Quantum Physics Notes PDF

Schrodinger Equation and Hydrogen Atom (Bohr’s model of the hydrogen atom) Schrodinger Equation: Well, Step, Barrier Part 3 Schrodinger Equation...

Schrodinger Equation and Hydrogen Atom (Bohr’s model of the hydrogen atom) Schrodinger Equation: Well, Step, Barrier Part 3 Schrodinger Equation Part 2 Electron microscopy Part 4 Heisenberg Uncertainty Relationships De Broglie’s Hypothesis Wave-Par(cle Duality Wave-like particle of electron Part 1 Concepts from Part 1 &2 1  1801: Young’s double-slit experiment: wave behavior of light through interference patterns. 1887: Hertz discovered that UV light can cause sparks between metal electrodes: a particle-like behavior of light. 1905: Following Hertz's discovery, Einstein explained the photoelectric effect and photons, contributing to the understanding of wave-particle duality of light. 1924: Broglie proposed the concept of wave-particle duality of matter : electrons have the wave-like behavior as shown in a de Broglie wavelength. 1926: Schrödinger formulated the Schrödinger Eq., highlighting the wave-like behavior of electrons through wave functions. 1927: Davisson and Germer showed that electrons are scattered by a crystal in their electron diffraction experiments, confirming a wave-like behavior of electrons. 1927: Heisenberg formulates the Uncertainty Principle, highlighting fundamental limits in measuring quantum systems, i.e. , position and momentum cannot be simultaneously measured with precision. 1950s: Electron microscopy utilizing the wave-like behavior of electrons was developed. 2 เกณฑ์ การให้ คะแนน 50% 20% - มีสว่ นร่วมในชันเรี + ยน (เช่น quiz ก่อนเรี ยน /ระหว่างเรี ยน) - การบ้ าน ไม่คดั ลอกมาจากเพื>อน – เขียน และ upload (pdf) ใน LEB2 ภายใน 1 สัปดาห์หลังจาก มอบหมายงาน 30 % (Final Exam) - ผิด-ถูก / ปรนัย/ อัตนัย - 90% ของข้ อสอบจะดัดแปลงมาจากในสไลด์ที>สอนในระดับความยาก-ง่ายที>ใกล้ เคียงกัน 10% ของข้ อสอบจะดัดแปลงมาจากเนื +อหาที>สอนแต่จะมีความยาก-ง่ายที>สงู กว่าโจทย์ตวั อย่าง 3 Part 1 4 Outline 1.1 Wave-Par+cle Duality 1.2 Uncertainty Rela+onships 1.3 Heisenberg Uncertainty Rela+onships 5 1.1 Wave-Particle Duality 6 1.1 Wave-Particle Duality èthe concept that light and electrons (including other quantum particles) can exhibit both wave-like and particle-like properties. (depending on the experimental setup) à The evidence for light behaving as wave: interference, diffraction, polarization, and reflection. à UV light could cause sparks to jump between metal electrodes. Einstein (1905) explained the effect and demonstrated the particle-like behavior of light. (Photoelectric effect) à X-rays were scattered by electrons in a material, the scattered X-rays exhibited a longer wavelength than the incident X-rays, indicating a transfer of energy and a loss of energy in the process. (Compton effect) à Electron appears as a localized point with mass and charge (classical particles). However, it can exhibit interference patterns (wave behavior) in double-slit experiment. à In classical, neutrons contribute to the mass and stability of atoms. However, it exhibits interference patterns in neutron diffraction experiments. 