Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus E2.1 to E2.6考纲 E2.1 至 E2.6PHYSICS HL
Light on a clean metal emits photoelectrons. (a) define threshold frequency; (b) two observations a classical wave model cannot explain, with the failed classical prediction in each case.光照射洁净金属发射光电子。(a) 定义截止频率；(b) 经典波动模型无法解释的两个观察，并写出每种情形下失败的经典预言。
The threshold frequency $f_0$ is the minimum frequency of incident light that will just cause electrons to be emitted from the metal surface; below it no electrons are emitted at any intensity. (A1)
Threshold frequency. Below $f_0$ no electrons are emitted however bright the light, whereas a classical wave predicts that a sufficiently intense beam of any frequency should eventually free electrons. (A1)
Instant emission. Electrons appear with no measurable delay even at very low intensity, whereas a classical wave predicts a build-up time while the electron slowly accumulates enough energy. (A1)
(Either of the above, or: maximum KE depends only on frequency and not on intensity, whereas a classical wave predicts a brighter beam should give faster electrons.) (A1)
截止频率 $f_0$ 是恰能使电子从金属表面发射的入射光最低频率；低于它时，任何强度都不发射电子。(A1)
截止频率。低于 $f_0$ 时无论光多亮都不发射电子，而经典波动预言只要足够强，任何频率最终都应释放电子。(A1)
瞬时发射。即使强度极低电子也几乎无可测延迟地出现，而经典波动预言需要累积时间让电子缓慢积攒足够能量。(A1)
（以上任一，或：最大动能只依赖频率而与强度无关，而经典波动预言更亮的光应给出更快的电子。）(A1)
Metal $\Phi = 2.5\ \mathrm{eV}$, light $\lambda = 380\ \mathrm{nm}$. (a) threshold frequency; (b) show photon energy $\approx 3.3\ \mathrm{eV}$; (c) maximum KE in eV.金属 $\Phi = 2.5\ \mathrm{eV}$，光 $\lambda = 380\ \mathrm{nm}$。(a) 截止频率；(b) 证明光子能量 $\approx 3.3\ \mathrm{eV}$；(c) 最大动能（eV）。
At threshold $E_{max} = 0$, so $\Phi = h f_0$. Convert $\Phi$ to joules: $\Phi = 2.5 \times 1.60\times 10^{-19} = 4.0\times 10^{-19}\ \mathrm{J}$. (M1)
$$ f_0 = \frac{\Phi}{h} = \frac{4.0\times 10^{-19}}{6.63\times 10^{-34}} \approx 6.0\times 10^{14}\ \mathrm{Hz}. $$(A1)
Use $E = hc/\lambda$ with $\lambda = 380\times 10^{-9}\ \mathrm{m}$: (M1)
$$ E = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{380\times 10^{-9}} = 5.23\times 10^{-19}\ \mathrm{J} = \frac{5.23\times 10^{-19}}{1.60\times 10^{-19}} \approx 3.3\ \mathrm{eV}. $$(A1)
Einstein's equation in electronvolts: $E_{max} = hf - \Phi = 3.27 - 2.5$. (M1)
$$ E_{max} \approx 0.77\ \mathrm{eV}. $$(A1)
阈值处 $E_{max} = 0$，故 $\Phi = h f_0$。把 $\Phi$ 换算为焦耳：$\Phi = 2.5 \times 1.60\times 10^{-19} = 4.0\times 10^{-19}\ \mathrm{J}$。(M1)
$$ f_0 = \frac{\Phi}{h} = \frac{4.0\times 10^{-19}}{6.63\times 10^{-34}} \approx 6.0\times 10^{14}\ \mathrm{Hz}. $$(A1)
用 $E = hc/\lambda$，$\lambda = 380\times 10^{-9}\ \mathrm{m}$：(M1)
