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Chemistry-140 Lecture 17

Chemistry-140 Lecture 17. Chapter Highlights electromagnetic radiation photons & Planck’s constant Bohr model of the atom Rydberg equation quantum mechanics orbitals Heisenberg uncertainty principle quantum numbers. Chapter 7: Atomic Structure. Chemistry-140 Lecture 17.

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Chemistry-140 Lecture 17

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  1. Chemistry-140 Lecture 17 • Chapter Highlights • electromagnetic radiation • photons & Planck’s constant • Bohr model of the atom • Rydberg equation • quantum mechanics • orbitals • Heisenberg uncertainty principle • quantum numbers Chapter 7: Atomic Structure

  2. Chemistry-140 Lecture 17 • Electronic structure of an atom: detailed description of the arrangement of electrons in the atom • Electromagnetic radiation: electrical and magnetic waves traveling at 2.9979 x 108 m/s (speed of light, c). Includes visible light, radio waves, microwaves, infrared (heat),ultraviolet, X-ray, and g-ray radiation…. Electronic Structure & Electromagnetic Radiation

  3. Chemistry-140 Lecture 17 • Wavelength, l: distance between two successive peaks or troughs of a wave. Units are length (m). • Frequency, n: number of complete waveforms that pass through a point in one second. Units of s-1, /s, or hertz (Hz). • Relationship: • speed of light = (wavelength) x (frequency) Electromagnetic Radiation c = ln

  4. Chemistry-140 Lecture 17

  5. Chemistry-140 Lecture 17 Electromagnetic Radiation Question: Yellow light of a sodium vapour lamp has a wavelength of 589 nm. What is the frequency of this light?

  6. Chemistry-140 Lecture 17 Answer: Since we know: then n = = = c = ln 5.09 x 1014 s-1

  7. Chemistry-140 Lecture 17 • Max Planck proposed that radiation is not continuous, but rather consists of small pieces known as quanta (a quantum). • Frequencies, n, of these quanta were whole-number multiples of a fundamental frequency. • Energies E = hn, 2hn, 3hn,... • where h = Planck's constant: • h = 6.626 x 10-34 J-s. • and Quantum Effects & Photons E = hn

  8. Chemistry-140 Lecture 17 Quantum Effects & Photons Question: A laser emits light energy in short pulses with frequency 4.69 x 1014 Hz and deposits 1.3 x 10-2 J for each pulse. How many quanta of energy does each pulse deposit ?

  9. Chemistry-140 Lecture 17 Answer: Step 1: Determine the energy of one quantum (photon). E = hn = (6.63 x 10-34 J-s) (4.69 x 1014 s-1) = 3.11 x 10-19 J

  10. Chemistry-140 Lecture 17 Step 2: Determine how many quanta are in a laser pulse. Number = = 4.18 x 1016 quanta

  11. Chemistry-140 Lecture 17 • Photoelectric effect: metallic surfaces produce electricity (electrons are ejected) when exposed to light. • There is a minimum frequency below which • no electricity is produced. • Above the minimum frequency: • i) number of electrons ejected depends only on light intensity, • ii) energy of the ejected electrons depends only on the frequency of the light. Photoelectric Effect

  12. Chemistry-140 Lecture 17

  13. Chemistry-140 Lecture 17 The packet of energy sufficient to eject an electron is called a photon. The kinetic energy EK of the electrons is given by EB = binding energy EP = photon energy = hn Photoelectric Effect EK = EP - EB

  14. Chemistry-140 Lecture 17 Question: Potassium metal must absorb radiation with a minimum frequency of 5.57 x 1014 Hz before it can emit electrons from its surface via the photoelectric effect. If K(s) is irradiated with light of wavelength 510 nm, what is the maximum possible velocity of an emitted electron? Photoelectric Effect

  15. Chemistry-140 Lecture 17 Answer: Step 1: Convert threshold frequency to binding energy. Eb = hn = (6.63 x 10-34 J-s) (5.57 x 1014 s-1) = 3.69 x 10-19 J

