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HKU PHYS2265 Test1复习(论述)

  • 2026-10-04 01:22:37
HKU PHYS2265 Test1复习(论述)
本人英文表述很差,所以专门做了这个专题,用于个人复习。本文所有问题答案除特殊标注之外,均为参考资料提供的答案。

目录

一,黑体辐射与普朗克假说

二,光电效应与经典波动理论的冲突

三,X射线产生(轫致辐射)与康普顿散射

四,对产生与对湮灭

五,波粒二象性和物质波干涉

六,海森堡不确定性原理

七,波函数物理意义

一,黑体辐射与普朗克假说
  • 瑞利-金斯定律(Rayleigh-Jeans Law):
经典电磁理论预测高频下的能量密度趋于无穷大,出现“紫外灾变”(Ultraviolet Catastrophe)。
  • 维恩位移律 (Wien's Displacement Law):
峰值波长与绝对温度成反比。
  • 斯特藩-玻尔兹曼定律 (Stefan-Boltzmann Law):
总辐射强度与绝对温度的四次方成正比。
  • 普朗克假说 :
振子能量是量子化的(E = nhf);能量吸收与发射以离散的“量子”形式进行。打破了经典物理中能量连续可分的观念。

1, (Conceptual Questions 2 Q2) What assumptions did Planck make in dealing with the problem of blackbody radiation? Discuss the consequences of these assumptions.

Assumptions: 
(1) Energy of radiation oscillator is quantized.
(2) Oscillators emit or absorb energy in discrete irreducible packets.
(Here the “oscillator” refers to the atoms or molecules that made up the wall of the blackbody cavity.) 
Consequence: Contradicts the classical ideas that energy is continuously divisible.
2, (2016/17 Test 1 - Q1(b)) Explain why the statement is incorrect: The Rayleigh-Jeans law implies that the energy radiated by a blackbody per unit time is directly proportional to T^4.(答案由Gemini整理)
The Rayleigh-Jeans law leads to the ultraviolet catastrophe, where the total radiated energy integrated over all frequencies diverges to infinity rather than yielding a finite value proportional to T^4.  
3, (2024/25 Test 1 - Q1(I)) True or False: If the Celsius temperature of a blackbody is reduced by a factor of 2, then the total intensity emitted from the blackbody will decrease by a factor of 16.
FALSE. The total intensity is proportional to T^4 where T is the Kelvin temperature, not Celsius temperature.
二,光电效应与经典波动理论的冲突
光强的经典波动理论解释:
光是一种连续的电磁波,光强 I 正比于电磁波振幅(Electric Field Amplitude, E_0)的平方。光强增大仅仅代表电磁波的振幅变大、连续传输的能量流密度增大,与频率无关。
Light with a larger intensity
→Magnitude of the E-field of the light is larger
→Greater electric force is applied on the electrons
→Electrons will absorb more amount of energy
光强的量子解释:
频率不变时,光强增加 = 单位时间入射的光子数量增加 → 光电流上限(最大光电流)增大。
若光强固定而频率增加 → 单光子能量增加,单位时间光子数减少 → 遏止电压增加,但饱和光电流会下降。

1, (Conceptual Questions 2 Q3) Consider the following characteristics of the photoelectric effect. Which of these cannot be explained by classical physics? (答案由Gemini整理)

