they bind the nucleus; (b) discrete gamma energies and the nuclear vs atomic energy scale.质子互相排斥但原子核不散。(a) 强核力三条性质及其如何束缚原子核；(b) 分立 γ 能量与核/原子能量尺度比较。
It is attractive between all pairs of nucleons (proton-proton, proton-neutron, neutron-neutron), so it acts on every nucleon and binds protons and neutrons alike. (A1)
It is very short range, of order $10^{-15}\ \mathrm{m}$ (about one nucleon diameter), and is essentially zero beyond a few femtometres. (A1)
At separations inside that range it is much stronger than the Coulomb repulsion between protons, so within the nucleus the net force is attractive and the nucleus stays bound; it also becomes repulsive at very small separations, which stops the nucleus collapsing. (A1)
Gamma photons are emitted with only certain discrete energies, so the energy differences inside the nucleus are discrete, which means the nucleus itself has discrete (quantised) energy levels. (A1·A1)
Nuclear transitions are MeV-scale, roughly a million times larger than the eV-scale energies of atomic electron transitions, which is why nuclear de-excitation produces gamma photons rather than visible light. (A1)
它对所有核子对（质子-质子、质子-中子、中子-中子）都吸引，故作用于每个核子，对质子与中子一视同仁地束缚。(A1)
它的程极短，约 $10^{-15}\ \mathrm{m}$（约一个核子直径），超过几飞米基本为零。(A1)
在该程内的间距上它远强于质子间的库仑斥力，故核内净力为吸引、原子核保持束缚；它在极小间距处还会变为排斥，从而阻止原子核坍缩。(A1)
γ 光子仅以某些分立能量发射，故核内的能级差是分立的，这意味着原子核本身具有分立（量子化）能级。(A1·A1)
核跃迁为 MeV 级，约为原子电子跃迁 eV 级能量的一百万倍，这正是核退激产生 γ 光子而非可见光的原因。(A1)
Three levels: $E_1 = -6.00$, $E_2 = -3.40$, $E_3 = -1.50\ \mathrm{eV}$. (a) longest-wavelength transition and its $\lambda$; (b) shortest-wavelength photon's $\lambda$; (c) ionisation energy from the ground state in J; (d) which transitions give absorption lines from the cool ground-state gas.三能级：$E_1 = -6.00$、$E_2 = -3.40$、$E_3 = -1.50\ \mathrm{eV}$。(a) 波长最长的跃迁及其 $\lambda$；(b) 波长最短光子的 $\lambda$；(c) 从基态电离能（J）；(d) 低温基态气体的吸收线来自哪些跃迁。
Longest wavelength means smallest photon energy, so the smallest gap. The three downward gaps are $E_3\to E_2 = 1.90\ \mathrm{eV}$, $E_2\to E_1 = 2.60\ \mathrm{eV}$, $E_3\to E_1 = 4.50\ \mathrm{eV}$. The smallest is $E_3 \to E_2 = 1.90\ \mathrm{eV}$. (M1·A1)
$$ \lambda = \frac{hc}{\Delta E} = \frac{1.989\times10^{-25}}{1.90 \times 1.60\times10^{-19}} = \frac{1.989\times10^{-25}}{3.04\times10^{-19}} \approx 6.54\times10^{-7}\ \mathrm{m} \approx 654\ \mathrm{nm}. $$(A1)
Shortest wavelength means largest energy gap: $E_3 \to E_1 = (-1.50) - (-6.00) = 4.50\ \mathrm{eV}$. (M1·A1)
$$ \lambda = \frac{1.989\times10^{-25}}{4.50 \times 1.60\times10^{-19}} = \frac{1.989\times10^{-25}}{7.20\times10^{-19}} \approx 2.76\times10^{-7}\ \mathrm{m} \approx 276\ \mathrm{nm}. $$(A1)
Ionisation lifts the electron from $E_1 = -6.00\ \mathrm{eV}$ to $E = 0$, so the energy needed is $6.00\ \mathrm{eV}$: (M1)
$$ E = 6.00 \times 1.60\times10^{-19} = 9.60\times10^{-19}\ \mathrm{J}. $$(A1)