$$ E = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{380\times 10^{-9}} = 5.23\times 10^{-19}\ \mathrm{J} = \frac{5.23\times 10^{-19}}{1.60\times 10^{-19}} \approx 3.3\ \mathrm{eV}. $$(A1)
以电子伏特写爱因斯坦方程：$E_{max} = hf - \Phi = 3.27 - 2.5$。(M1)
$$ E_{max} \approx 0.77\ \mathrm{eV}. $$(A1)
Cathode $f_0 = 5.0\times 10^{14}\ \mathrm{Hz}$, light $f = 8.0\times 10^{14}\ \mathrm{Hz}$. (a) show $E_{max} = h(f - f_0)$; (b) the stopping voltage.阴极 $f_0 = 5.0\times 10^{14}\ \mathrm{Hz}$，光 $f = 8.0\times 10^{14}\ \mathrm{Hz}$。(a) 证明 $E_{max} = h(f - f_0)$；(b) 遏止电压。
Einstein's equation is $E_{max} = hf - \Phi$, and the work function is fixed by the threshold through $\Phi = h f_0$. (M1)
$$ E_{max} = hf - h f_0 = h(f - f_0). $$(A1)
The stopping voltage just removes the maximum KE, so $e V_s = E_{max} = h(f - f_0)$. (M1)
$$ V_s = \frac{h(f - f_0)}{e} = \frac{(6.63\times 10^{-34})(3.0\times 10^{14})}{1.60\times 10^{-19}} = \frac{1.989\times 10^{-19}}{1.60\times 10^{-19}} \approx 1.2\ \mathrm{V}. $$(A1)
爱因斯坦方程为 $E_{max} = hf - \Phi$，逸出功由阈频率确定：$\Phi = h f_0$。(M1)
$$ E_{max} = hf - h f_0 = h(f - f_0). $$(A1)
遏止电压恰好抵消最大动能，故 $e V_s = E_{max} = h(f - f_0)$。(M1)
$$ V_s = \frac{h(f - f_0)}{e} = \frac{(6.63\times 10^{-34})(3.0\times 10^{14})}{1.60\times 10^{-19}} = \frac{1.989\times 10^{-19}}{1.60\times 10^{-19}} \approx 1.2\ \mathrm{V}. $$(A1)
Laser light $\lambda = 500\ \mathrm{nm}$. (a) photon momentum; (b) photon energy in joules via an energy-momentum relation.激光 $\lambda = 500\ \mathrm{nm}$。(a) 光子动量；(b) 用能量-动量关系求光子能量（焦耳）。
Use $p = h/\lambda$ with $\lambda = 500\times 10^{-9}\ \mathrm{m}$: (M1)
$$ p = \frac{6.63\times 10^{-34}}{500\times 10^{-9}} \approx 1.3\times 10^{-27}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
A photon is massless, so $E = pc$: (M1)
$$ E = pc = (1.326\times 10^{-27})(3.00\times 10^{8}) \approx 4.0\times 10^{-19}\ \mathrm{J}. $$(A1)
用 $p = h/\lambda$，$\lambda = 500\times 10^{-9}\ \mathrm{m}$：(M1)
$$ p = \frac{6.63\times 10^{-34}}{500\times 10^{-9}} \approx 1.3\times 10^{-27}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
光子无质量，故 $E = pc$：(M1)
$$ E = pc = (1.326\times 10^{-27})(3.00\times 10^{8}) \approx 4.0\times 10^{-19}\ \mathrm{J}. $$(A1)
Electron accelerated from rest through $100\ \mathrm{V}$. (a) show $p = \sqrt{2 m_e e V}$; (b) momentum; (c) de Broglie wavelength and why it suits crystal study.电子从静止经 $100\ \mathrm{V}$ 加速。(a) 证明 $p = \sqrt{2 m_e e V}$；(b) 动量；(c) 德布罗意波长及为何适合研究晶体。
The work done by the accelerating field becomes kinetic energy: $eV = \tfrac{1}{2}m_e v^{2}$. Since $p = m_e v$, write $E_k = \dfrac{p^{2}}{2 m_e}$, so $eV = \dfrac{p^{2}}{2 m_e}$. (M1)