(i) The generation of photoelectrons.
Yes. Classical wave theory predicts that the electric field of an incoming light wave exerts a force on electrons in the metal, doing work and transferring energy until electrons gain enough kinetic energy to escape.
(ii) The existence of a threshold frequency.
No. Classical physics assumes light energy depends solely on light intensity, not frequency. It predicts that light of any frequency should eventually eject electrons if the intensity is high enough or given enough time. However, experiments show no electrons are emitted below a specific threshold frequency, regardless of intensity.
(iii) The photoelectric current increases with increasing light intensity.
Yes. Classical theory holds that higher light intensity means a greater rate of energy transfer, which frees more electrons per unit time and thus produces a larger photoelectric current.
(iv) The photoelectric current is independent of the anode-cathode potential difference ΔV for ΔV > 0.
Yes. Once the positive accelerating potential ΔV is large enough, all photoelectrons emitted per second are collected by the anode. The current reaches saturation and cannot increase further with higher potential. 
(v) The photoelectric current decreases slowly as ΔV becomes much more negative.
Yes.A negative potential creates a retarding electric field that opposes the motion of electrons. Because emitted electrons possess different initial velocities and directions, fewer electrons have enough kinetic energy to reach the anode as the potential becomes increasingly negative, resulting in a gradual decrease in current.
(vi) The stopping potential is independent of the light intensity.
No. Classical theory predicts that higher light intensity supplies more energy per electron, which should increase the maximum initial kinetic energy and require a larger stopping potential. Experimentally, stopping potential depends only on light frequency, not on intensity.
(vii) The photoelectric current appears almost instantly when the light is turned on.
No. In classical wave theory, at very low light intensity, the energy rate received by an individual electron is extremely small, predicting a long time lag (hours or days) for the electron to accumulate enough energy to escape. Experimentally, photoelectrons are emitted almost instantaneously.
2, (Conceptual Questions 2 Q3)
3, (2016/17 Test 1 - Q1(a)) Explain why incorrect: "When shining radio wave on a block of sodium metal, photo-electrons come out only after about a few seconds."(答案由Gemini整理)
Radio wave frequencies are below the work function cutoff frequency of sodium, so no photoelectrons can be emitted regardless of time; if above cutoff, emission is instantaneous.
4, (2017/18 Test 1 - Q1(a)) State using no more than four sentences the implications of Photoelectric effect.
The maximum k.e. of emitted electrons is independent of light intensity (1), the existence of a cutoff frequency of light below which no photo-electron is found (2), and instantaneous emission of photo-electrons when the light is turned on (3) all point to the particle nature of light, which classical wave theory cannot explain.
三,X射线产生(轫致辐射)与康普顿散射
轫致辐射 (Bremsstrahlung):
高速电子撞击靶核减速释放 X 射线光子,可以看作逆光电效应:
光电效应——光子能量转化为电子动能
轫致辐射——电子动能转化为光子能量
  • 短波极限仅取决于加速电压 V,与粒子质量无关。
康普顿散射 (Compton Effect):
  • 双峰成因:移动峰(Shifted peak)来自于光子与近乎自由的碰撞电子散射;未移动峰(Unshifted peak)来自于光子与束缚极紧的原子/整个原子散射,因为原子质量极大,波长偏移几乎为零。
  • 适用范围:康普顿波长偏移最大约为 0.00485 nm (ceta取180°,2h/(m_e·c))。对于可见光、无线电波或电视信号,这一偏移量相对其本身波长微不足道。
1, (Conceptual Questions 2 Q9) An isolated atom can emit a photon and the atom’s internal energy drops. This process is called spontaneous emission. Can an isolated electron emit a photon? Why or why not?
No. This is the reverse of a completely inelastic collision. Kinetic energy would have to increase, meaning that internal energy would have to decrease. The photon has zero internal energy, and the electron’s internal energy cannot be changed as it is a fundamental particle.
1, (2024/25 Test 1 - Q1(II)) True or False: The production of X rays by the bremsstrahlung process can be regarded as an inverse photoelectric effect.
True. According to the photon theory, an X-ray photon is produced in the
bremsstrahlung process (by the bombardment of electrons on a target) due to part of kinetic energy of incoming electrons is converted into photon energy. So this process is equivalent to an inverse photoelectric effect in which the energy of the incident photon is delivered to the metal to release an electron from the metal.
2, (2024/25 Test 1 - Q3(a)) Explain why two peaks would be observed in the spectrum of scattered X-ray photons for Compton scattering at any angle θ ≠ 0 ?
(图片来自 Chapter 2 Lecture note)
For Compton scattering at any angle θ ≠ 0 , two peaks would be observed in the spectrum of scattered X-ray photons: one is near the wavelength of the incident X-ray photons and the other is at a longer wavelength. 
The shifted peak is due to Compton scattering of X-ray photons by the nearly free electrons in the target while the unshifted peak is due to Compton scattering of X-ray photons by the tightly bound electrons in the target. For the latter case, the whole atom recoils from the X-ray photon which results in a nearly zero wavelength shift of the photon.
四,对产生与对湮灭
  • 重原子核的作用:维持动量守恒,吸收的能量可忽略不计。
1, (2016/17 Test 1 - Q2(a)) Argue why the reaction Q^+ + Q^- → γ can never occur.(答案由Gemini整理)
In the center-of-mass frame, total momentum is 0. Since a single photon always travels at speed c and carries non-zero momentum E/c, total momentum cannot be conserved.
2, (2023/24 Test 1 - Q1(III)) True or False: It is impossible for pair annihilation to occur with the creation of only one photon.
False. Pair annihilation occurs if both the relativistic energy and momentum are conserved during the process. Conservation of energy and momentum allows the creation of only one photon if there is a nearby heavy atom which carries away the recoil momentum but absorbs negligible amount of energy from the process.
五,波粒二象性和物质波干涉
  • 德布罗意波长 (de Broglie Wavelength):λ = h/p
  • 互补原理 (Principle of Complementarity & Measurement Failure):在电子/光子双缝干涉实验中,无法同时观测到波动性与粒子性。
  • 低强度干涉 (Low-intensity Double-Slit):即使极低光强/粒子流(一次只穿过一个光子/电子/中子),随着时间积累依然会形成干涉图样。这说明每个粒子都是与自己发生干涉,单个粒子具有波函数的概率分布特征)。
1, (Conceptual Questions 3 Q4) Why it is not reasonable to do crystallographic studies with protons?
Proton's de Broglie wavelength (~10^-12 m) is much smaller than crystal plane spacing (~10^-10 m}). Diffraction effects would be unobservable.
2, (2016/17 Test 1 - Q1(d)) Explain why incorrect: "No interference pattern can be seen when light is replaced by proton in the Young double-slit experiment."(答案由Gemini整理)
Protons have wave nature with de Broglie wavelength = h/p. Interference occurs if slit separation is comparable to this wavelength.
3, (2017/18 Test 1 - Q1(c)) State the implications of Young's double slit experiment with very low intensity light.
It shows dual nature of light and validity of wavefunction. 
On one hand, photons are recorded one by one points to its particle nature. 
On the other hand, interference pattern is found after gathering sufficient statistics points to both wave and probabilistic natures.
4, (2024/25 Test 1 - Q1(V)) True or False: In an electron double-slit experiment, no interference pattern would be observed if a detector is installed to detect which slit each electron passes through.
True. According to Bohr's principle of complementarity, it is impossible to simultaneously describe physical observables in terms of both particles and waves.
When we examine the interference pattern, we are studying the wave-like behavior of the electron. Besides, when we examine which slit each electron passes through, we are studying the particle-like behavior of electron. Hence, no interference pattern would be observed on the screen if a detector is installed to detect which slit each electron passes through in an electron double-slit experiment.
六,海森堡不确定性原理