Cool atoms sit in the ground state $E_1$, so they can only absorb photons that raise them to a higher level: $E_1 \to E_2$ ($2.60\ \mathrm{eV}$) and $E_1 \to E_3$ ($4.50\ \mathrm{eV}$). (A1)
The $E_3 \to E_2$ wavelength does not appear in absorption because no atoms start in level $E_2$ to make the upward $E_2 \to E_3$ jump. (R1)
波长最长意味着光子能量最小，即能隙最小。三个向下能隙为 $E_3\to E_2 = 1.90\ \mathrm{eV}$、$E_2\to E_1 = 2.60\ \mathrm{eV}$、$E_3\to E_1 = 4.50\ \mathrm{eV}$。最小者为 $E_3 \to E_2 = 1.90\ \mathrm{eV}$。(M1·A1)
$$ \lambda = \frac{hc}{\Delta E} = \frac{1.989\times10^{-25}}{1.90 \times 1.60\times10^{-19}} = \frac{1.989\times10^{-25}}{3.04\times10^{-19}} \approx 6.54\times10^{-7}\ \mathrm{m} \approx 654\ \mathrm{nm}. $$(A1)
波长最短意味着能隙最大：$E_3 \to E_1 = (-1.50) - (-6.00) = 4.50\ \mathrm{eV}$。(M1·A1)
$$ \lambda = \frac{1.989\times10^{-25}}{4.50 \times 1.60\times10^{-19}} = \frac{1.989\times10^{-25}}{7.20\times10^{-19}} \approx 2.76\times10^{-7}\ \mathrm{m} \approx 276\ \mathrm{nm}. $$(A1)
电离把电子从 $E_1 = -6.00\ \mathrm{eV}$ 提升到 $E = 0$，故所需能量为 $6.00\ \mathrm{eV}$：(M1)
$$ E = 6.00 \times 1.60\times10^{-19} = 9.60\times10^{-19}\ \mathrm{J}. $$(A1)
低温原子处于基态 $E_1$，故只能吸收使其升到更高能级的光子：$E_1 \to E_2$（$2.60\ \mathrm{eV}$）与 $E_1 \to E_3$（$4.50\ \mathrm{eV}$）。(A1)
$E_3 \to E_2$ 的波长不出现在吸收谱中，因为没有原子起始于能级 $E_2$ 来做向上的 $E_2 \to E_3$ 跃迁。(R1)
$^{4}_{2}\mathrm{He}$ from 2 p and 2 n. $m_p = 1.00728$, $m_n = 1.00867$, $m(^{4}_{2}\mathrm{He}) = 4.00150\ \mathrm{u}$; $1\ \mathrm{u} = 931.5\ \mathrm{MeV\,c^{-2}}$. (a) define mass defect and compute it in u; (b) binding energy in MeV; (c) binding energy per nucleon; (d) use the binding-energy-per-nucleon curve to explain fusion and fission release.$^{4}_{2}\mathrm{He}$ 由 2 质子 2 中子组成。$m_p = 1.00728$、$m_n = 1.00867$、$m(^{4}_{2}\mathrm{He}) = 4.00150\ \mathrm{u}$；$1\ \mathrm{u} = 931.5\ \mathrm{MeV\,c^{-2}}$。(a) 定义质量亏损并以 u 计算；(b) 结合能（MeV）；(c) 每核子结合能；(d) 用每核子结合能曲线解释聚变与裂变释能。
The mass defect is the difference between the total mass of the separated, free nucleons and the mass of the assembled nucleus; the missing mass corresponds to the energy released when the nucleus forms. (A1)
$\Delta m = 2m_p + 2m_n - m(^{4}_{2}\mathrm{He})$: (M1)
$$ \Delta m = 2(1.00728) + 2(1.00867) - 4.00150 = 4.03190 - 4.00150 = 0.03040\ \mathrm{u}. $$(A1)
Convert the mass defect to energy using $1\ \mathrm{u} = 931.5\ \mathrm{MeV\,c^{-2}}$: (M1)
$$ E_b = \Delta m \times 931.5\ \mathrm{MeV} = 0.03040 \times 931.5. $$(M1 for substitution)
$$ E_b \approx 28.3\ \mathrm{MeV}. $$(A1)
Divide by the 4 nucleons: (M1)
$$ \frac{E_b}{A} = \frac{28.3}{4} \approx 7.08\ \mathrm{MeV\ per\ nucleon}. $$(A1)
Binding energy per nucleon measures how tightly each nucleon is held; a higher value means a more stable nucleus. (M1)
Light nuclei lie low on the rising part of the curve. When they fuse, the product sits higher up (nearer the iron peak), so its nucleons are more tightly bound; the gain in binding energy per nucleon is released. (A1)
