1. ATOMIC STRUCTURE
CHAPTER 7
All Bold Numbered Problems

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

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

4. ELECTROMAGNETIC RADIATION

5. Electromagnetic Radiation
Most subatomic particles behave as PARTICLES and obey the physics of waves.
Define properties of waves
Figure 7.1 and7.2.
Wavelength, 
Node
Amplitude

6. Figures 7.1
Electromagnetic Frequency

7. Electromagnetic Radiation
wavelength
Visible light
Wavelength ()
Ultraviolet radiation
Amplitude
Node
There are no
LIMITS
to ...
there are
an  .

8. Electromagnetic Radiation
wavelength
Visible light
Ultaviolet radiation
Amplitude
Node
Node in a standing wave

9. Electromagnetic Radiation
Waves have a frequency
Use the Greek letter “nu”, , for frequency, and units are “cycles per sec”
All radiation:  •  = c
where c = velocity of light = 3.00x108 m/sec
Long wavelength ----> small frequency
Short wavelength ----> high frequency

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

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

12. Electromagnetic Radiation
Red light has  = 700. nm.
Calculate the frequency.
700 nm •
1 x 10 -9 m
1 nm =
7.00 x 10 -7
Freq =
3.00 x 10 8
m/s
7.00 x 10 -7 m
4.29 x 10 14
sec -1
Examples

13. Standing Waves 1st vibration 2nd vibration 1st vibration = ½
3rd vibration = 3(½)
2nd vibration
See Figure 7.4

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

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

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

17. From Planck on to Einstein

18. Photoelectric Effect A. Einstein (1879-1955)
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!
No e- observed until light of a certain minimum E is used &
Number of e- ejected depends on light intensity.

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 mv2 term.
It would take some minimum energy i.e. critical energy to knock that electron away from it’s atom.

20. Energy of Radiation
PROBLEM: Calculate the energy of 1.00 mole of photons of red light.
 = nm ( c = l n )
 = x 1014 sec-1
Ep = h•
= (6.63 x J•s)(4.29 x 1014 sec-1)
= x J/photon
Notice Einstein's use of Planck's formula.

21. Energy of Radiation Energy of 1.00 mol of photons of red light.
Ep = h•
= (6.63 x J•s)(4.29 x 1014 sec-1)
= x J per photon
E per mol =
(2.85 x J/ph)(6.02 x 1023 ph/mol)
= kJ/mol
This is in the range of energies that can break bonds.

22. Photoelectric Effect
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.
EK = 1/2 meve2 = Einput - Eminimum
Einstein finds:
Ep = h• = 1/2 meve2, evidence that photons have both wave/particle properties

23. Photoelectric Effect
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.
Step by step!!

24. The Final Crack in Classical, Newtonian Physics MONUMENTAL Edifice
Planck---Energy is NOT Continuous like waves
Einstein---Energy comes in packets or is Quantized and energy also has some wave and particle behavior
Bohr---Applies Quantized idea to atomic particles….the H1 Atom to explain…..

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.

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

27. Figure 7.7

28. Figure 7.8

29. Figure 7.9

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

31. Shells or Levels!!
Why??

32. Figure 7.12

33. Excited Atoms Emit Light

34. Atomic Spectra and Bohr
One view of atomic structure in early 20th century was that an electron (e-) traveled about the nucleus in an orbit.
Electron +
Orbit
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!

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, En = - C/n2
where n = quantum no. = 1, 2, 3, 4, ....
this describes the potential energy of an electron

36. Atomic Spectra and Bohr
Energy of quantized state, En = - C/n2
Only orbits where n = integral numbers are permitted.
Radius of allowed orbitals, Rn, Rn= n2 R0 with Ro = nm
Note the same equations come from modern wave mechanics approach.
Results can be used to explain atomic spectra.

37. Atomic Spectra and Bohr
If e-’s are in quantized energy states, then DE 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

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

39. Atomic Spectra and Bohr= 1 2 E - C ( / )
ENERGY
DE = - (3/4)C
C has been found from experiment and is proportional to RH, the Rydberg constant.
RHhc = C = kJ/mole.
n of emitted light = (3/4)C = x 1015 sec-1
and l = c/n = nm
This is exactly in agreement with experiment!

40. Line Spectra of Excited Atoms
DE = Efinal - Einitial = - RHhc [ (1/nfinal2) - (1/ninitial2) ]
A photon of light with frequency 8.02 x 1013 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.