  16. Chemistry-140 Lecture 17 Step 2: Determine the photon energy of 510 nm light. EP = = = 3.90 x 10-19 J

  17. Chemistry-140 Lecture 17 Step 3: Determine the kinetic energy of the emitted electrons EK = EP - EB = (3.90 x 10-19 J) - (3.69 x 10-19 J) = 2.10 x 10-20 J

  18. Chemistry-140 Lecture 17 Step 4: Calculate the velocity of the emitted electrons EK = mv2 = 2.10 x 10-20 J v = = 2.15 x 105 m/s

  19. Chemistry-140 Lecture 17 • A spectrum is produced when radiation from a source is separated into its component wavelengths. • Bohr used Planck's quantum theory to interpret the line spectrum of hydrogen. • Bohr's model of the hydrogen atom described a nucleus surrounded orbits of fixed (quantized) radius, • numbered n = 1, 2, 3,...¥ Bohr’s Model of the Hydrogen Atom

  20. Chemistry-140 Lecture 17

  21. Chemistry-140 Lecture 17

  22. Chemistry-140 Lecture 17 • Bohr concluded: • the energy of the electron in an orbit of hydrogen is quantized • the energy difference between two orbits must also be quantized • The frequency of a line in the spectrum corresponds to the energy difference between two orbits; Bohr’s Model of the Hydrogen Atom DE = hn

  23. En = -RH Chemistry-140 Lecture 17 • The energy of a Bohr orbit (and an electron in it) is given by • where RH is the Rydberg constant = 2.179 x 10-18 J Bohr’s Model of the Hydrogen Atom

  24. Chemistry-140 Lecture 17 Transitions in the Bohr Hydrogen Atom

  25. DE = hn = RH Chemistry-140 Lecture 17 • An electron in the lowest energy orbit, n = 1, is in the ground state • An electron in any orbit other than n = 1, is in an • excited state • The energy of a line is the difference in the energies of the two orbits involved in the transition • DE = Efinal - Einitial Transitions and the Rydberg Equation

  26. Chemistry-140 Lecture 17 • The radius of a Bohr orbit is given by: • r = n2(5.30 x 10-11 m) • The ionization energy of hydrogen is the energy required to remove the electron from the atom, that is; • the energy of the n = 1 to n = ¥ transition Transitions in the Bohr Hydrogen Atom

  27. Chemistry-140 Lecture 17 Question (similar to example 7.4): Calculate the wavelength of light that corresponds to the transition of the electron from the n = 4 to the n = 2 state of the hydrogen atom. Is the light absorbed or emitted by the atom? Transitions in the Bohr Hydrogen Atom

  28. DE = hn = RH Chemistry-140 Lecture 17 Answer: Step 1: Use the Rydberg equation with ni = 4 and nf = 2. n = = = -6.17 x 1014 s-1

  29. Chemistry-140 Lecture 17 Step 2: Convert to wavelength of light l = = = 4.86 x 10-7 m = c = ln 486 nm

  30. Chemistry-140 Lecture 18 • Chapter Highlights • electromagnetic radiation • photons & Planck’s constant • Bohr model of the atom • Rydberg equation • quantum mechanics • Heisenberg uncertainty principle • orbitals • quantum numbers Chapter 7: Atomic Structure

  31. En = -RH Chemistry-140 Lecture 18 • The energy of a Bohr orbit (and an electron in it) is given by • where RH is the Rydberg constant = 2.179 x 10-18 J Bohr’s Model of the Hydrogen Atom

  32. DE = hn = RH Chemistry-140 Lecture 18 • An electron in the lowest energy orbit, n = 1, is in the ground state • An electron in any orbit other than n = 1, is in an • excited state • The energy of a line is the difference in the energies of the two orbits involved in the transition • DE = Efinal - Einitial Transitions and the Rydberg Equation

  33. Chemistry-140 Lecture 18 Transitions in the Bohr Hydrogen Atom

  34. Chemistry-140 Lecture 18 • The radius of a Bohr orbit is given by: • r = n2(5.30 x 10-11 m) • The ionization energy of hydrogen is the energy required to remove the electron from the atom, that is; • the energy of the n = 1 to n = ¥ transition Transitions in the Bohr Hydrogen Atom