Heavy nuclei lie on the slowly falling part beyond iron. When a heavy nucleus splits, the two fragments sit higher on the curve than the original, so again binding energy per nucleon increases and energy is released. (A1)
In both cases the system moves toward the iron peak, the most stable region, and the increase in total binding energy appears as released energy. (R1)
质量亏损是分离的自由核子的总质量与组装成原子核后质量之差；缺失的质量对应原子核形成时释放的能量。(A1)
$\Delta m = 2m_p + 2m_n - m(^{4}_{2}\mathrm{He})$：(M1)
$$ \Delta m = 2(1.00728) + 2(1.00867) - 4.00150 = 4.03190 - 4.00150 = 0.03040\ \mathrm{u}. $$(A1)
用 $1\ \mathrm{u} = 931.5\ \mathrm{MeV\,c^{-2}}$ 把质量亏损换为能量：(M1)
$$ E_b = \Delta m \times 931.5\ \mathrm{MeV} = 0.03040 \times 931.5. $$(代入得 M1)
$$ E_b \approx 28.3\ \mathrm{MeV}. $$(A1)
除以 4 个核子：(M1)
$$ \frac{E_b}{A} = \frac{28.3}{4} \approx 7.08\ \mathrm{MeV\ 每核子}. $$(A1)
每核子结合能衡量每个核子被束缚的紧密程度；数值越大原子核越稳定。(M1)
轻核位于曲线上升段的低处。它们聚变时，产物位于更高处（更接近铁峰），其核子束缚更紧；每核子结合能的增量被释放。(A1)
重核位于铁之后的缓降段。重核裂变时，两块碎片在曲线上比原核更高，故每核子结合能同样增大，能量被释放。(A1)
两种情形下系统都朝铁峰这一最稳定区移动，总结合能的增加表现为释放的能量。(R1)
Bohr model: $E_n = -13.6/n^2\ \mathrm{eV}$. (a) $E_2$ and $E_3$; (b) photon energy for $n=3\to n=2$ in J; (c) wavelength and colour; (d) ground-state ionisation energy in J; (e) why the spectrum is discrete.玻尔模型：$E_n = -13.6/n^2\ \mathrm{eV}$。(a) $E_2$ 与 $E_3$；(b) $n=3\to n=2$ 的光子能量（J）；(c) 波长与颜色；(d) 基态电离能（J）；(e) 为何谱是分立的。
(A1 each)
$\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \mathrm{eV}$. (M1·A1)
$$ \Delta E = 1.89 \times 1.60\times10^{-19} \approx 3.02\times10^{-19}\ \mathrm{J}. $$(A1)
$\lambda = hc/\Delta E$: (M1)
$$ \lambda = \frac{1.989\times10^{-25}}{3.02\times10^{-19}} \approx 6.58\times10^{-7}\ \mathrm{m} \approx 658\ \mathrm{nm}. $$(A1)
A wavelength near $658\ \mathrm{nm}$ is red light; this is the H-$\alpha$ line of the Balmer series. (A1)
Ionisation takes the electron from $E_1 = -13.6\ \mathrm{eV}$ to $E = 0$, needing $13.6\ \mathrm{eV}$: (M1)
$$ E = 13.6 \times 1.60\times10^{-19} \approx 2.18\times10^{-18}\ \mathrm{J}. $$(A1)
The electron can only occupy the discrete energies $E_n$, so any transition releases a photon of a fixed energy difference $\Delta E = E_{n_i} - E_{n_f}$. (R1)
Only certain $\Delta E$ values exist, so only certain frequencies $f = \Delta E / h$ appear: the spectrum is a set of sharp lines rather than a continuous band. (R1)
(各 A1)
$\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \mathrm{eV}$。(M1·A1)
$$ \Delta E = 1.89 \times 1.60\times10^{-19} \approx 3.02\times10^{-19}\ \mathrm{J}. $$(A1)
$\lambda = hc/\Delta E$：(M1)
$$ \lambda = \frac{1.989\times10^{-25}}{3.02\times10^{-19}} \approx 6.58\times10^{-7}\ \mathrm{m} \approx 658\ \mathrm{nm}. $$(A1)
约 $658\ \mathrm{nm}$ 的波长为红光；即巴尔末系的 H-$\alpha$ 线。(A1)
电离把电子从 $E_1 = -13.6\ \mathrm{eV}$ 提升到 $E = 0$，需 $13.6\ \mathrm{eV}$：(M1)
