Unit E2 · Nuclear and Quantum PhysicsUnit E2 · 核物理与量子物理
Quantum Physics量子物理
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus E2.1 to E2.6考纲 E2.1 至 E2.6PHYSICS HL
Name:姓名：Date:日期：
PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · HL only · 30 marks短结构题 · 仅 HL · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. State whether you are working in joules or electronvolts before substituting, and convert with $1\ \mathrm{eV} = 1.60\times 10^{-19}\ \mathrm{J}$. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。代入前先写明使用焦耳还是电子伏特，并用 $1\ \mathrm{eV} = 1.60\times 10^{-19}\ \mathrm{J}$ 换算。数值答案保留适当的有效数字。
Monochromatic light is shone on a clean metal surface and photoelectrons are emitted. A classical wave model of light is used to predict the behaviour of the surface.单色光照射在洁净金属表面，发射出光电子。有人用光的经典波动模型来预言该表面的行为。
(a)State what is meant by the threshold frequency of the metal.说明该金属截止频率的含义。[1]
(b)Identify two experimental observations of the photoelectric effect that a classical wave model cannot explain, and state the classical prediction that fails in each case.指出光电效应中经典波动模型无法解释的两个实验观察，并写出每种情形下失败的经典预言。[3]
Q2MEDIUMPaper 1HL ONLYEinstein equation, work function, threshold爱因斯坦方程、逸出功、阈频率[6 marks]
A metal surface has a work function of $\Phi = 2.5\ \mathrm{eV}$. Ultraviolet light of wavelength $\lambda = 380\ \mathrm{nm}$ is incident on the surface.某金属表面逸出功为 $\Phi = 2.5\ \mathrm{eV}$。波长 $\lambda = 380\ \mathrm{nm}$ 的紫外光入射到该表面。
(a)Calculate the threshold frequency of the metal.计算该金属的截止频率。[2]
(b)Show that the energy of one incident photon is about $3.3\ \mathrm{eV}$.证明一个入射光子的能量约为 $3.3\ \mathrm{eV}$。[2]
(c)Hence determine the maximum kinetic energy of the emitted photoelectrons, in electronvolts.由此求发射光电子的最大动能（以电子伏特表示）。[2]
Q3HARDPaper 1HL ONLYstopping voltage from threshold由阈频率求遏止电压[4 marks]
A photocell has a metal cathode of threshold frequency $f_0 = 5.0\times 10^{14}\ \mathrm{Hz}$. Light of frequency $f = 8.0\times 10^{14}\ \mathrm{Hz}$ illuminates the cathode.某光电管的金属阴极截止频率为 $f_0 = 5.0\times 10^{14}\ \mathrm{Hz}$。频率 $f = 8.0\times 10^{14}\ \mathrm{Hz}$ 的光照射阴极。
(a)Show that the maximum kinetic energy of the photoelectrons can be written $E_{max} = h(f - f_0)$.证明光电子的最大动能可写为 $E_{max} = h(f - f_0)$。[2]
(b)Determine the stopping voltage needed to reduce the photocurrent to zero.求使光电流降为零所需的遏止电压。[2]
A laser emits monochromatic light of wavelength $\lambda = 500\ \mathrm{nm}$.某激光器发出波长 $\lambda = 500\ \mathrm{nm}$ 的单色光。
(a)Calculate the momentum of a single photon of this light.计算这种光中单个光子的动量。[2]
(b)Hence determine the energy of the photon, in joules, using a relation between energy and momentum.由此用能量与动量的关系求该光子的能量（焦耳）。[2]
Q5HARDPaper 1HL ONLYde Broglie wavelength of an accelerated electron加速电子的德布罗意波长[6 marks]
An electron, initially at rest, is accelerated from rest through a potential difference of $100\ \mathrm{V}$.一个最初静止的电子从静止经 $100\ \mathrm{V}$ 电势差加速。
(a)Show that the momentum of the electron is given by $p = \sqrt{2 m_e e V}$.证明电子动量为 $p = \sqrt{2 m_e e V}$。[2]
(b)Calculate the momentum of the electron.计算电子的动量。[2]
(c)Hence calculate its de Broglie wavelength, and state why this electron beam is suitable for studying the structure of a crystal.由此计算其德布罗意波长，并说明为何这束电子适合研究晶体结构。[2]