7 1.1 Wave-Particle Duality Photoelectric effect 5. The energy of a photon depends on its frequency (f). 1. A phenomenon in which electrons are emitted from a metal surface when it is exposed to light (photon) of enough frequency. 𝐸 = ℎ𝑓 2. Light is expressed in term of photon. h = Planck's equation 3. The emitted electron is called photoelectron. 6. Light intensity (I) is a measure of the number of photons per 4. This effect was first observed and explained by unit area and per unit (me. Einstein and is a fundamental principle in the field of quantum mechanics. 7. So, with a higher light intensity, there are more photons available to interact with the metal, leading to a larger number of photoelectrons being emiSed. Example, 8 1.1 Wave-Particle Duality Photoelectric effect - Escape the material entirely. - Free from the material’s potential. Vacuum level (eV) - Not the same for all metals; 1. Threshold frequency (𝒇𝟎 ): (e.g., electronic structure, surface Solid-state physics conditions ) The minimum frequency of incident light (photon) required to initiate the emission of photoelectrons (A) CB from a metal surface. 2. Work func the highest energy level in a material at absolute zero temperature (0 K or -273.15°C) at which electrons can occupy. => It serves as a reference point in energy band theory for the distribution of electrons in a solid. => above absolute zero (0 K), the Fermi level indicates the energy state at which the probability of à Minimum amount of energy finding an electron is 50%. needed to remove an electron from the metal. In metals, work function and ionization energy are the same. 10 (h#ps://en.wikipedia.org/wiki/Fermi_level#/media/File:Band_filling_diagram.svg) 1.1 Wave-Particle Duality Photoelectric effect 𝐸 = ℎ𝑓 (1) ℎ𝑓 = 𝑊 + 𝐸' (3) (energy of photon) (conservation of energy in photoelectric effect) ℎ𝑓& = 𝑊 (2) 1 (specific for each metal) 𝐸! = 𝑚𝑣 " (4) 2 (Here, it is a maximum value) 11 1.1 Wave-Particle Duality Photoelectric effect UV UV ℎ𝑓 = 𝑊 + 𝐸! (5) 𝐸 = ℎ𝑓 (1) 1 𝒇𝒓𝒐𝒎, 𝑚𝑣 # = 𝑒𝑉$%% (8) 1 2 ℎ𝑓# = 𝑊 (2) 𝑚𝑣 # = 𝑒𝑉" (6) 2 𝟖 → (𝟓) (kinetic energy of the electrons and ℎ𝑓 = 𝑊 + 𝐸! (3) potential energy due to the electric field) ℎ𝑓 = 𝑊 + 𝑒𝑉$%%. (9) 1 𝐸! = 𝑚𝑣 " (4) 𝟔 → (𝟓) e = charge of the electron, 1.6×10−19 C. 2 ℎ𝑓 = 𝑊 + 𝑒𝑉" (7) 12  1.1 Wave-Particle Duality Photoelectric effect In a photoelectric effect experiment, UV light of wavelength 200 nm is incident on a