Download presentation

Presentation is loading. Please wait.

Published bySydni Downard Modified about 1 year ago

1
1 ATOMIC STRUCTURE All Bold Numbered Problems

2
2 Chapter 7 Outline Events leading to Quantum MechanicsEvents leading to Quantum Mechanics –Newton –Planck –Einstein –Bohr –de Broglie –Schrödinger –Heisenberg Using Quantum NumbersUsing Quantum Numbers

3
ATOMIC STRUCTURE From the ERA of Newtonian Physics to Quantum Physics

4
4 ELECTROMAGNETIC RADIATION

5
5 Electromagnetic Radiation Most subatomic particles behave as PARTICLES and obey the physics of waves.Most subatomic particles behave as PARTICLES and obey the physics of waves. Define properties of wavesDefine properties of waves – Figure 7.1 and7.2. Wavelength,Wavelength, NodeNode AmplitudeAmplitude

6
6 Figures 7.1 Electromagnetic Frequency

7
7 Electromagnetic Radiation wavelength Visible light Wavelength ( ) Ultraviolet radiation Amplitude Node There are noLIMITS to to... there are an

8
8 Electromagnetic Radiation Node in a standing wave wavelengt h Visible light wavelengt h Ultaviolet radiation Amplitude Nod e

9
9 Electromagnetic Radiation Waves have a frequencyWaves have a frequency Use the Greek letter “nu”,, for frequency, and units are “cycles per sec”Use the Greek letter “nu”,, for frequency, and units are “cycles per sec” All radiation: = cAll radiation: = c where c = velocity of light = 3.00x10 8 m/sec Long wavelength ----> small frequencyLong wavelength ----> small frequency Short wavelength ----> high frequencyShort wavelength ----> high frequency

10
10 increasing wavelength increasing frequency Electromagnetic Radiation Long wavelength -----> small frequency Short wavelength -----> high frequency See Figure 7.3

11
11 Figure 7.3 Long wavelength -----> small frequency Short wavelength -----> high frequency

12
12 Electromagnetic Radiation Red light has = 700. nm. Calculate the frequency. Freq= 3.00 x 10 8 m/s 7.00 x m 4.29 x sec 700 nm 1 x m 1 nm = 7.00 x m Examples

13
13 Standing Waves See Figure st vibration 2 nd vibration 1 st vibration = ½ 1 st vibration = ½ 2 nd vibration = 2(½ ) 3 rd vibration = 3(½ )

14
14 Newtonian Physics Breakdown -Quantization of Energy- Max Planck ( ) Solved the “ultraviolet catastrophe” It was believed that like wave theory, energy was also continuous.

15
15 Figure 7.5 Objects can gain or lose energy by absorbing or emitting radiant energy in QUANTA. Intensity should Increase with Decreasing. As you add more energy, atoms should vibrate with a higher energy, in a continuous fashion.

16
16 Energy of a vibrating system (electro- magnetic radiation) is proportional to frequency. Energy of a vibrating system (electro- magnetic radiation) is proportional to frequency. E p = h E p = h h = Planck’s constant = x Js Quantization of Energy We now MUST abandon the idea that Energy acts as a continuous wave!

17
17 From Planck on to Einstein

18
18 Photoelectric Effect A. Einstein ( ) Experiment demonstrates the particle nature of light. (Figure 7.6)Experiment demonstrates the particle nature of light. (Figure 7.6) Classical theory said that E of ejected electron should increase with increase in light intensity—not observed!Classical theory said that E of ejected electron should increase with increase in light intensity—not observed! No e - observed until light of a certain minimum E is used &No e - observed until light of a certain minimum E is used & Number of e - ejected depends on light intensity.Number of e - ejected depends on light intensity.

19
19 Photoelectric Effect Experimental observations says that light consists of particles called PHOTONS having discrete energy. It takes a high energy particle to bump into an atom to knock it’s electron out, hence the use of a ½ mv 2 term.It takes a high energy particle to bump into an atom to knock it’s electron out, hence the use of a ½ mv 2 term. It would take some minimum energy i.e. critical energy to knock that electron away from it’s atom.It would take some minimum energy i.e. critical energy to knock that electron away from it’s atom.