$$ p^{2} = 2 m_e e V \quad\Rightarrow\quad p = \sqrt{2 m_e e V}. $$(A1)
Substitute $V = 100\ \mathrm{V}$: (M1)
$$ p = \sqrt{2(9.11\times 10^{-31})(1.60\times 10^{-19})(100)} \approx 5.4\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
$\lambda = \dfrac{h}{p} = \dfrac{6.63\times 10^{-34}}{5.40\times 10^{-24}} \approx 1.2\times 10^{-10}\ \mathrm{m}$. (A1)
This is comparable to the spacing between atoms in a crystal ($\sim 10^{-10}\ \mathrm{m}$), so the electrons will diffract strongly off the lattice and reveal its structure. (R1)
加速电场做的功转化为动能：$eV = \tfrac{1}{2}m_e v^{2}$。由 $p = m_e v$ 写出 $E_k = \dfrac{p^{2}}{2 m_e}$，故 $eV = \dfrac{p^{2}}{2 m_e}$。(M1)
$$ p^{2} = 2 m_e e V \quad\Rightarrow\quad p = \sqrt{2 m_e e V}. $$(A1)
代入 $V = 100\ \mathrm{V}$：(M1)
$$ p = \sqrt{2(9.11\times 10^{-31})(1.60\times 10^{-19})(100)} \approx 5.4\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
$\lambda = \dfrac{h}{p} = \dfrac{6.63\times 10^{-34}}{5.40\times 10^{-24}} \approx 1.2\times 10^{-10}\ \mathrm{m}$。(A1)
这与晶体原子间距（约 $10^{-10}\ \mathrm{m}$）相当，故电子会在晶格上强烈衍射，揭示其结构。(R1)
Electron confined to $\Delta x = 5.0\times 10^{-11}\ \mathrm{m}$, $\Delta x\,\Delta p \geq h/4\pi$. (a) minimum $\Delta p$; (b) minimum $\Delta v$ and comment; (c) define wave-particle duality with one wave and one particle example.电子约束于 $\Delta x = 5.0\times 10^{-11}\ \mathrm{m}$，$\Delta x\,\Delta p \geq h/4\pi$。(a) 最小 $\Delta p$；(b) 最小 $\Delta v$ 并评论；(c) 定义波粒二象性并各举一波动、一粒子例子。
Take the equality limit $\Delta p = \dfrac{h}{4\pi\,\Delta x}$: (M1)
$$ \Delta p = \frac{6.63\times 10^{-34}}{4\pi (5.0\times 10^{-11})} \approx 1.1\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
Divide by the electron mass: $\Delta v = \dfrac{\Delta p}{m_e} = \dfrac{1.06\times 10^{-24}}{9.11\times 10^{-31}}$. (M1)
$$ \Delta v \approx 1.2\times 10^{6}\ \mathrm{m\,s^{-1}}. $$This is a large fraction of typical electron speeds in atoms, so confining the electron makes its velocity hugely uncertain. (A1)
Wave-particle duality is the principle that light and matter each show wave behaviour and particle behaviour, with the face displayed depending on the experiment. (A1)
For an electron: wave behaviour is shown by diffraction off a crystal; particle behaviour is shown when it is detected as a single localized impact (a dot on a screen) carrying definite charge and mass. (A1)
取等号极限 $\Delta p = \dfrac{h}{4\pi\,\Delta x}$：(M1)
$$ \Delta p = \frac{6.63\times 10^{-34}}{4\pi (5.0\times 10^{-11})} \approx 1.1\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