metal surface. The work function of the metal is 3.0 eV. (h = 6.626×10−34J s and 1 eV = 1.602×10−19 J) (a) What is the maximum kinetic energy of the ejected electrons? (b) What is the stopping potential required to prevent these electrons from reaching the detector? (a) What is the maximum kinetic (b) What is the stopping potential required to prevent these energy of the ejected electrons? electrons from reaching the detector? ℎ𝑓 = 𝑊 + 𝐸! (1) 1 𝑚𝑣 # = 𝑒𝑉" 2 𝑐 𝐸! = ℎ − 𝑊 𝜆 (3×108) 3.2 𝑒𝑉 = 𝑒𝑉" 𝐸! = (6.626×10−34 ) −𝑊 (200×10-9) 9.939 × 10−19 𝑉" = 3.2 𝑉 𝐸! = 𝑒𝑉 − 3 𝑒𝑉 (1.602 × 10-19) 𝐸! = 3.2 𝑒𝑉 13 1.1 Wave-Particle Duality Photoelectric effect An electron is accelerated from rest through a potential difference (accelerating voltage) of 5 kV. (a) What is the kinetic energy of the electron after being accelerated? (b) What is the speed of the electron after being accelerated? Ignore relativistic effects. (a) What is the kinetic energy of the electron (b) What is the speed of the electron after being accelerated? after being accelerated? Ignore relativistic effects. 1 1 𝑚𝑣 # = 𝑒𝑉$%% 𝐸. 𝑘 = 𝑒𝑉$%% 𝑚𝑣 # = 𝑒𝑉$%% 2 2 = (1.602 ×10−19C)(5,000V) 1 𝑚𝑣 # = 8.01 × 10−16 J 2 𝐸. 𝑘 = 8.01 × 10−16 J # 2 (8.01 × 10−16 ) 𝑣 = To convert this energy to electron volts (eV), 9.109 ×10−31 8.01 × 10−16 𝐸. 𝑘 (eV) = 1.602 × 10−19 𝑣 ≈ 1.3 𝑥 10 & 𝑚/𝑠 𝐸. 𝑘 eV = 5,000 𝑒𝑉 14 1.1 Wave-Particle Duality Compton effects => Phenomenon that X-ray (or gamma-ray) collide with electrons in atom, resulting in a change in the 𝜆 (and hence E) of the scattered photons. => Scattering of photons from charged particles is called Compton scattering => Evidence for the particle nature of X-ray. => Total energy before and after the collision is conserved. => Total momentum before and after the collision is conserved. => During the interaction, the photon transfers part of its energy to the electron, causing the electron to recoil and the photon to be scattered at a longer wavelength (lower energy). 15 1.1 Wave-Particle Duality Compton effects (2.1) (2.2) (2.3) (1) (2.4) Δλ = change in photon’s wavelength Note: me= is the mass of the electron Scattering from inner carbon electron produces little Θ = the scattering angle of photon wavelength shift because they are h = Planck's constant tightly bound. 16 1.1 Wave-Particle Duality Compton effects A photon with an initial wavelength of 4 nm undergoes Compton scattering. Calculate the change in wavelength Δλ if the photon is scattered at θ=90∘. How does the change in wavelength Δλ vary for different scattering angles, such as θ=0∘and θ=180∘? Sol. At θ=0o , cos 0o = 1 Δλ = 0 m (no change) At θ=90o , cos 90o =0 At θ=180o , cos 180o = -1 Δλ = Δλ = 2 x (0.00243 nm) Δλ = 0.00243 nm Δλ = 0.00486 nm 17 1.1 Wave-Particle Duality De Broglie’s Hypothesis