20
20 Energy of Radiation PROBLEM: Calculate the energy of 1.00 mole of photons of red light. = 700. nm ( c = ) = 700. nm ( c = ) = 4.29 x sec -1 = 4.29 x sec -1 E p = h E p = h = (6.63 x Js)(4.29 x sec -1 ) = (6.63 x Js)(4.29 x sec -1 ) = 2.85 x J/photon = 2.85 x J/photon Notice Einstein's use of Planck's formula.

21
21 Energy of Radiation Energy of 1.00 mol of photons of red light. E p = h E p = h = (6.63 x Js)(4.29 x sec -1 ) = (6.63 x Js)(4.29 x sec -1 ) = 2.85 x J per photon = 2.85 x J per photon E per mol = (2.85 x J/ph)(6.02 x ph/mol) (2.85 x J/ph)(6.02 x ph/mol) = kJ/mol = kJ/mol This is in the range of energies that can break bonds.

22
22 A minimum frequency is required to cause any current flow. Above that frequency, the current is related to the intensity of the light used. The ejected electrons (since we are talking about collisions between photons and electrons) also have more kinetic energy when higher frequencies are used. A minimum frequency is required to cause any current flow. Above that frequency, the current is related to the intensity of the light used. The ejected electrons (since we are talking about collisions between photons and electrons) also have more kinetic energy when higher frequencies are used. E K = 1/2 m e v e 2 = E input - E minimum Einstein finds: E p = h = 1/2 m e v e 2, evidence that photons have both wave/particle properties Photoelectric Effect

23
23 Light is used to eject an electron from a metal. Calculate the velocity of the ejected electron if the photon used to eject the electron has a wavelength of 2.35 x m and the minimum frequency required to eject an electron is 8.45 x s -1. Light is used to eject an electron from a metal. Calculate the velocity of the ejected electron if the photon used to eject the electron has a wavelength of 2.35 x m and the minimum frequency required to eject an electron is 8.45 x s -1. Photoelectric Effect Step by step!!

24
24 The Final Crack in Classical, Newtonian Physics MONUMENTAL Edifice Planck---Energy is NOT Continuous like wavesPlanck---Energy is NOT Continuous like waves Einstein---Energy comes in packets or is Quantized and energy also has some wave and particle behaviorEinstein---Energy comes in packets or is Quantized and energy also has some wave and particle behavior Bohr---Applies Quantized idea to atomic particles….the H 1 Atom to explain…..Bohr---Applies Quantized idea to atomic particles….the H 1 Atom to explain…..

25
25 Atomic Line Spectra and Niels Bohr Bohr’s greatest contribution to science was BUILDING a SIMPLE MODEL of the ATOM. It was based on an understanding of the LINE SPECTRA of excited atoms and it’s relationship to quantized energy. Niels Bohr ( )

26
26 Line Spectra of Excited Atoms Excited atoms emit light of only certain wavelengths (Planck).Excited atoms emit light of only certain wavelengths (Planck). The wavelengths of emitted light depend on the element.The wavelengths of emitted light depend on the element.

27
27 Figure 7.7

28
28 Figure 7.8

29
29 Figure 7.9

30
30 Visible lines in H atom spectrum are called the BALMER series. Visible lines in H atom spectrum are called the BALMER series. High E Short High Low E Long Low Line Spectra of Excited Atoms

31
31 Shells or Levels!! Why??

32
32 Figure 7.12

33
33 Excited Atoms Emit Light

34
34 Atomic Spectra and Bohr 1. Any orbit (like a wave-see slide 3) should be possible and so should any energy. 2. But a charged particle would always be accelerating from the nucleus (vector velocity is always changing) and since it is moving in an electric field would continuously emit energy. End result should be destruction since the energy mentioned in the previous step is finite! End result should be destruction since the energy mentioned in the previous step is finite! + Electron Orbit One view of atomic structure in early 20th century was that an electron (e-) traveled about the nucleus in an orbit.