$$ E = 13.6 \times 1.60\times10^{-19} \approx 2.18\times10^{-18}\ \mathrm{J}. $$(A1)
电子只能占据分立能量 $E_n$，故任何跃迁都释放一个固定能级差的光子 $\Delta E = E_{n_i} - E_{n_f}$。(R1)
仅存在某些 $\Delta E$，故仅出现某些频率 $f = \Delta E / h$：谱为一组锐线而非连续带。(R1)
Geiger-Marsden. (a) outline the apparatus and the two observations with their conclusions; (b) why plum pudding cannot explain large-angle deflections; (c) gamma photon between levels $1.33\ \mathrm{MeV}$ apart, find its frequency; (d) what the discrete gamma energy shows.盖革-马斯登。(a) 概述装置与两个观测及结论；(b) 为何葡萄干布丁无法解释大角度偏转；(c) 相距 $1.33\ \mathrm{MeV}$ 两能级间的 γ 光子，求频率；(d) 分立 γ 能量说明什么。
A narrow beam of alpha particles from a radioactive source is directed at a very thin gold foil in an evacuated chamber, and a movable detector (scintillation screen) counts alphas scattered through each angle. (M1)
Observation 1: the great majority of alphas pass through almost undeflected, showing the atom is mostly empty space. (A1)
Observation 2: a very small fraction are deflected through large angles, a few by more than $90^{\circ}$, showing the positive charge and almost all the mass are concentrated in a tiny dense nucleus. (A1·A1)
If positive charge were spread uniformly through the whole atom, the electric field inside it would be weak everywhere, so it could only nudge a fast alpha by a fraction of a degree. (M1)
Such a smeared charge cannot produce the strong, close-range repulsion needed to turn an alpha through a large angle, so the observed back-scattering forces the charge into a tiny concentrated region. (R1)
Convert the energy gap to joules: $1.33\ \mathrm{MeV} = 1.33\times10^{6} \times 1.60\times10^{-19} = 2.13\times10^{-13}\ \mathrm{J}$. (M1)
Use $\Delta E = hf$, so $f = \Delta E / h$: (M1)
$$ f = \frac{2.13\times10^{-13}}{6.63\times10^{-34}} \approx 3.21\times10^{20}\ \mathrm{Hz}. $$(A1)
Because the gamma photon carries one of only certain fixed energies, the nucleus has discrete (quantised) energy levels, just as discrete atomic spectra reveal discrete electron levels. (A1)
来自放射源的一束细 α 粒子在真空室中射向极薄金箔，可移动的探测器（闪烁屏）按各角度计数被散射的 α。(M1)
观测 1：绝大多数 α 几乎不偏转地穿过，说明原子大部分是空的。(A1)
观测 2：极少数被大角度偏转，少数超过 $90^{\circ}$，说明正电荷与几乎全部质量集中于微小致密的原子核。(A1·A1)
若正电荷均匀弥散于整个原子，其内部电场处处很弱，至多使快速 α 偏转零点几度。(M1)
这种弥散电荷无法产生使 α 大角度偏转所需的近距离强斥力，故观测到的背散射迫使电荷集中到极小区域。(R1)
把能隙换为焦耳：$1.33\ \mathrm{MeV} = 1.33\times10^{6} \times 1.60\times10^{-19} = 2.13\times10^{-13}\ \mathrm{J}$。(M1)
用 $\Delta E = hf$，故 $f = \Delta E / h$：(M1)
$$ f = \frac{2.13\times10^{-13}}{6.63\times10^{-34}} \approx 3.21\times10^{20}\ \mathrm{Hz}. $$(A1)
由于 γ 光子只携带某些固定能量之一，原子核具有分立（量子化）能级，正如分立的原子光谱揭示分立的电子能级。(A1)