除以电子质量：$\Delta v = \dfrac{\Delta p}{m_e} = \dfrac{1.06\times 10^{-24}}{9.11\times 10^{-31}}$。(M1)
$$ \Delta v \approx 1.2\times 10^{6}\ \mathrm{m\,s^{-1}}. $$这是原子中电子典型速率的相当大一部分，故约束电子使其速度极其不确定。(A1)
波粒二象性指光与物质都既显示波动行为又显示粒子行为，显示哪一面取决于实验。(A1)
对电子：波动行为体现于在晶体上的衍射；粒子行为体现于它被探测为一次定域的单次撞击（屏上一个点）、带有确定的电荷与质量。(A1)
$E_{max}$ vs $f$: $(6.0\times 10^{14}\ \mathrm{Hz},\,0.50\ \mathrm{eV})$ and $(12.0\times 10^{14}\ \mathrm{Hz},\,2.99\ \mathrm{eV})$. (a) show line, name gradient and intercepts; (b) $h$ from gradient; (c) work function; (d) threshold frequency; (e) effect of a larger-$\Phi$ metal.$E_{max}$ 对 $f$：$(6.0\times 10^{14}\ \mathrm{Hz},\,0.50\ \mathrm{eV})$ 与 $(12.0\times 10^{14}\ \mathrm{Hz},\,2.99\ \mathrm{eV})$。(a) 证明为直线并说明斜率与截距；(b) 由斜率求 $h$；(c) 逸出功；(d) 截止频率；(e) 逸出功更大的金属的影响。
Einstein's equation $E_{max} = hf - \Phi$ has the form $y = mx + c$ with $y = E_{max}$, $x = f$. (M1)
So the plot is a straight line of gradient $h$ and vertical intercept $-\Phi$. (A1)
The line crosses the $f$-axis ($E_{max} = 0$) at the threshold frequency $f_0 = \Phi/h$. (A1)
Convert both energies to joules: $0.50\ \mathrm{eV} = 8.0\times 10^{-20}\ \mathrm{J}$ and $2.99\ \mathrm{eV} = 4.784\times 10^{-19}\ \mathrm{J}$. (M1)
Gradient $= h = \dfrac{\Delta E_{max}}{\Delta f}$: (M1)
$$ h = \frac{4.784\times 10^{-19} - 8.0\times 10^{-20}}{(12.0 - 6.0)\times 10^{14}} = \frac{3.984\times 10^{-19}}{6.0\times 10^{14}} \approx 6.6\times 10^{-34}\ \mathrm{J\,s}. $$(A1)
Use one data point in $\Phi = hf - E_{max}$ at $f = 6.0\times 10^{14}\ \mathrm{Hz}$: (M1)
$$ \Phi = (6.63\times 10^{-34})(6.0\times 10^{14}) - 8.0\times 10^{-20} = 3.18\times 10^{-19}\ \mathrm{J} \approx 2.0\ \mathrm{eV}. $$(A1)
$f_0 = \dfrac{\Phi}{h} = \dfrac{3.18\times 10^{-19}}{6.63\times 10^{-34}}$. (M1)
$$ f_0 \approx 4.8\times 10^{14}\ \mathrm{Hz}. $$(A1)
The new line is parallel to the original (same gradient $h$, which is a universal constant), but shifted: it has a more negative vertical intercept and a higher threshold frequency, so it lies to the right and below the original. (A1)
Reason: the gradient depends only on $h$, never on the metal; only the intercept $-\Phi$ (and hence $f_0 = \Phi/h$) changes with the material. (R1)
爱因斯坦方程 $E_{max} = hf - \Phi$ 形如 $y = mx + c$，其中 $y = E_{max}$、$x = f$。(M1)
故图像为直线，斜率为 $h$，纵轴截距为 $-\Phi$。(A1)
直线在 $f$ 轴（$E_{max} = 0$）处与之相交于截止频率 $f_0 = \Phi/h$。(A1)
把两个能量换算为焦耳：$0.50\ \mathrm{eV} = 8.0\times 10^{-20}\ \mathrm{J}$，$2.99\ \mathrm{eV} = 4.784\times 10^{-19}\ \mathrm{J}$。(M1)
斜率 $= h = \dfrac{\Delta E_{max}}{\Delta f}$：(M1)
