1 The Quantum Mechanical Model of the Atom Chapter 7

2 The nature of light Light is electromagnetic radiation: a wave of oscillating electric & magnetic fields.

3 Wave properties A wave has wavelength, frequency, and amplitude

4 Wave properties Wavelength = (lambda), in m Frequency = (nu), in cycles/sec or s -1 Wavelength & frequency related by wave speed: Speed of light c = 3.00 x 10 8 m s -1

5 Example1 Calculate the frequency of light with = 589 nm

6 Example 1 Calculate the frequency of light with = 589 nm  = cc = 3.00 x 10 8 m s -1  Must convert nm to m so units agree (n = )

7 Example 2 Calculate the wavelength in nm of light with = 9.83 x s -1

8 Electromagnetic spectrum = all wavelengths of electromagnetic radiation

9 Photoelectric effect Light shining on metal surface can cause metal to emit e - (measured as electric current)

10 Photoelectric effect Classical theory  more e - emitted if light either brighter (amplitude) or more energetic (shorter )  e – still emitted in dim light if given enough time for e - to gather enough energy to escape

11 Photoelectric effect Observations do not match classical predictions!  A threshold frequency exists for e - emission: no e - emitted below that regardless of brightness  Above threshold, e - emitted even with dim light  No lag time for e - emission in high, dim light

12 Albert Einstein explains photoelectric effect Light comes in packets or particles called photons Amount of energy in a photon related to its frequency h = x J s

13 Example 3 What is the energy of a photon with wavelength nm?

14 Example 4 A Cl 2 molecule has bond energy = 243 kJ/mol. Calculate the minimum photon frequency required to dissociate a Cl 2 molecule.

15 Example 4

16 Example 4 What color is this photon? A photon with wavelength 492 nm is blue

17 Example 4 A Cl 2 molecule has bond energy = 243 kJ/mol. Calculate minimum photon frequency to dissociate a Cl 2 molecule. What color is this photon? A photon with wavelength 492 nm is blue

18 Emission spectra When atom absorbs energy it may re-emit the energy as light

19 Emission spectra White light spectrum is continuous Atomic emission spectrum is discontinuous  Each substance has a unique line pattern

20 Hydrogen emission spectrum Visible lines at  410 nm (far violet)  434 nm (violet)  486 nm (blue-green)  656 nm (red)

21 Emission spectra Classical theories could not explain  Why atomic emission spectra were not continuous  Why electron doesn’t continuously emit energy as it spirals into the nucleus

22 Bohr model Niels Bohr’s model to explain atomic spectra  electron = particle in circular orbit around nucleus  Only certain orbits (called stationary states) can exist r n = orbit radius, n = positive integer, a 0 = 53 pm  Electron in stationary state has constant energy R H = x 10 –18 J  Bohr model is quantized

23 Bohr model e – can pass only from one allowed orbit to another When making a transition, a fixed quantum of energy is involved

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25 Bohr model Bohr model Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2 Photon energy is an absolute amount of energy  Electron absorbs photon, ∆E electron is +  Electron emits photon, ∆E electron is –

26 Bohr model Bohr model Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2 486 nm corresponds to the blue-green line

27 Example Example What wavelength of light will cause the H electron to jump from n=1 to n=3? To what region of the electromagnetic spectrum does this photon belong?

28 Example Example What wavelength of light will cause the H electron to pass from n=1 to n=3?

29 Example Example What wavelength of light will cause the H electron to pass from n=1 to n=3? The atom must absorb an ultraviolet photon with = 103 nm

30 Light has both particle & wave behaviors Wave nature shown by diffraction

31 Light has both particle & wave behaviors Particle nature shown by photoelectric effect

32 Electrons also have wave properties Individual electrons exhibit diffraction, like waves How can e – be both particle & wave?

33 Complementarity Without laser, single e – produces diffraction pattern (wave-like) With laser, single e – makes a flash behind one slit or the other, indicating which slit it went through – – and diffraction pattern is gone (particle-like) We can never simultaneously see the interference pattern and know which slit the e – goes through

34 Complementarity Complementary properties exclude each other  If you know which slit the e – passes through (particle), you lose the diffraction pattern (wave)  If you see interference (wave), you lose information about which slit the e – passes through (particle)  Heisenberg uncertainty principle sets limit on what we can know

35 Indeterminancy Classical outcome is predictable from starting conditions Quantum-mechanical outcome not predictable but we can describe probability region

36 Electrons & probability Schrodinger applied wave mechanics to electrons  Equation (wave function,  ) describe e – energy  Equation requires 3 integers (quantum numbers)  Plot of  2 gives a probability distribution map of e – location = orbital Schrodinger wave functions successfully predict energies and spectra for all atoms

37 Quantum numbers Principal quantum number, n  Determines size and overall energy of orbital  Positive integer 1, 2, 3...  Corresponds to Bohr energy levels

38 Quantum numbers Angular momentum quantum number, l  Determines shape of orbital  Positive integer 0, 1, 2... (n–1)  Corresponds to sublevels l letter 0s 1p 2d 3f

39 Quantum numbers Magnetic quantum number, m l  Determines number of orbitals in a sublevel and orientation of each orbital in xyz space  integers –l l

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41 Shapes of orbitals s orbital (l = 0, m l = 0) p orbitals (l = 1, m l = –1, 0, +1)

42 Shapes of orbitals d orbitals (l = 2, m l = –2, –1, 0, +1, +2)

43 What type of orbital is designated by each set of quantum numbers?  n = 5, l = 1, m l = 0  n = 4, l = 2, m l = –2  n = 2, l = 0, m l = 0 Write a set of quantum numbers for each orbital  4s  3d  5p

44 What type of orbital is designated by each set of quantum numbers?  n = 5, l = 1, m l = 05p  n = 4, l = 2, m l = –24d  n = 2, l = 0, m l = 02s Write a set of quantum numbers for each orbital  4s n = 4, l = 0, m l = 0  3d n = 3, l = 2, m l = –2, –1, 0, +1, or +2  5p n = 5, l = 1, m l = –1, 0, or +1

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46 Electron configurations Aufbau principle: e – takes lowest available energy Hund’s rule: if there are 2 or more orbitals of equal energy (degenerate orbitals), e – will occupy all orbitals singly before pairing Pauli principle:  Adds a 4th quantum number, m s (spin)  No two e – in an atom can have the same set of 4 quantum numbers ⇒ 2 e – per orbital