Chapter 7 The Quantum–Mechanical Model of the Atom

1. Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro Chapter 7The Quantum–Mechanical Model of the Atom Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Electromagnetic Radiation Electromagnetic Radiation or “Light” is composed of two orthogonal vectors: An electric waveand a magnetic wave.

3. Electromagnetic Radiation • Light is characterized by its wavelength and frequency.

4. Electromagnetic Radiation wavelength Visible Light Amplitude wavelength Node Ultraviolet radiation The intensity of light is a function of the wave’s amplitude. A point of zero amplitude is called a “node”.

5. Electromagnetic Radiation • The frequency of light is represented by the Greek letter “nu”,  •  has units of “cycles per sec” or Hertz (s-1) • All radiation obeys the relationship:  = c • Long wavelength, low frequency • Short wavelength, high frequency • Wavelength and frequency are inversely proportional.

6. Electromagnetic Radiation or Wavelength has units of length: m… m… nm… pm Frequency has units of inverse time: s-1 or Hz (hertz) •  (m)   (s–1) = c (m s–1) c, the speed of electromagnetic radiation (light) moving through a vacuum is: 2.99792458  108 m/s

7. Electromagnetic Radiation increasing frequency increasing wavelength Long wavelength = low frequency Short wavelength = high frequency

8. Electromagnetic Radiation The visible region of the electromagnetic spectrum is only a small portion of the entire spectrum.

9. Electromagnetic Radiation Problem: Visible red light has a wavelength () of 685 nm, calculate the frequency.

10. Quantization of Energy Max Planck (1858-1947) proposed that light waves existed as discrete packets of energy, “quanta” in order to account for the “ultraviolet catastrophe” predicted by classical physics. The “ultraviolet catastrophe” arises from the classical theory for the energy emitted by an ideal black-body governed by the Rayleigh-Jeans law. According to classical physics, the intensity of emitted light approaches infinity as the wavelength of the light approaches zero, hence the term catastrophe.

11. Quantization of Energy Energy of radiation is proportional to frequency An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA. A quanta of energy is the smallest unit of energy that may be exchanged between oscillators or emitted as radiation. It is too small to be observed in the classical world in which we live. E = h ·  h = Planck’s constant = 6.6262 x 10-34 J·s

12. Quantization of Energy E = h ·  Light with long  (low ) has a low Energy Light with a short  (high ) has a high Energy

13. n(MHz) n(s−1) l(m) Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz Given: Find: n = 100.7 MHz l, (m) Conceptual Plan: Relationships: l∙n = c, 1 MHz = 106 s−1 Solve: Check: the unit is correct, the wavelength is appropriate for radiowaves Tro: Chemistry: A Molecular Approach, 2/e 13

14. Planck’s Law and yields: • As the frequency of light increases, the energy of the photon increases • As the wavelength of light increases, the energy of the photon decreases. Blue Light,(higher frequency)has more energy than Red Light,with a lower frequency.

15. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. The energy of a photon in units of kJ/mol can be determined by converting units of wavelength to frequency to energy using Planck’s law and Avogadro's number via dimensional analysis.

16. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol.

17. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol.

18. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol.

19. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol.

20. Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. The units cancel leaving J/mol of photon. This is in the range of energies that can break chemical bonds!

21. number photons Ephoton n(s−1) Practice – What is the frequency of radiation required to supply 1.0 x 102 J of energy from 8.5 x 1027 photons? Given: Find: Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027 n Conceptual Plan: Relationships: E=hn, Etotal = Ephoton∙# photons Solve: Tro: Chemistry: A Molecular Approach, 2/e 21

22. Photoelectric Effects • Certain metals will release (eject) electrons when light strikes the metal surface. • The energy of the light must exceed a minimum or “threshold energy” for this to occur. • Any excess energy beyond this minimum goes into the kinetic energy of the ejected electron. (They fly away with greater velocity) A. Einstein (1879-1955)

23. Photoelectric Effect • Classical theory suggests that energy of an ejected electron should increase with an increase in light intensity. • This however is not experimentally observed! • No ejected electrons were observed until light of a certain minimum energy is applied. • Number of electrons ejected depends on light intensity so long as the light is above a minimum energy. (This “minimum energy” is also the ionization energy of the metal.) A. Einstein (1879-1955)

24. Photoelectric Effect Experiment demonstrates the particle like nature of light.

25. Photoelectric Effect • Conclusion: There is a one-to-one correspondence between ejected electrons and light waves. • This can only occur if light consists of individual units called “PHOTONS”. • A Photon is a packet of light of discrete energy.

26. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.710-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not?

27. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.710-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Solution: Determine if the energy of the light is greater than the minimum threshold energy. If so, then electrons will be ejected, if not, then the switch will not work.

28. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.710-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not?

29. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.710-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? = 3.71019J 540 nm

30. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.710-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? 3.710-19J < 6.7 10-19J Conclusion: The energy of the light is below the minimum threshold. No ejection of electrons will occur. The incident light must have a   297 nm to eject electrons. (Confirm this on your own.)

31. Atomic Line Emission Spectra and Niels Bohr Bohr is credited with the first modern model of the hydrogen atom based on the “line spectra” of atomic emission sources. He proposed a “planetary” structure for the atom where the electrons circled the nucleus in defined orbits. In this model, the attractive electrostatic forces of the electron and nucleus were balanced by the centripetal forces of the orbiting electron. Niels Bohr (1885-1962)

32. Spectrum of White Light When white light passes through a prism, all the colors of the rainbow are observed.

33. Spectrum of Excited Hydrogen Gas When the light from a discharge tube containing a pure element (hydrogen in this case) is passed through the same prism, only certain colors (lines) are observed. Recall that color (wavelength) is related to energy via Planck’s law.

34. Line Emission Spectra of Excited Atoms • Excited atoms emit light of only certain wavelengths • The wavelengths of emitted light are unique to each individual element.

35. Atomic Spectra & Bohr Model + Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states calledorbits. The orbits or energy levels are “quantized” such that only certain levels are allowed. n = 1, 2, 3...  The Bohr Model: rn = n2ao ao = Bohr radius (53 pm)

36. Atomic Spectra & Bohr Model + Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states calledorbits. The lines (colors) corresponded to “jumps” or transitions between the levels.

37. Quantum Mechanical Explanation of Atomic Spectra • Each wavelength in the spectrum of an atom corresponds to an electron transition between orbitals • When an electron is excited, it transitions from an orbital in a lower energy level to an orbital in a higher energy level • When an electron relaxes, it transitions from an orbital in a higher energy level to an orbital in a lower energy level • When an electron relaxes, a photon of light is released whose energy equals the energy difference between the orbitals Tro: Chemistry: A Molecular Approach, 2/e 37

38. Origin of Line Spectra The “Balmer” series for the hydrogen atom is in the visible region of the spectrum. A “series” of transitions end with a common lower level.

39. Line Spectra of Other Elements • Each element has a unique line spectrum. • The lines indicate that the electrons can only make “jumps” between allowed energy levels. • Knowing the color (wavelength) on can determine the magnitude of the energy gaps using Planck's Law.

40. Line Emission Spectra of Excited Atoms Visible lines in H atom spectrum are called the BALMER series. High E Short  High  Low E Long  Low 

41. The Bohr Model of the Atom The energy of each level is given by: R = Rydberg constant (1.097  107 m-1) c = speed of light (2.997  108 ms-1) h = Planck’s constant (6.626 10-34Js) n = the quantum level of the electron (1, 2, 3…) The sign of En is negative because the potential energy between the electron and the nucleus is attractive.

42. Energy Levels • The location of electrons in an energy level is indicated by assigning a number n. The value of n can be 1, 2, 3, 4, etc. • The higher the n value, the higher is the energy of the “shell”that that the electrons occupy. n =  n = 3 n = 2 Each shell can be thought of as a step on a ladder n = 1

43. Energy Levels n =  n = 4 n = 3 Energy n = 2 n = 1 The spacing between adjacent levels is given by: virtual continuum of levels between n = 1 and 2: between n = 2 and 3: (as n increases, the levels get closer together)

44. Energy Levels Key terms and Vocabulary: Ground State: The lowest energy level (n = 1) Excited State: A subsequently higher energy level. n = 2 is the “first excited state” and so on. Absorption: An electron moving from a lower energy level to a higher energy level via excitation. Emission: An electron moving from a higher to a lower energy level accompanied by the release of a photon.

45. Energy Absorption/Emission Since the gaps between states get closer and closer together with increasing n, the frequency of the light emitted changes.

46. Problem: Determination the photon wavelength of a transition between two energy levels:

47. Problem: Determination the photon wavelength of a transition between two energy levels:

48. Problem: Determination the photon wavelength of a transition between two energy levels:

49. Problem: Determination the photon wavelength of a transition between two energy levels:

50. Problem: Determination the photon wavelength of a transition between two energy levels: