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Chapter 7

Chapter 7. Quantum Theory and Atomic Structure. 제 7 장 양자론과 원자 구조. 7.1 빛의 파동성 , 빛의 입자성 고전물리학에서 양자론까지 광전효과 7.2 원자 스펙트럼 , 수소 원자의 보어 모형 Bohr 수소 7.3 물질과 에너지의 파동 - 입자 이중성 하이젠베르크의 불확정성 원리 전자의 이중성 양자역학 7.4 원자의 양자역학적 모형 , 원자 궤도함수 양자수 원자궤도함수. Quantum Theory and Atomic Structure.

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Chapter 7

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  1. Chapter7 Quantum Theory and Atomic Structure

  2. 제 7장 양자론과 원자 구조 7.1 빛의 파동성, 빛의 입자성 • 고전물리학에서 양자론까지 • 광전효과 7.2 원자 스펙트럼, 수소 원자의 보어 모형 • Bohr 수소 7.3 물질과 에너지의 파동-입자 이중성 하이젠베르크의 불확정성 원리 • 전자의 이중성 • 양자역학 7.4 원자의 양자역학적 모형, 원자 궤도함수 • 양자수 • 원자궤도함수

  3. Quantum Theory and Atomic Structure 7.1 The Nature of Light 7.2 Atomic Spectra 7.3 The Wave-Particle Duality of Matter and Energy 7.4 The Quantum-Mechanical Model of the Atom

  4. Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electric field electromagnetic waves. Amplitude Wavelength(l) Direction of wave Magnetic field Speed of light (c) in vacuum = 299 792 458 m/s (exact) = 3.00* × 108 m/s All electromagnetic radiation l×n = c 7.1

  5. Atom Human Bacteria Ǻ FM Radio 88-108 MHz AM Radio 0.6-1.6 MHz Medical X 10-0.1Ǻ microwave oven 2.4 GHz PET Imaging 0.1-0.01Ǻ

  6. λ×n = c λ = c/n λ= 3.00 ×108 m/s / 6.0 ×104 Hz λ= 5.0 ×103 m AM Radio wave λ= 5.0 ×1012 nm Convert frequency to wavelength A photon has a frequency of 6.0× 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? 7.1

  7. Different behaviors of waves and particles. from Silberberg Figure 7.4

  8. The diffraction pattern caused by light passing through two adjacent slits.

  9. Mystery #1, “Black Body Problem” Stefan–Boltzmann law Wien’s law Ultraviolet catastrophe! Rayleigh-Jeans formula 7.1

  10. Mystery #1, “Black Body Problem”Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h× n Planck’s constant (h) h = 6.626×10-34J•s 7.1

  11. When (hc/kT) << 1(long wavelength range) 7.1

  12. When (hc/kT) >> 1(short wavelength range) 7.1

  13. Mystery #2, “Photoelectric Effect” 7.2 from:http://www.mtmi.vu.lt/pfk/funkc_dariniai/quant_mech/wave_f.htm

  14. Mystery #2, “Photoelectric Effect” hn Photoelectric Effect on Sodium Plate KE e- Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html 7.2

  15. KE = hn - BE KE n 0 -BE Mystery #2, “Photoelectric Effect” Photoelectric Effect on Sodium Plate = 1.25 eV × 1.60217646 × 10-19J/eV = 2.0025 × 10-19J = 6.6 × 10-34J s = 1.82 eV = 1.82 × 1.60217646 × 10-19J/eV = 2.91 × 10-19J 7.2

  16. Work function(Binding Energy) Work Functions for Photoelectric Effect CRC handbook on Chemistry and Physics version 2008, p. 12-114

  17. Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905 hn • Light has both: • wave nature • particle nature KE e- Photon is a “particle” of light hn = KE + BE KE = hn - BE 7.2

  18. Convert wavelength to energy When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. E = h ×n E = h ×c / l E = 6.63×10-34(J•s) × 3.00×10 8 (m/s) / 0.154×10-9(m) E = 1.29×10 -15 J 7.2

  19. x-ray diffraction of aluminum foil electron diffraction of aluminum foil Comparing the diffraction patterns of x-rays and electrons. from Silberberg Figure 7.14

  20. The diffraction pattern caused by electron passing through two adjacent slits.

  21. The line spectra of several elements. from Silberberg Figure 7.7

  22. Line Emission Spectrum of Hydrogen Atoms Rydberg equation RHis the Rydberg constant = 1.096776×107m-1 Lyman Balmer Paschen • n1 = 1 and n2 = 2, 3, 4, ... n1= 3 and n2 = 4, 5, 6, ... n1= 2 and n2 = 3, 4, 5, ...