An electron is confined within a region of width $\Delta x = 5.0\times 10^{-11}\ \mathrm{m}$, comparable to the size of an atom. The Heisenberg uncertainty principle may be taken as $\Delta x\,\Delta p \geq \dfrac{h}{4\pi}$.一个电子被约束在宽度 $\Delta x = 5.0\times 10^{-11}\ \mathrm{m}$ 的区域内，与原子尺度相当。海森堡不确定性原理可取为 $\Delta x\,\Delta p \geq \dfrac{h}{4\pi}$。
(a)Estimate the minimum uncertainty in the momentum of the electron.估算该电子动量的最小不确定度。[2]
(b)Hence estimate the corresponding minimum uncertainty in its velocity, and comment on its size.由此估算其速度的相应最小不确定度，并对其大小作出评论。[2]
(c)State what is meant by wave-particle duality, giving one example of wave behaviour and one example of particle behaviour for the electron.说明波粒二象性的含义，并各举一个电子表现出波动行为与粒子行为的例子。[2]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · HL only · 22 marks图像 · 数据 · 不确定度 · 仅 HL · 22 分
Graph and Data Questions图像与数据题
These items reward extracting $h$ and $\Phi$ from a gradient and intercepts, and correct propagation of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查从斜率与截距中提取 $h$ 与 $\Phi$，以及不确定度的正确传递。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q7HARDPaper 1BHL ONLY$E_{max}$-vs-$f$ line: $h$ from gradient, $\Phi$ from intercept$E_{max}$-$f$ 直线：斜率求 $h$，截距求 $\Phi$[12 marks]
In a photoelectric experiment the maximum kinetic energy of the photoelectrons is measured at several frequencies of incident light. The two data points below lie on the best-fit straight line of $E_{max}$ against frequency $f$:在一次光电实验中，于若干入射光频率下测量光电子的最大动能。下表两个数据点位于 $E_{max}$ 对频率 $f$ 的最佳拟合直线上：
$f\ /\ 10^{14}\ \mathrm{Hz}$
$6.0$
$12.0$
$E_{max}\ /\ \mathrm{eV}$
$0.50$
$2.99$
(a)Starting from Einstein's photoelectric equation, show that a graph of $E_{max}$ against $f$ should be a straight line, and state what the gradient and the two intercepts represent.从爱因斯坦光电方程出发，证明 $E_{max}$ 对 $f$ 的图应为直线，并说明斜率与两个截距各代表什么。[3]
(b)Using the two data points, determine a value for Planck's constant from the gradient. Give your answer in $\mathrm{J\,s}$.用这两个数据点，由斜率求普朗克常数的值。以 $\mathrm{J\,s}$ 给出答案。[3]
(c)Determine the work function $\Phi$ of the metal, in electronvolts.求该金属的逸出功 $\Phi$（电子伏特）。[2]
(d)Determine the threshold frequency of the metal.求该金属的截止频率。[2]
(e)The same experiment is repeated with a different metal of larger work function. State, with a reason, how the new line on the graph compares with the original.用逸出功更大的另一种金属重做同一实验。说明图上的新直线与原直线相比如何，并给出理由。[2]
In an electron-diffraction tube, electrons are accelerated from rest through a measured potential difference $V = (200 \pm 5)\ \mathrm{V}$ before striking a thin graphite target. The de Broglie wavelength of the electrons is $\lambda = \dfrac{h}{\sqrt{2 m_e e V}}$.在电子衍射管中，电子从静止经测得的电势差 $V = (200 \pm 5)\ \mathrm{V}$ 加速，随后射向薄石墨靶。电子的德布罗意波长为 $\lambda = \dfrac{h}{\sqrt{2 m_e e V}}$。
(a)Calculate the de Broglie wavelength of the electrons.计算电子的德布罗意波长。[3]
(b)The percentage uncertainty in $V$ is the dominant source of error. Calculate the percentage uncertainty in $V$.$V$ 的百分比不确定度是主要误差来源。计算 $V$ 的百分比不确定度。[1]
(c)Since $\lambda \propto V^{-1/2}$, determine the percentage uncertainty in $\lambda$, and hence the absolute uncertainty in $\lambda$.由于 $\lambda \propto V^{-1/2}$，求 $\lambda$ 的百分比不确定度，并由此求 $\lambda$ 的绝对不确定度。[3]
(d)The electrons produce a ring diffraction pattern on a fluorescent screen. Explain what this pattern shows about the nature of electrons, and name the type of experiment that first demonstrated this.电子在荧光屏上产生环状衍射图样。解释该图样揭示了电子的什么本性，并说出首次证实这一点的实验类型。[3]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · HL only · 30 marks长结构题 · 仅 HL · 30 分