  35. Chemistry-140 Lecture 18 Question: Calculate the wavelength of light that corresponds to the transition of the electron from the n = 4 to the n = 2 state of the hydrogen atom. Is the light absorbed or emitted by the atom? Transitions in the Bohr Hydrogen Atom

  36. DE = hn = RH Chemistry-140 Lecture 18 Answer: Step 1: Use the Rydberg equation with ni = 4 and nf = 2. n = = = -6.17 x 1014 s-1

  37. Chemistry-140 Lecture 18 Step 2: Convert to wavelength of light l = = = 4.86 x 10-7 m = c = ln 486 nm

  38. l = Chemistry-140 Lecture 18 • De Broglie proposed that particles may behave as if they were waves. Similar to the idea that light may behave as if it was a particle. Matter wave is the term used by de Broglie where: • where momentum = mv, (m is mass & v is velocity) Dual Nature of the Electron

  39. Chemistry-140 Lecture 18 Dual Nature of the Electron Question: What is the characteristic wavelength of an electron with velocity 5.97 x 106 m/s ? (mass of an electron is 9.11 x 10-28 g)

  40. l = Chemistry-140 Lecture 18 Answer: Use de Broglie's equation for matter waves. = = 1.22 x 10-10 m = 0.122 nm

  41. Chemistry-140 Lecture 18 Quantum Mechanics • Heisenberg Uncertainty Principle: Werner Heisenberg proposed the uncertainty principle, which states that it is impossible for us to know, simultaneously, both the exact momentum of an electron and its exact location in space

  42. Chemistry-140 Lecture 18 • Schrödinger Wave Equation: Erwin Schrödinger proposed a mathematical model of the atom using measured energies and known forces rather than a preconceived "picture" of the atom's structure. This is called quantum mechanics or wave mechanics. Quantum Mechanics

  43. Chemistry-140 Lecture 18 • Solutions to the wave equation are called wave functions, symbolized y. Wave functions cannot describe the exact position of an electron only the probability of finding it in a given location. Wave Functions & Probability • The probability of finding the electron in a given location is the electron density and is given by the square of the wave function for that location,y2.

  44. Chemistry-140 Lecture 18 The Wave Equation & Orbitals • Solutions to the wave equation are called orbitals…..

  45. Chemistry-140 Lecture 19 • Chapter Highlights • electromagnetic radiation • photons & Planck’s constant • Bohr model of the atom • Rydberg equation • quantum mechanics • Heisenberg uncertainty principle • quantum numbers • orbitals Chapter 7: Atomic Structure

  46. Chemistry-140 Lecture 19 • Solutions to the wave equation are called orbitals and • each has a characteristic energy. • An orbital is a region for which there is a high probability of finding the electron; it is not a path or trajectory. The Wave Equation & Orbitals

  47. Chemistry-140 Lecture 19 • Variables in the wave equation are called quantum numbers. The Bohr model used only one variable or quantum number, n. The quantum mechanical model uses three quantum numbers, n,l& ml to describe each orbital The Wave Equation & Quantum Numbers yn, l, m (r, q, f) = Rnl(r)Clm(q, f)

  48. Chemistry-140 Lecture 19 • The principal quantum number (n) has possible values of: • It describes the relative size of the orbital Quantum Numbers n = 1, 2, 3,...¥

  49. Chemistry-140 Lecture 19 • The angular momentum quantum number (l) • has possible values of: • It describes the shape of the orbital. • The value of l is often referred to by a letter equivalent; • 0 = s, 1 = p, 2 = d, 3 = f, .... (the rest are alphabetical) Quantum Numbers l = 0, 1, 2, ...n-1

  50. Chemistry-140 Lecture 19 • The magnetic quantum number ( ml) has values: • It describes the orientation of the orbital in space. Quantum Numbers ml = -l,... -1, 0, 1, ...l

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