$$ h = \frac{4.784\times 10^{-19} - 8.0\times 10^{-20}}{(12.0 - 6.0)\times 10^{14}} = \frac{3.984\times 10^{-19}}{6.0\times 10^{14}} \approx 6.6\times 10^{-34}\ \mathrm{J\,s}. $$(A1)
在 $f = 6.0\times 10^{14}\ \mathrm{Hz}$ 处用一个数据点代入 $\Phi = hf - E_{max}$：(M1)
$$ \Phi = (6.63\times 10^{-34})(6.0\times 10^{14}) - 8.0\times 10^{-20} = 3.18\times 10^{-19}\ \mathrm{J} \approx 2.0\ \mathrm{eV}. $$(A1)
$f_0 = \dfrac{\Phi}{h} = \dfrac{3.18\times 10^{-19}}{6.63\times 10^{-34}}$。(M1)
$$ f_0 \approx 4.8\times 10^{14}\ \mathrm{Hz}. $$(A1)
新直线与原直线平行（斜率同为 $h$，是普适常数），但发生平移：纵轴截距更负、截止频率更高，故位于原直线的右下方。(A1)
理由：斜率只依赖 $h$，与金属无关；只有截距 $-\Phi$（从而 $f_0 = \Phi/h$）随材料改变。(R1)
Electrons accelerated through $V = (200 \pm 5)\ \mathrm{V}$; $\lambda = h/\sqrt{2 m_e e V}$. (a) $\lambda$; (b) percentage uncertainty in $V$; (c) percentage and absolute uncertainty in $\lambda$ (using $\lambda\propto V^{-1/2}$); (d) what the ring pattern shows and the experiment that first demonstrated it.电子经 $V = (200 \pm 5)\ \mathrm{V}$ 加速；$\lambda = h/\sqrt{2 m_e e V}$。(a) $\lambda$；(b) $V$ 的百分比不确定度；(c) $\lambda$ 的百分比与绝对不确定度（用 $\lambda\propto V^{-1/2}$）；(d) 环状图样揭示什么及首次证实的实验。
First the momentum: $p = \sqrt{2 m_e e V} = \sqrt{2(9.11\times 10^{-31})(1.60\times 10^{-19})(200)}$. (M1)
$$ p \approx 7.64\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$Then $\lambda = h/p$: (M1)
$$ \lambda = \frac{6.63\times 10^{-34}}{7.64\times 10^{-24}} \approx 8.7\times 10^{-11}\ \mathrm{m}. $$(A1)
(A1)
Since $\lambda \propto V^{-1/2}$, the power of $V$ is $-\tfrac{1}{2}$, so the percentage uncertainty in $\lambda$ is half that in $V$: (M1)
$$ \frac{\Delta\lambda}{\lambda} = \tfrac{1}{2}\times 2.5\% = 1.25\% \approx 1.3\%. $$(A1)
Absolute uncertainty: $\Delta\lambda = 0.0125\times 8.7\times 10^{-11} \approx 0.1\times 10^{-11}\ \mathrm{m}$, so $\lambda = (8.7 \pm 0.1)\times 10^{-11}\ \mathrm{m}$. (A1)
Diffraction rings are an interference effect, which only waves produce. (M1)
So electrons, although particles with definite charge and mass, also have a wave nature with the de Broglie wavelength $\lambda = h/p$. (A1)
This wave nature of electrons was first demonstrated by the Davisson-Germer experiment (electrons scattered off a nickel crystal). (B1)
先求动量：$p = \sqrt{2 m_e e V} = \sqrt{2(9.11\times 10^{-31})(1.60\times 10^{-19})(200)}$。(M1)
$$ p \approx 7.64\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$再求 $\lambda = h/p$：(M1)
$$ \lambda = \frac{6.63\times 10^{-34}}{7.64\times 10^{-24}} \approx 8.7\times 10^{-11}\ \mathrm{m}. $$(A1)
(A1)
由 $\lambda \propto V^{-1/2}$，$V$ 的幂为 $-\tfrac{1}{2}$，故 $\lambda$ 的百分比不确定度是 $V$ 的一半：(M1)
$$ \frac{\Delta\lambda}{\lambda} = \tfrac{1}{2}\times 2.5\% = 1.25\% \approx 1.3\%. $$(A1)
绝对不确定度：$\Delta\lambda = 0.0125\times 8.7\times 10^{-11} \approx 0.1\times 10^{-11}\ \mathrm{m}$，故 $\lambda = (8.7 \pm 0.1)\times 10^{-11}\ \mathrm{m}$。(A1)
衍射环是干涉效应，只有波才能产生。(M1)
故电子虽是带确定电荷与质量的粒子，也具有波动本性，其德布罗意波长为 $\lambda = h/p$。(A1)
电子的这一波动本性最早由戴维孙-革末实验（电子被镍晶体散射）证实。(B1)
Sodium $\Phi = 2.3\ \mathrm{eV}$, light $\lambda = 350\ \mathrm{nm}$. (a) show photon energy $\approx 3.6\ \mathrm{eV}$; (b) $E_{max}$ in J and stopping voltage; (c) longest wavelength causing emission; (d) photons/s for $1.5\ \mathrm{mW}$; (e) current if $1.0\%$ of photons free an electron.钠 $\Phi = 2.3\ \mathrm{eV}$，光 $\lambda = 350\ \mathrm{nm}$。(a) 证明光子能量 $\approx 3.6\ \mathrm{eV}$；(b) $E_{max}$（焦耳）与遏止电压；(c) 仍能发射的最长波长；(d) $1.5\ \mathrm{mW}$ 时每秒光子数；(e) 若 $1.0\%$ 光子释放一电子时的电流。
$E = hc/\lambda$ with $\lambda = 350\times 10^{-9}\ \mathrm{m}$: (M1)
$$ E = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{350\times 10^{-9}} = 5.68\times 10^{-19}\ \mathrm{J} \approx 3.6\ \mathrm{eV}. $$(A1)
$E_{max} = hf - \Phi$. Convert $\Phi = 2.3\ \mathrm{eV} = 3.68\times 10^{-19}\ \mathrm{J}$: (M1)
$$ E_{max} = 5.68\times 10^{-19} - 3.68\times 10^{-19} = 2.0\times 10^{-19}\ \mathrm{J}. $$(A1)
Stopping voltage: $V_s = \dfrac{E_{max}}{e} = \dfrac{2.0\times 10^{-19}}{1.60\times 10^{-19}} \approx 1.3\ \mathrm{V}$. (A1)
Emission just stops at the threshold, where the photon energy equals $\Phi$: $\dfrac{hc}{\lambda_0} = \Phi$. (M1)
$$ \lambda_0 = \frac{hc}{\Phi} = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{3.68\times 10^{-19}}. $$(M1)
$$ \lambda_0 \approx 5.4\times 10^{-7}\ \mathrm{m} = 540\ \mathrm{nm}. $$(A1)
Each photon carries $E = 5.68\times 10^{-19}\ \mathrm{J}$, and the power is energy per second: $N = P/E$. (M1)
$$ N = \frac{1.5\times 10^{-3}}{5.68\times 10^{-19}} \approx 2.6\times 10^{15}\ \mathrm{s^{-1}}. $$(A1)
Only $1.0\%$ of photons free an electron, and each electron carries charge $e$: $I = 0.010\,N\,e$. (M1)
$$ I = (0.010)(2.64\times 10^{15})(1.60\times 10^{-19}) \approx 4.2\times 10^{-6}\ \mathrm{A} = 4.2\ \mathrm{\mu A}. $$(A1)
$E = hc/\lambda$，$\lambda = 350\times 10^{-9}\ \mathrm{m}$：(M1)
$$ E = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{350\times 10^{-9}} = 5.68\times 10^{-19}\ \mathrm{J} \approx 3.6\ \mathrm{eV}. $$(A1)
$E_{max} = hf - \Phi$。换算 $\Phi = 2.3\ \mathrm{eV} = 3.68\times 10^{-19}\ \mathrm{J}$：(M1)
$$ E_{max} = 5.68\times 10^{-19} - 3.68\times 10^{-19} = 2.0\times 10^{-19}\ \mathrm{J}. $$(A1)
遏止电压：$V_s = \dfrac{E_{max}}{e} = \dfrac{2.0\times 10^{-19}}{1.60\times 10^{-19}} \approx 1.3\ \mathrm{V}$。(A1)