$^{7}_{3}\mathrm{Li}$ (3 p, 4 n). $m_p = 1.00728$, $m_n = 1.00867$, $m(^{7}_{3}\mathrm{Li}) = 7.01436\ \mathrm{u}$; $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$, $c = 3.00\times10^{8}$. (a) define binding energy and write the mass-defect expression; (b) mass defect in kg; (c) binding energy in J via $E = \Delta m c^2$; (d) show $\approx 39\ \mathrm{MeV}$ and find binding energy per nucleon; (e) which of Li-7 and He-4 is more tightly bound.$^{7}_{3}\mathrm{Li}$（3 质子 4 中子）。$m_p = 1.00728$、$m_n = 1.00867$、$m(^{7}_{3}\mathrm{Li}) = 7.01436\ \mathrm{u}$；$1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$、$c = 3.00\times10^{8}$。(a) 定义结合能并写质量亏损式；(b) 质量亏损（kg）；(c) 由 $E = \Delta m c^2$ 求结合能（J）；(d) 证明约 $39\ \mathrm{MeV}$ 并求每核子结合能；(e) Li-7 与 He-4 哪个束缚更紧。
The binding energy is the energy that must be supplied to separate a nucleus completely into its individual free nucleons; equivalently it is the energy released when those nucleons come together to form the nucleus. (A1)
$$ \Delta m = 3m_p + 4m_n - m(^{7}_{3}\mathrm{Li}). $$(A1)
In u: (M1)
$$ \Delta m = 3(1.00728) + 4(1.00867) - 7.01436 = 7.05652 - 7.01436 = 0.04216\ \mathrm{u}. $$(A1)
Convert to kilograms with $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$:
$$ \Delta m = 0.04216 \times 1.66\times10^{-27} \approx 7.00\times10^{-29}\ \mathrm{kg}. $$(A1)
Use $E = \Delta m\, c^2$: (M1)
$$ E_b = (7.00\times10^{-29})(3.00\times10^{8})^2 = (7.00\times10^{-29})(9.00\times10^{16}). $$(M1 for substitution)
$$ E_b \approx 6.30\times10^{-12}\ \mathrm{J}. $$(A1)
Convert to eV by dividing by $1.60\times10^{-19}$: $E_b = 6.30\times10^{-12} / 1.60\times10^{-19} \approx 3.9\times10^{7}\ \mathrm{eV} \approx 39.3\ \mathrm{MeV}$ (consistent with $0.04216 \times 931.5 = 39.3\ \mathrm{MeV}$). (M1)
$$ \frac{E_b}{A} = \frac{39.3}{7} \approx 5.6\ \mathrm{MeV\ per\ nucleon}. $$(A1)
$^{4}_{2}\mathrm{He}$ is more tightly bound. (A1)
Tightness of binding is measured by binding energy per nucleon, and helium-4 has about $7.1\ \mathrm{MeV}$ per nucleon against lithium-7's $5.6\ \mathrm{MeV}$, so each nucleon in helium-4 is held more strongly. (R1)
结合能是把原子核完全分离为各个自由核子所必须供给的能量；等价地，它是这些核子结合成原子核时释放的能量。(A1)
$$ \Delta m = 3m_p + 4m_n - m(^{7}_{3}\mathrm{Li}). $$(A1)
以 u 计：(M1)
$$ \Delta m = 3(1.00728) + 4(1.00867) - 7.01436 = 7.05652 - 7.01436 = 0.04216\ \mathrm{u}. $$(A1)
用 $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$ 换为千克：
$$ \Delta m = 0.04216 \times 1.66\times10^{-27} \approx 7.00\times10^{-29}\ \mathrm{kg}. $$(A1)
用 $E = \Delta m\, c^2$：(M1)
$$ E_b = (7.00\times10^{-29})(3.00\times10^{8})^2 = (7.00\times10^{-29})(9.00\times10^{16}). $$(代入得 M1)
$$ E_b \approx 6.30\times10^{-12}\ \mathrm{J}. $$(A1)
除以 $1.60\times10^{-19}$ 换为 eV：$E_b = 6.30\times10^{-12} / 1.60\times10^{-19} \approx 3.9\times10^{7}\ \mathrm{eV} \approx 39.3\ \mathrm{MeV}$（与 $0.04216 \times 931.5 = 39.3\ \mathrm{MeV}$ 一致）。(M1)
$$ \frac{E_b}{A} = \frac{39.3}{7} \approx 5.6\ \mathrm{MeV\ 每核子}. $$(A1)
$^{4}_{2}\mathrm{He}$ 束缚更紧。(A1)
束缚紧密程度由每核子结合能衡量，氦-4 约 $7.1\ \mathrm{MeV}$ 每核子，而锂-7 为 $5.6\ \mathrm{MeV}$，故氦-4 中每个核子被束缚得更强。(R1)