位置-动量不确定性:被限制在微观尺寸r内的粒子(如原子核内的中子、势阱中的粒子),位置不确定度 ▲x ≈ r,导致必有最小动量不确定度

,因此最低动能不能为 0(基态能量不为零)。且空间受限越小(r越小),最低能量越高。

1, (2017/18 Test 1 - Q2(a))
Δx = 0 implies Δp is infinite by the Heisenberg uncertainty principle. Hence, the particle cannot be at rest.
2, (2023/24 Test 1 - Q1(V)) True or False: If lifetime of an excited atom increases, the minimum fractional uncertainty(相对不确定度) of the emitted photon's wavelength increases.
思路:先用海森堡不确定性原理(能量-时间)推出能量不确定度,再将能量不确定度与波长不确定度联系起来(找到能量和波长的关系式后求导)。
3, (2024/25 Test 1 - Q1(IV)) True or False: A neutron trapped within an atom has a non-zero minimum energy which increases as radius r increases.
七,波函数物理意义
1, (2016/17 Test 1 - Q1(e)) Explain why incorrect: 

(答案由Gemini整理)

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  1. CONNECT:[ UseTime:0.000487s ] mysql:host=172.18.0.4;port=3306;dbname=www_sjds;charset=utf8mb4
  2. SHOW FULL COLUMNS FROM `fenlei` [ RunTime:0.000875s ]
  3. SELECT * FROM `fenlei` WHERE `fid` = 0 [ RunTime:0.000447s ]
  4. SELECT * FROM `fenlei` WHERE `fid` = 63 [ RunTime:0.000345s ]
  5. SHOW FULL COLUMNS FROM `set` [ RunTime:0.000647s ]
  6. SELECT * FROM `set` [ RunTime:0.000297s ]
  7. SHOW FULL COLUMNS FROM `article` [ RunTime:0.000685s ]
  8. SELECT * FROM `article` WHERE `id` = 519336 LIMIT 1 [ RunTime:0.000466s ]
  9. UPDATE `article` SET `lasttime` = 1791069825 WHERE `id` = 519336 [ RunTime:0.002407s ]
  10. SELECT * FROM `fenlei` WHERE `id` = 65 LIMIT 1 [ RunTime:0.000399s ]
  11. SELECT * FROM `article` WHERE `id` < 519336 ORDER BY `id` DESC LIMIT 1 [ RunTime:0.000524s ]
  12. SELECT * FROM `article` WHERE `id` > 519336 ORDER BY `id` ASC LIMIT 1 [ RunTime:0.000436s ]
  13. SELECT * FROM `article` WHERE `id` < 519336 ORDER BY `id` DESC LIMIT 10 [ RunTime:0.000695s ]
  14. SELECT * FROM `article` WHERE `id` < 519336 ORDER BY `id` DESC LIMIT 10,10 [ RunTime:0.000736s ]
  15. SELECT * FROM `article` WHERE `id` < 519336 ORDER BY `id` DESC LIMIT 20,10 [ RunTime:0.000602s ]
0.085456s