发射恰在阈值处停止，此处光子能量等于 $\Phi$：$\dfrac{hc}{\lambda_0} = \Phi$。(M1)
$$ \lambda_0 = \frac{hc}{\Phi} = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{3.68\times 10^{-19}}. $$(M1)
$$ \lambda_0 \approx 5.4\times 10^{-7}\ \mathrm{m} = 540\ \mathrm{nm}. $$(A1)
每个光子携带 $E = 5.68\times 10^{-19}\ \mathrm{J}$，功率即每秒能量：$N = P/E$。(M1)
$$ N = \frac{1.5\times 10^{-3}}{5.68\times 10^{-19}} \approx 2.6\times 10^{15}\ \mathrm{s^{-1}}. $$(A1)
只有 $1.0\%$ 的光子释放一个电子，每个电子带电荷 $e$：$I = 0.010\,N\,e$。(M1)
$$ I = (0.010)(2.64\times 10^{15})(1.60\times 10^{-19}) \approx 4.2\times 10^{-6}\ \mathrm{A} = 4.2\ \mathrm{\mu A}. $$(A1)
Crystal spacing $d = 0.21\ \mathrm{nm}$; required electron wavelength $\lambda = 0.10\ \mathrm{nm}$. (a) electron momentum; (b) accelerating voltage; (c) photon of same wavelength: its energy, ratio to the electron KE, which probe deposits more energy; (d) first-order Bragg angle.晶体间距 $d = 0.21\ \mathrm{nm}$；所需电子波长 $\lambda = 0.10\ \mathrm{nm}$。(a) 电子动量；(b) 加速电压；(c) 同波长光子：能量、与电子动能之比、哪种探针沉积更多能量；(d) 一级布拉格角。
From de Broglie, $p = h/\lambda$ with $\lambda = 0.10\times 10^{-9}\ \mathrm{m}$: (M1)
$$ p = \frac{6.63\times 10^{-34}}{1.0\times 10^{-10}} \approx 6.6\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
Energy gained equals KE: $eV = \dfrac{p^{2}}{2 m_e}$, so $V = \dfrac{p^{2}}{2 m_e e}$. (M1)
$$ V = \frac{(6.63\times 10^{-24})^{2}}{2(9.11\times 10^{-31})(1.60\times 10^{-19})}. $$(M1)
$$ V \approx 1.5\times 10^{2}\ \mathrm{V}. $$(A1)
For the photon, $E_\gamma = hc/\lambda$: (M1)
$$ E_\gamma = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{1.0\times 10^{-10}} \approx 2.0\times 10^{-15}\ \mathrm{J}. $$The electron KE is $eV \approx (1.60\times 10^{-19})(150.8) = 2.4\times 10^{-17}\ \mathrm{J}$, so the photon energy is larger by a factor $\dfrac{2.0\times 10^{-15}}{2.4\times 10^{-17}} \approx 82$. (A1)
So at equal wavelength the photon deposits far more energy than the electron; the electron is the gentler probe and damages a delicate sample less. (R1)
$\lambda = 2 d \sin\theta$ with $n = 1$: $\sin\theta = \dfrac{\lambda}{2 d} = \dfrac{0.10}{2(0.21)} = 0.238$. (M1)
$$ \theta = \sin^{-1}(0.238) \approx 14^{\circ}. $$(A1)
由德布罗意，$p = h/\lambda$，$\lambda = 0.10\times 10^{-9}\ \mathrm{m}$：(M1)
$$ p = \frac{6.63\times 10^{-34}}{1.0\times 10^{-10}} \approx 6.6\times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}. $$(A1)
获得的能量等于动能：$eV = \dfrac{p^{2}}{2 m_e}$，故 $V = \dfrac{p^{2}}{2 m_e e}$。(M1)
$$ V = \frac{(6.63\times 10^{-24})^{2}}{2(9.11\times 10^{-31})(1.60\times 10^{-19})}. $$(M1)
$$ V \approx 1.5\times 10^{2}\ \mathrm{V}. $$(A1)