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+, Li2+.
introduced quantum idea artificially.
However, 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.
For light: E = mc2
E = h = hc / 
Therefore, mc = h / 
and for particles
(mass)(velocity) = h /  ,
the wave-nature of matter.
L. de Broglie
( ) l = h mv

43. Quantum or Wave Mechanics
Baseball (115 g) at 100 mph
 = 1.3 x cm
e- with velocity = x 108 cm/sec
 = nm
Experimental proof of wave properties of electrons

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

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

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

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.
W. Heisenberg

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

49. Symbol Values Description
QUANTUM NUMBERS
Symbol Values Description
n (major) 1, 2, 3, .. Orbital size and energy where E = - RHhc(1/n2)
l (angular) 0, 1, 2, .. n-1 Orbital shape or type (subshell)
ml (magnetic) - l l Orbital orientation
# of orbitals in subshell = 2 l + 1

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

51. Atomic Orbitals the result of Quantum Mechanics Calculations

52. Shells and Subshells When n = 1, then l = 0 and ml = 0 .
Therefore, if n = 1, there is 1 type of subshell and that subshell has a single orbital.
(ml 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. 1s Orbital

54. 2s Orbital

55. 3s Orbital

57. Atomic Orbitals the result of Quantum Mechanics Calculations

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

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

61. 2px Orbital

62. 2py Orbital

63. 2pz Orbital

64. 3px Orbital

65. 3py Orbital

66. 3pz Orbital

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

68. d Orbitals typical d orbital
planar node
See Figure 7.16
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 planar nodes

69. 3dxy Orbital

70. 3dxz Orbital

71. 3dyz Orbital

72. 3dz2 Orbital

73. 3dx2- y2 Orbital

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

75. Orbitals and Quantum Numbers
n l ml s s 2p

76. Orbitals and Quantum Numbers
n l ml s 3p 3d

77. Sample Problems No Yes One 9 5d 3d 6f
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, ml = 1
b) n = 3, l = 2, ml =-1
c) n = 6, l = 3, ml = -3 No
Yes
One 9 5d 3d 6f

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, > 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. Practice Problems Answers
x > 2
x m 5. a) 9 b) 7
4. 4p n l ml
Problem 4 continued on next slide.

80. Practice Problems Answers
3d n l ml 5s

81.  •  = c Sample Problem 1 x 10-7m . n = 3.00 x 108 m/s
1. Calculate the frequency of light having a wavelength of 1 x 10-7m.
 •  = c
1 x 10-7m . n = 3.00 x 108 m/s
n = 3 x 1015 /s

82.  •  = c . 1.5 x 108 /s = 3.00 x 108 m/s Sample Problem  = 2.0 m
2. Calculate the wavelength of light having a frequency of 1.5 x 108 hz.
 •  = c
. 1.5 x 108 /s = 3.00 x 108 m/s
 = 2.0 m

83.  •  = c Sample Problem 1 x 10-6m . n = 3.00 x 108 m/s
3. Calculate the frequency of light having a wavelength of 1 x 103nm.
 •  = c
1 x 10-6m . n = 3.00 x 108 m/s
n = 3 x 1014 /s

84. Practice Problem Ep = h• Ep = 6.63 x 10-34 Js • 3 x 1015/s
1. Calculate the energy of a photon having a frequency of 3 x 1015/s.
Ep = h•
Ep = 6.63 x Js • 3 x 1015/s
= 2 x J

85. Practice Problem Ep = h•  = 5.0 x 1014 /s
2. Calculate the frequency of light having an energy of 2.0 x 105 J/mole.
Ep = h•
2.0 X 105 J mole
mole x 1023 photon
3.3 x J = 6.63 x Js • 
 = 5.0 x 1014 /s

86.  •  = c Practice Problem Ep = h•
3. Calculate the energy of a photon with a wavelength of 575 nm.
 •  = c
5.75 x 10-7 m •  = x 108 m/s
 = 5.22 x 1014/s
Ep = h•
Ep = 6.63 x Js • 5.22 x 1014/s
= 3.46 x J

87.  •  = c Calculate the energy of the photon: Ep = h•
Photon wavelength = 2.35 x m
 •  = c
2.35 x 10-7 m •  = x 108 m/s
 = 1.28 x 1015/s
Ep = h•
Ep = 6.63 x Js • 1.28 x 1015/s
= 8.49 x J

88. Ep = h• Calculate the min. energy to eject an electron:
Min. n = 8.45 x s-1.
Ep = h•
Ep = 6.63 x Js • 8.45 x 1014/s
= 5.60 x J

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 v2
2.89 x J = (1/2) 9.11 x kg v2
= 7.96 x 105 m/s

90. Calculate the energy of the photon:
Ep = 6.63 x Js • 8.02 x 1013/s
5.32 X J x 1023 photon
photon mole
= 3.20 x 104 J = 32.0 kJ

91. Calculate the level number :
Ep = h•
DE = -C [(1/n)2 - (1/n)2]
-32. kJ = kJ [(1/n)2 - (1/8)2]
n = 5