35
35 Atomic Spectra and Bohr Bohr said classical (Newtonian) view is wrong !!!. Need a new theory — now called QUANTUM or WAVE MECHANICS. e - can only exist in certain discrete orbits — called stationary states. e - is restricted to QUANTIZED energy states. Energy of state, E n = - C/n 2 where n = quantum no. = 1, 2, 3, 4,.... this describes the potential energy of an electron

36
36 Atomic Spectra and Bohr Only orbits where n = integral numbers are permitted.Only orbits where n = integral numbers are permitted. Radius of allowed orbitals, R n, R n = n 2 R 0 with R o = nmRadius of allowed orbitals, R n, R n = n 2 R 0 with R o = nm Note the same equations come from modern wave mechanics approach.Note the same equations come from modern wave mechanics approach. Results can be used to explain atomic spectra.Results can be used to explain atomic spectra. Energy of quantized state, E n = - C/n 2

37
37 Atomic Spectra and Bohr If e - ’s are in quantized energy states, then E of states can have only certain values. This explain sharp line spectra. If e - ’s are in quantized energy states, then E of states can have only certain values. This explain sharp line spectra. n = 1 n = 2 E = -C ( 1/2 2 ) E = -C ( 1/1 2 )

38
38 Atomic Spectra and Bohr Calculate E for e - “falling” from high energy level (n = 2) to low energy level (n = 1). E = E final - E initial = - C [ (1/1) 2 - (1/2) 2 ] E = - (3/4) C Note that the process is exothermic! n = 1 n = 2 E = -C (1/2 2 ) E = -C (1/1 2 ) ENERGYENERGYENERGYENERGY

39
39 Atomic Spectra and Bohr E = - (3/4)C C has been found from experiment and is proportional to R H, the Rydberg constant. C has been found from experiment and is proportional to R H, the Rydberg constant. R H hc = C = 1312 kJ/mole. of emitted light = (3/4)C = 2.47 x sec -1 and = c/ = nm This is exactly in agreement with experiment! n = 1 n = 2 E = -C (1/2 2 ) E = -C (1/1 2 ) ENERGYENERGYENERGYENERGY

40
40 E = E final - E initial = - R H hc [ (1/n final 2 ) - (1/n initial 2 ) ] A photon of light with frequency 8.02 x s -1 is emitted from a hydrogen atom when it de- excites from the n = 8 level to the n = ? level. Calculate the final quantum number state of the electron. A photon of light with frequency 8.02 x s -1 is emitted from a hydrogen atom when it de- excites from the n = 8 level to the n = ? level. Calculate the final quantum number state of the electron. Line Spectra of Excited Atoms

41
41 Atomic Line Spectra and Niels Bohr Bohr’s theory was a great accomplishment. Rec’d Nobel Prize, 1922 Problems with theory — theory only successful for H and only 1e - systems He +, Li 2+.theory only successful for H and only 1e - systems He +, Li 2+. introduced quantum idea artificially.introduced quantum idea artificially. However, Bohr’s model does not explain many e - systems….So, we go on to QUANTUM or WAVE MECHANICSHowever, Bohr’s model does not explain many e - systems….So, we go on to QUANTUM or WAVE MECHANICS Niels Bohr ( )

42
Quantum or Wave Mechanics de Broglie (1924) proposed that all moving objects have wave properties. de Broglie (1924) proposed that all moving objects have wave properties. For light: E = mc 2 E = h = hc / E = h = hc / Therefore, mc = h / Therefore, mc = h / and for particles (mass)(velocity) = h / (mass)(velocity) = h / the wave-nature of matter. de Broglie (1924) proposed that all moving objects have wave properties. de Broglie (1924) proposed that all moving objects have wave properties. For light: E = mc 2 E = h = hc / E = h = hc / Therefore, mc = h / Therefore, mc = h / and for particles (mass)(velocity) = h / (mass)(velocity) = h / the wave-nature of matter. L. de Broglie ( ) = hmv

43
43 Quantum or Wave Mechanics Baseball (115 g) at 100 mph Baseball (115 g) at 100 mph = 1.3 x cm = 1.3 x cm e - with velocity = 1.9 x 10 8 cm/sec e - with velocity = 1.9 x 10 8 cm/sec = nm = nm Experimental proof of wave properties of electrons

44
44 Schrödinger applied idea of e - behaving as a wave to the problem of electrons in atoms. Schrödinger applied idea of e - behaving as a wave to the problem of electrons in atoms. He developed the WAVE EQUATION. He developed the WAVE EQUATION. E. Schrödinger Quantum or Wave Mechanics

45
45 Solution of the wave equation give a set of mathematical expressions called Solution of the wave equation give a set of mathematical expressions called WAVE FUNCTIONS, Each describes an allowed energy state of an e -. Each describes an allowed energy state of an e -. Quantization is introduced naturally. Quantization is introduced naturally. E. Schrodinger Quantum or Wave Mechanics

46
46 WAVE FUNCTIONS, is a function of distance and two angles. is a function of distance and two angles. Each corresponds to an ORBITAL — the region of space within which an electron is found. Each corresponds to an ORBITAL — the region of space within which an electron is found. does NOT describe the exact location of the electron. does NOT describe the exact location of the electron. 2 is proportional to the probability of finding an e - at a given point. 2 is proportional to the probability of finding an e - at a given point.