Double-slit experiment à Light behaves as a wave. Photoelectric and Compton effects à Light behaves as a particle. Is this dual particle-wave nature a property only of light or of objects as well? De Broglie’s Hypothesis : any material particle moving with momentum p there is a wave of wavelength λ, related to p according to (1) (1) 18 1.1 Wave-Particle Duality De Broglie’s Hypothesis Calculate the de Broglie wavelength of the following : (a) A 1000-kg automobile traveling at 100 m/s. Note: ℎ = 4.1357 × 10−15 eV 0 s ℎ = 6.626 × 10−34 𝑱 0 s (b) A 10-g bullet traveling at 500 m/s. 1 𝑒𝑉 = 1.6 × 10−19 𝑱 (c) A smoke particle of mass 10-9 g moving at 1 cm/s. 19 1.1 Wave-Particle Duality De Broglie’s Hypothesis Quantum mechanics Note: Kinetic energy of electron Wave-particle duality. Classical physics The concept of kinetic energy is more complex. According to de Broglie's hypothesis, particles, including electrons, exhibit both particle-like and wave-like properties. The kinetic energy of an electron is the energy it possesses due to its motion. 1 1 𝐸. 𝐾 = 𝑚𝑣 # 𝐸. 𝐾 = 𝑚𝑣 # 2 2 1 𝑚 1 (dealing with macroscopic objects, e.g. cars, = 𝑚𝑣 # I = I (𝑚𝑣)# 2 𝑚 2𝑚 sports, basic mechanics) 1 = I (𝑝)# 2𝑚 𝑝# 𝐸. 𝐾 = 2𝑚 (dealing with microscopic par(cles, collisions, or quantum systems) 20 1.1 Wave-Particle Duality De Broglie’s Hypothesis (Nonrelativistic mechanics) Calculate the de Broglie wavelength of the following: Sol 2: (eV) An electron with a kinetic energy of 1 eV. 𝑝 = 2𝑚𝐾 ℎ 𝜆= 𝑐# 1 𝑝 𝑝= 2𝑚𝐾( # ) 𝑝= 2𝑚𝑐 # 𝐾 𝑐 𝑐 Sol 1: (eV -> J) 𝑚𝑐 # = 0.511 𝑀𝑒𝑉 𝑝# 𝐾= 𝑝𝑐 = 2(0.511 𝑀𝑒𝑉)(1 𝑒𝑉) = 1.01 𝑥 10' 𝑒𝑉 2𝑚 ℎ ℎ 𝑐 𝜆= 𝜆= I 𝑝 𝑝 𝑐 ℎ𝑐 = 1240 𝑒𝑉𝑥 10() 𝑚 1240 𝑒𝑉 𝑥 10() 𝑚 𝜆= = 1.2 𝑛𝑚 1010 𝑒𝑉 21 1.1 Wave-Particle Duality De Broglie’s Hypothesis (RelaGvisGc mechanics) Calculate the de Broglie wavelength of the following: Relaงในแนวแกน y So, at the beginning, a) we know nothing about the x coordinates of electron ∆𝑥 = ∞ (𝑎) b) we know ∆𝒑𝒙 = 𝟎 (𝑏) 54 1.3 Heisenberg Uncertainty Relationships A Statistical Interpretation of Uncertainty A diffraction pattern in double slit exp, shown in the figures and in Slide 28, are the result of the passage of The posi(on of an electron going through a slit is not fixed like a classical par(cle's posi(on. many particles (electrons or photons) through the slit. à They are just spread out. X So, What is the posi(on of an electron? Here, the answer lies in how you measure the posi(on of an electron. If we set up exactly the same situa(on and measure it again, we will find the electron in a different loca(on. Repeated measurements will help and the process displays a staOsOcal distribuOon of locaOons that appears wavelike. 