Extended Structured Problems长结构问题
Carry intermediate values to extra figures and round only the final answer. Keep a clear running note of whether each energy is in joules or electronvolts; mixing the two is the most common error in this unit.中间值多保留几位，仅在最终答案处取舍有效数字。清楚记录每个能量用焦耳还是电子伏特；混用二者是本单元最常见的错误。
Q9HARDPaper 2HL ONLYfull photoelectric cell: energy, $V_s$, photon and electron counts完整光电管：能量、$V_s$、光子与电子计数[12 marks]
A clean sodium surface has a work function $\Phi = 2.3\ \mathrm{eV}$. It is illuminated with ultraviolet light of wavelength $\lambda = 350\ \mathrm{nm}$.洁净的钠表面逸出功为 $\Phi = 2.3\ \mathrm{eV}$。用波长 $\lambda = 350\ \mathrm{nm}$ 的紫外光照射它。
(a)Show that the energy of an incident photon is about $3.6\ \mathrm{eV}$.证明一个入射光子的能量约为 $3.6\ \mathrm{eV}$。[2]
(b)Calculate the maximum kinetic energy of the photoelectrons, in joules, and the stopping voltage required to halt the most energetic electrons.计算光电子的最大动能（焦耳），以及阻止最高能电子所需的遏止电压。[3]
(c)Determine the longest wavelength of light that would still cause emission from this surface.求仍能使该表面发射电子的最长光波长。[3]
(d)The light delivers a power of $1.5\ \mathrm{mW}$ to the surface. Calculate the number of photons striking the surface per second.该光向表面输送 $1.5\ \mathrm{mW}$ 的功率。计算每秒打到表面的光子数。[2]
(e)In practice only $1.0\%$ of incident photons release a photoelectron. Calculate the resulting photoelectric current.实际上只有 $1.0\%$ 的入射光子释放一个光电子。计算由此产生的光电流。[2]
Q10HARDPaper 2HL ONLYmatter waves: electron vs photon of equal wavelength物质波：等波长的电子与光子[10 marks]
A beam of electrons is to be used to probe a crystal in which the atomic spacing is $d = 0.21\ \mathrm{nm}$. For clear diffraction the de Broglie wavelength of the electrons should be $\lambda = 0.10\ \mathrm{nm}$.要用一束电子探测原子间距 $d = 0.21\ \mathrm{nm}$ 的晶体。为产生清晰衍射，电子的德布罗意波长应为 $\lambda = 0.10\ \mathrm{nm}$。
(a)Calculate the momentum of an electron of wavelength $0.10\ \mathrm{nm}$.计算波长 $0.10\ \mathrm{nm}$ 的电子的动量。[2]
(b)Hence determine the accelerating voltage through which the electrons must be accelerated from rest to reach this wavelength.由此求电子从静止须经多大加速电压才能达到此波长。[3]
(c)A photon is now chosen to have the same wavelength, $0.10\ \mathrm{nm}$. Calculate its energy, and state how many times larger it is than the kinetic energy of the electron in (b). Comment on which probe deposits more energy in the sample.现选取一个波长同为 $0.10\ \mathrm{nm}$ 的光子。计算其能量，并说明它是 (b) 中电子动能的多少倍。评论哪种探针在样品中沉积更多能量。[3]
(d)Using the first-order Bragg condition $\lambda = 2 d \sin\theta$, determine the angle $\theta$ at which the first diffraction maximum occurs for the electrons.用一级布拉格条件 $\lambda = 2 d \sin\theta$，求电子的一级衍射极大出现的角度 $\theta$。[2]
A helium-neon laser emits a continuous beam of wavelength $\lambda = 633\ \mathrm{nm}$ carrying a power of $5.0\ \mathrm{mW}$.一台氦氖激光器发出波长 $\lambda = 633\ \mathrm{nm}$、功率 $5.0\ \mathrm{mW}$ 的连续光束。
(a)Calculate the energy and the momentum of a single photon in the beam.计算光束中单个光子的能量与动量。[2]
(b)Calculate the number of photons emitted by the laser each second.计算激光器每秒发出的光子数。[2]
(c)The beam is completely absorbed by a small black target. Show that the force exerted on the target equals $P/c$, and calculate its value.光束被一个小黑色靶完全吸收。证明施加在靶上的力等于 $P/c$，并计算其值。[2]
(d)The target is replaced by a perfect mirror that reflects the beam straight back. State, with a reason, the new force on the target.把靶换成把光束原路反射回去的理想镜面。写出靶上的新作用力，并给出理由。[2]