47
47 Uncertainty Principle Problem of defining nature of electrons in atoms solved by W. Heisenberg. Cannot simultaneously define the position and momentum (= m v) of an electron. We define e - energy exactly but accept limitation that we do not know exact position. Problem of defining nature of electrons in atoms solved by W. Heisenberg. Cannot simultaneously define the position and momentum (= m v) of an electron. We define e - energy exactly but accept limitation that we do not know exact position. W. Heisenberg

48
48 QUANTUM NUMBERS Each orbital is a function of 3 quantum numbers: n, l, and m l Electrons are arranged in shells(levels) and subshells(sublevels). n --> shell l --> subshell m l --> designates an orbital within a subshell

49
49 SymbolValuesDescription n (major)1, 2, 3,..Orbital size and energy where E = - R H hc(1/n 2 ) l (angular)0, 1, 2,.. n-1Orbital shape or type (subshell) m l (magnetic)- l lOrbital orientation # of orbitals in subshell = 2 l + 1 # of orbitals in subshell = 2 l + 1 QUANTUM NUMBERS

50
50 All 4 Quantum Numbers Principle quantum number (n)Principle quantum number (n) Azimuthal quantum number (l)Azimuthal quantum number (l) Magnetic quantum number (m)Magnetic quantum number (m) Spin quantum number (s)Spin quantum number (s)

51
51 Atomic Orbitals the result of Quantum Mechanics Calculations

52
52 Shells and Subshells When n = 1, then l = 0 and m l = 0. Therefore, if n = 1, there is 1 type of subshell and that subshell has a single orbital. Therefore, if n = 1, there is 1 type of subshell and that subshell has a single orbital. (m l has a single value ---> 1 orbital) This subshell is labeled s Each shell has 1 orbital labeled s, and it is SPHERICAL in shape.

53
53 1s Orbital

54
54 2s Orbital

55
55 3s Orbital

56
56

57
57 Atomic Orbitals the result of Quantum Mechanics Calculations

58
58 p Orbitals When n = 2, then l = 0 and 1 Therefore, in the n = 2 shell there are 2 types of orbitals — 2 subshells For l = 0m l = 0 this is an s subshell this is an s subshell For l = 1 m l = -1, 0, +1 this is a p subshell with 3 orbitals this is a p subshell with 3 orbitals When n = 2, then l = 0 and 1 Therefore, in the n = 2 shell there are 2 types of orbitals — 2 subshells For l = 0m l = 0 this is an s subshell this is an s subshell For l = 1 m l = -1, 0, +1 this is a p subshell with 3 orbitals this is a p subshell with 3 orbitals planar node Typical p orbital See Figure 7.16 When l = 1, there is a PLANAR NODE PLANAR NODE thru the nucleus.

59
59

60
60 p Orbitals A p orbital The three p orbitals lie 90 o apart in space

61
61 2p x Orbital

62
62 2p y Orbital

63
63 2p z Orbital

64
64 3p x Orbital

65
65 3p y Orbital

66
66 3p z Orbital

67
67 d Orbitals When n = 3, what are the values of l? l = 0, 1, 2 and so there are 3 subshells in the shell. For l = 0, m l = 0 ---> s subshell with a single orbital ---> s subshell with a single orbital For l = 1, m l = -1, 0, > p subshell with 3 orbitals ---> p subshell with 3 orbitals For l = 2, m l = -2, -1, 0, +1, > d subshell with 5 orbitals ---> d subshell with 5 orbitals

68
68 d Orbitals s orbitals have no planar node (l = 0) and so are spherical. p orbitals have l = 1, and have 1 planar node, and so are “dumbbell” shaped. This means d orbitals, ( l = 2) have 2 planar nodes typical d orbital planar node See Figure 7.16