  23. 7.3

  24. Chemistry Mystery: Discovery of Helium In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines Mystery element was named Helium (from helios) In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Helium 587.49 nm Hydrogen

  25. Bohr’s Model of the Atom (1913) • e- can only have specific (quantized) energy values • light is emitted as e- moves from one energy level to a lower energy level +Ze n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.17987209 ×10-18J = 1.096776×107m-1 7.3

  26. ni= 3 nf = 2 656.3 nm 7.3

  27. Calculate a hydrogen atomic emission line wavelength Ephoton = -DE= RH i f ( ) 1 1 n2 n2 Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. Ephoton = - 2.18×10-18 J ×(1/25 - 1/9) = 1.55 ×10-19 J 1/ l = -1.10×107 m-1×(1/25 - 1/9) = 7.82 × 10 5 m-1 l = 1.28×10-6 m = 1280 nm Ephoton = h×c/ l l = h×c/ Ephoton l= 6.63×10-34(J•s)× 3.00×108 (m/s)/1.55×10-19J = 1.28×10-6 m l=1280 nm 7.3

  28. De Broglie (1924) reasoned that e- is both particle and wave. u= velocity of e- m = mass of e- 7.4

  29. Calculate the de Broglie wavelength What is the de Broglie wavelength (in nm) associated with an electron traveling at 1.00× 106m/s? λ= h/mu h in J•s m in kg u in (m/s) λ= 6.63×10-34/ (9.11×10-31× 1.00×106) λ= 7.27×10-10m = 0.727 nm 7.4

  30. Bohr 이론 전자는 원운동을 함 전자 궤도 상에 존재함 전자는 핵에 가까이 갈 수 없음 다전자 원자 적용 불가능 Bohr 수소 원자와 양자역학 양자역학 s 전자는 원운동 하지 않음 전자의 위치는 정할 수 없고 확률 분포함수 s 전자는 핵에서 발견될 확률이 최대 다전자 원자 및 분자에 적용 가능

  31. Electron Imaging Electron Microscopy STM's ability to image the wave patterns of electrons on the surface of a metal le = 0.004 nm 1993 - IBM's Quantum Corral

  32. Classical Theory to QUANTUM THEORY

  33. Heisenberg Uncertainty Principle 위치와운동량과의 관계 위치와운동량을 동시에 정확히 알 수 없다.

  34. Estimate uncertainty of a position 양성자의 위치를 1×10-11m의 정확도로 측정하였다. 1초 후의 양성자 위치의 불확정성은?

  35. Estimate minimum uncertainty of momentum and speed 원자핵의 반경은 5×10-15m이다. 전자가 원자핵의 일부가 되기 위해 필요한 운동에너지는? 너무 큰 에너지로 전자가 핵 안에 있을 확률이 거의 없다.

  36. Estimate minimum uncertainty of momentum and speed 수소원자의 반경이 5.3×10-11m이다. 전자의 최소 운동에너지는?

  37. Schrodinger Wave Equation • In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- • Wave function (Y) describes: • . energy of e- with a given Y • . probability of finding e- in a volume of space • Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5

  38. wave function total quantized energy of the atomic system mass of electron kinetic energy potential energy Hamiltonian Operator The Schrödinger Equation HY = EY

  39. Schrodinger Wave Equation Ψ = f(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus n=1 n=2 n=3 7.6

  40. Where 90% of the e- density is found for the 1s orbital 7.6

  41. Schrodinger Wave Equation Ψ= f(n, l, ml, ms) angular momentum quantum number l for a given value of n, l= 0, 1, 2, 3, … n-1 l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e- occupies 7.6

  42. l = 0 (sorbitals) l = 1 (porbitals) 7.6

  43. l = 2 (dorbitals) 7.6

  44. Schrodinger Wave Equation Ψ= f(n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or2 orientation of the orbital in space 7.6

  45. ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2 7.6

  46. Schrodinger Wave Equation Ψ= f(n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½ 7.6

  47. Schrodinger Wave Equation Ψ= f(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Ψ. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Electronisa fermion. Only one fermion can occupy a quantum state at a given time; this is the Pauli Exclusion Principle. 7.6

  48. Schrodinger Wave Equation Ψ = f (n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of nandl Orbital – electrons with the same values of n, l, andml How many electrons can an orbital hold? If n, l, and mlare fixed, then ms = ½ or - ½ Ψ = f (n, l, ml, ½) orΨ = f (n, l, ml, -½) An orbital can hold 2 electrons 7.6

  49. n=2 n=3 If l = 1, then ml = -1, 0, or +1 3 orbitals l = 1 l = 2 If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e- How many 2porbitals are there in an atom? 2p How many electrons can be placed in the 3dsubshell? 7.6

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