对光子，$E_\gamma = hc/\lambda$：(M1)
$$ E_\gamma = \frac{(6.63\times 10^{-34})(3.00\times 10^{8})}{1.0\times 10^{-10}} \approx 2.0\times 10^{-15}\ \mathrm{J}. $$电子动能为 $eV \approx (1.60\times 10^{-19})(150.8) = 2.4\times 10^{-17}\ \mathrm{J}$，故光子能量大出约 $\dfrac{2.0\times 10^{-15}}{2.4\times 10^{-17}} \approx 82$ 倍。(A1)
故在等波长下光子沉积的能量远多于电子；电子是更温和的探针，对脆弱样品的损伤更小。(R1)
$\lambda = 2 d \sin\theta$，$n = 1$：$\sin\theta = \dfrac{\lambda}{2 d} = \dfrac{0.10}{2(0.21)} = 0.238$。(M1)
$$ \theta = \sin^{-1}(0.238) \approx 14^{\circ}. $$(A1)
He-Ne laser, $\lambda = 633\ \mathrm{nm}$, power $5.0\ \mathrm{mW}$. (a) energy and momentum of one photon; (b) photons per second; (c) show absorbed-beam force is $P/c$ and find it; (d) new force on a perfect mirror, with a reason.氦氖激光，$\lambda = 633\ \mathrm{nm}$，功率 $5.0\ \mathrm{mW}$。(a) 单光子能量与动量；(b) 每秒光子数；(c) 证明被吸收光束的力为 $P/c$ 并求之；(d) 理想镜面上的新作用力及理由。
$E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times 10^{-34})(3.00\times 10^{8})}{633\times 10^{-9}} \approx 3.1\times 10^{-19}\ \mathrm{J}$. (A1)
$p = \dfrac{h}{\lambda} = \dfrac{6.63\times 10^{-34}}{633\times 10^{-9}} \approx 1.0\times 10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ (or $p = E/c$). (A1)
$N = P/E$: (M1)
$$ N = \frac{5.0\times 10^{-3}}{3.14\times 10^{-19}} \approx 1.6\times 10^{16}\ \mathrm{s^{-1}}. $$(A1)
Force is the rate of momentum delivery. Each second $N$ photons deliver momentum $N p$, and since $p = E/c$ and $NE = P$: $F = N p = \dfrac{N E}{c} = \dfrac{P}{c}$. (M1)
$$ F = \frac{P}{c} = \frac{5.0\times 10^{-3}}{3.00\times 10^{8}} \approx 1.7\times 10^{-11}\ \mathrm{N}. $$(A1)
The new force is double, $F \approx 3.3\times 10^{-11}\ \mathrm{N}$. (A1)
On reflection each photon reverses its momentum from $+p$ to $-p$, a change of $2p$ rather than $p$, so the rate of momentum transfer, and hence the force, doubles. (R1)
$E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times 10^{-34})(3.00\times 10^{8})}{633\times 10^{-9}} \approx 3.1\times 10^{-19}\ \mathrm{J}$。(A1)
$p = \dfrac{h}{\lambda} = \dfrac{6.63\times 10^{-34}}{633\times 10^{-9}} \approx 1.0\times 10^{-27}\ \mathrm{kg\,m\,s^{-1}}$（或 $p = E/c$）。(A1)
$N = P/E$：(M1)
$$ N = \frac{5.0\times 10^{-3}}{3.14\times 10^{-19}} \approx 1.6\times 10^{16}\ \mathrm{s^{-1}}. $$(A1)
力是动量传递的速率。每秒 $N$ 个光子传递动量 $N p$，由 $p = E/c$ 与 $NE = P$：$F = N p = \dfrac{N E}{c} = \dfrac{P}{c}$。(M1)
$$ F = \frac{P}{c} = \frac{5.0\times 10^{-3}}{3.00\times 10^{8}} \approx 1.7\times 10^{-11}\ \mathrm{N}. $$(A1)
新作用力为两倍，$F \approx 3.3\times 10^{-11}\ \mathrm{N}$。(A1)
反射时每个光子的动量从 $+p$ 反向为 $-p$，变化量为 $2p$ 而非 $p$，故动量传递速率（即力）加倍。(R1)