69
69 3d xy Orbital

70
70 3d xz Orbital

71
71 3d yz Orbital

72
72 3d z 2 Orbital

73
73 3d x 2 - y 2 Orbital

74
74 f Orbitals When n = 4, l = 0, 1, 2, 3 so there are 4 subshells in the shell. For l = 0, m l = 0 ---> s subshell with single orbital ---> s subshell with single orbital For l = 1, m l = -1, 0, > p subshell with 3 orbitals ---> p subshell with 3 orbitals For l = 2, m l = -2, -1, 0, +1, > d subshell with 5 orbitals ---> d subshell with 5 orbitals For l = 3, m l = -3, -2, -1, 0, +1, +2, > f subshell with 7 orbitals ---> f subshell with 7 orbitals

75
75 Orbitals and Quantum Numbers n l m l 1001s 2002s p

76
76 Orbitals and Quantum Numbers n l m l 3003s p 3d

77
77 Sample Problems 1. Is it possible to have a d orbital in level 1? 2. Is it possible to have a 6s subshell? 3. How many orbitals are in a 7s sublevel? 4. How many orbitals are possible if n = 3? 5. What type of orbital has the quantum numbers a) n = 5, l = 2, m l = 1 b) n = 3, l = 2, m l =-1 c) n = 6, l = 3, m l = -3 NoYesOne 9 5d3d6f

78
78 Practice Problems 1. Calculate the wavelength of a photon having an energy of 2.58 x J. 2. In the hydrogen atom, which transition, 3 --> 2 or 2 --> 1, has the longer wavelength? 3. Calculate the wavelength of an object (mass = 545 lbs) with a speed of 45 miles/hour. 4. Give all possible sets of quantum numbers for 4p, 3d, and 5s. 5. How many orbitals are in the a. the third level?b. l = 3 sublevel?

79
79 Practice Problems Answers x > x m 5. a) 9b) p n l m l Problem 4 continued on next slide.

80
80 Practice Problems Answers 3d n l m l 3d n l m l s500

81
81 Sample Problem 1. Calculate the frequency of light having a wavelength of 1 x m. = c = c 1 x m. = 3.00 x 10 8 m/s = 3 x /s

82
82 Sample Problem 2. Calculate the wavelength of light having a frequency of 1.5 x 10 8 hz. = c = c. x /s = 3.00 x 10 8 m/s. x /s = 3.00 x 10 8 m/s = 2.0 m = 2.0 m

83
83 Sample Problem 3. Calculate the frequency of light having a wavelength of 1 x 10 3 nm. = c = c 1 x m. = 3.00 x 10 8 m/s = 3 x /s

84
84 Practice Problem 1. Calculate the energy of a photon having a frequency of 3 x /s. E p = h E p = h E p = 6.63 x Js 3 x /s = 2 x J

85
85 Practice Problem 2. Calculate the frequency of light having an energy of 2.0 x 10 5 J/mole. E p = h E p = h 3.3 x J = 6.63 x Js 3.3 x J = 6.63 x Js = 5.0 x /s = 5.0 x /s 2.0 X 10 5 J mole mole 6.02 x photon

86
86 Practice Problem 3. Calculate the energy of a photon with a wavelength of 575 nm. = c = c 5.75 x m = 3.00 x 10 8 m/s = 5.22 x /s = 5.22 x /s E p = h E p = h E p = 6.63 x Js 5.22 x /s = 3.46 x J

87
87 Calculate the energy of the photon: Photon wavelength = 2.35 x m = c = c 2.35 x m = 3.00 x 10 8 m/s = 1.28 x /s = 1.28 x /s E p = h E p = h E p = 6.63 x Js 1.28 x /s = 8.49 x J

88
88 Calculate the min. energy to eject an electron: Min. = 8.45 x s -1. Min. = 8.45 x s -1. E p = h E p = h E p = 6.63 x Js 8.45 x /s = 5.60 x J

89
89 Calculate the extra energy of the electron: 8.49 x J x J = 2.89 x J Calculate the velocity of the electron: E = 1/2 m v 2 E = 1/2 m v x J = (1/2) 9.11 x kg v 2 = 7.96 x 10 5 m/s

90
90 Calculate the energy of the photon: E p = h E p = h = 3.20 x 10 4 J = 32.0 kJ 5.32 X J 6.02 x photon photon mole E p = 6.63 x Js 8.02 x /s

91
91 Calculate the level number : E p = h E p = h E = -C [(1/n) 2 - (1/n) 2 ] -32. kJ = kJ [(1/n) 2 - (1/8) 2 ] n = 5 n = 5

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google