55 1.3 Heisenberg Uncertainty Relationships A Statistical Interpretation of Uncertainty For an experiment with a large number of particles passing (one at a time) through the slit, and we measure the x-component momentum of each particle after it passes through the slit. The y axis: the number of particles with momentum in each interval corresponding to different locations of the detector on the screen. The x axis: the distribution of momentum in x direction. The distribution is centered about zero, which indicates that the mean or average value of 𝑝0 is zero. This distribuAon is similar to the ‘staAsAcal distribuAon’. 56 1.3 Heisenberg Uncertainty Relationships A Statistical Interpretation of Uncertainty In the ‘statistical distribution’, the standard deviation (𝜎‘ ) of a quantity ‘A’ is 𝜎‘ = (𝐴ƒ)’“ −(𝐴’“ )ƒ (𝑎) 𝐴2" + 𝐴"" + 𝐴"-... +𝐴"4 ∑4 𝐴" 562 5 𝐴2 + 𝐴" + 𝐴-... +𝐴4 ∑4 562 𝐴5 (𝐴" )&1 = = (𝑏) 𝐴&1 = = (𝑐) 𝑁 𝑁 𝑁 𝑁 By analogy to Eq. (a), we can make a definition of the uncertainty in momentum as 𝐹𝑜𝑟 1𝐷, ∆𝑝ˆ = (𝑝ˆƒ)’“ −(𝑝ˆ,’“ )ƒ 𝑑 57 1.4 Heisenberg Uncertainty Relationships A Statistical Interpretation of Uncertainty ∆𝑝ˆ = (𝑝ˆƒ)’“ −(𝑝ˆ,’“ )ƒ (𝑎) 𝑝ˆ,’“ = 0 (𝑠𝑙𝑖𝑑𝑒 55) ∆𝑝ˆ = (𝑝ˆƒ)’“ (𝑏) (momentum distribution is centered zero) Thus, it is often said that ∆𝑝ˆ gives a measurement of the magnitude of the momentum of particle in x direction. ∆𝑝ˆ = 𝑝ˆ (magnitude) (𝑐) 58  Note: 1.3 Heisenberg Uncertainty Relationships ℎ = 4.1357 × 10−15 eV # s ℎ = 6.626 × 10−34 $ # s A Statistical Interpretation of Uncertainty Electron was trapped about the diameter of nucleus, thus (High energy) ∆𝑥 ≈ 10 𝑥 10(27 𝑚 = 10 𝑓𝑚 𝑓𝑟𝑜𝑚 ∆𝑥∆𝑝0 ≈ ℏ ℏ ℏ 𝑐 1 ℏ𝑐 ∆𝑝0 ≈ ∆𝑝ˆ ≈ _ = ∆𝑥 ∆𝑥 𝑐 𝑐 ∆𝑥 1 ℏ𝑐 1 197 𝑀𝑒𝑉 𝑥 10(27 𝑚 𝐾 𝑜𝑓 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = ? ∆𝑝0 ≈ = 0 𝑐 ∆𝑥 𝑐 10 𝑓𝑚 𝐾 + 𝑚𝑐 ƒ = (𝑝𝑐)ƒ+(𝑚𝑐 ƒ)ƒ (slide 21) = 19.7𝑀𝑒𝑉/𝑐 𝑚𝑐 # 𝑓𝑟𝑜𝑚 ∆𝑝0 = (𝑝0" )&1 ≈ 𝑝0 (𝑚𝑎𝑔𝑛𝑒𝑡𝑢𝑑𝑒) = 𝑟𝑒𝑠𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 0.511 𝑀𝑒𝑉 𝑝2 ≈ ? 𝐾= (19.7 𝑀𝑒𝑉/𝑐)# 𝑐 # − 0.511 𝑀𝑒𝑉 # − 0.511 𝑀𝑒𝑉 ∆𝑝2 = 𝑝2 (magnitude) 59 (slide 57) 𝐾 ≈ 19. 18 𝑀𝑒𝑉 (𝑚𝑢𝑐ℎ ℎ𝑖𝑔ℎ𝑒𝑟 𝑡ℎ𝑎𝑛 𝑖𝑛 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡) 1.3 Heisenberg Uncertainty Relationships A Statistical Interpretation of Uncertainty 𝑚𝑐 " = 𝑟𝑒𝑠𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 0.511 𝑀𝑒𝑉 Electron was trapped about the diameter of nucleus, thus ∆𝑥 ≈ 1 𝑥 10(27 𝑚 During radioactive decay of beta of Carbon-14, in 𝑓𝑟𝑜𝑚 ∆𝑥∆𝑝0 ~ ℏ which a beta particle (electron) is emitted. Suppose that electrons are previously trapped by the ℏ ℏ 𝑐 1 ℏ𝑐 ∆𝑝0 ~ ∆𝑝0 ≈ 0 = nucleus (1 x 10-15 m). ∆𝑥 ∆𝑥 𝑐 𝑐 ∆𝑥 Q: Use the uncertainty principle to estimate the 1 ℏ𝑐 1 197 𝑀𝑒𝑉 𝑥 10(27 𝑚 range of kinetic energy (K) of the electron. ∆𝑝0 ≈ = 0 𝑐 ∆𝑥 𝑐 1 𝑥 10(27 𝑚 𝐾 + 𝑚𝑐 ƒ = (𝑝𝑐)ƒ+(𝑚𝑐 ƒ)ƒ (see, slide 21) = 197𝑀𝑒𝑉/𝑐 𝑓𝑟𝑜𝑚 ∆𝑝2 = (𝑝2# )$3 ≈ 𝑝2 𝑝2 𝑐 = 197 𝑀𝑒𝑉 𝑝2 ≈ ? ∆𝑝2 = 𝑝2 (magnitude) 𝐾= (197 𝑀𝑒𝑉)# − 0.511 𝑀𝑒𝑉 # − 0.511 𝑀𝑒𝑉 60 (slide 57) 𝐾 ≈ 197 𝑀𝑒𝑉