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

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

3. Electromagnetic Radiation speed of light = (wavelength) x (frequency)
Chemistry Lecture 17
Electromagnetic Radiation
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)
c = ln

4. Chemistry Lecture 17

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

7. Quantum Effects & Photons where h = Planck's constant:
Chemistry Lecture 17
Quantum Effects & Photons
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 = x J-s.
and
E = hn

8. Quantum Effects & Photons
Chemistry 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. Step 1: Determine the energy of one quantum (photon). E = hn
Chemistry Lecture 17
Answer:
Step 1: Determine the energy of one quantum (photon).
E = hn
= (6.63 x J-s) (4.69 x 1014 s-1) =
3.11 x J

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

11. Chemistry Lecture 17
Photoelectric Effect
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.

12. Chemistry Lecture 17

13. Photoelectric Effect EK = EP - EB
Chemistry Lecture 17
Photoelectric Effect
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
EK = EP - EB

14. Photoelectric Effect Question:
Chemistry Lecture 17
Photoelectric Effect
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?

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

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

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

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

19. Bohr’s Model of the Hydrogen Atom
Chemistry Lecture 17
Bohr’s Model of the Hydrogen Atom
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,...¥

20. Chemistry Lecture 17

21. Chemistry Lecture 17

22. Bohr’s Model of the Hydrogen Atom
Chemistry Lecture 17
Bohr’s Model of the Hydrogen Atom
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;
DE = hn

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

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

25. Transitions and the Rydberg Equation
Chemistry Lecture 17
Transitions and the Rydberg Equation
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
DE = hn = RH

26. Transitions in the Bohr Hydrogen Atom
Chemistry Lecture 17
Transitions in the Bohr Hydrogen Atom
The radius of a Bohr orbit is given by:
r = n2(5.30 x 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

27. Transitions in the Bohr Hydrogen Atom
Chemistry Lecture 17
Transitions in the Bohr Hydrogen Atom
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?

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

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

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

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

32. Transitions and the Rydberg Equation
Chemistry Lecture 18
Transitions and the Rydberg Equation
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
DE = hn = RH

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

34. Transitions in the Bohr Hydrogen Atom
Chemistry Lecture 18
Transitions in the Bohr Hydrogen Atom
The radius of a Bohr orbit is given by:
r = n2(5.30 x 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

35. Transitions in the Bohr Hydrogen Atom
Chemistry Lecture 18
Transitions in the Bohr Hydrogen Atom
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?

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

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

38. Dual Nature of the Electron
Chemistry Lecture 18
Dual Nature of the Electron
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)
l =

39. Dual Nature of the Electron
Chemistry 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 g)

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

41. Chemistry 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 Lecture 18
Quantum Mechanics
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.

43. Wave Functions & Probability
Chemistry Lecture 18
Wave Functions & Probability
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.
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. The Wave Equation & Orbitals
Chemistry Lecture 18
The Wave Equation & Orbitals
Solutions to the wave equation are called orbitals…..

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

46. The Wave Equation & Orbitals
Chemistry Lecture 19
The Wave Equation & Orbitals
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.

47. y n, l, m (r, q, f) = Rnl(r)Clm(q, f)
Chemistry Lecture 19
The Wave Equation & Quantum Numbers
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
y n, l, m (r, q, f) = Rnl(r)Clm(q, f)

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

49. Chemistry Lecture 19
Quantum Numbers
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)
l = 0, 1, 2, ...n-1

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

51. 1s, 2s, 2p, 3s, 3p 3d, 3f etc Electronic Shells & Sub-shells
Chemistry Lecture 19
Electronic Shells & Sub-shells
A collection of orbitals with the same value of n
is called an electron shell
A collection of orbitals with the same values of n and l is called an electronic subshell. A subshell can be referred to using n and the letter equivalent of l,
1s, 2s, 2p, 3s, 3p 3d, 3f etc

53. Chemistry Lecture 19
s Orbitals
The s orbitals are those for which l = 0. All s orbitals are spherical. There is one s orbital in each s subshell.

54. Probability Density in Orbitals
Chemistry Lecture 19
Probability Density in Orbitals

55. Radial Distribution in Orbitals
Chemistry Lecture 19
Radial Distribution in Orbitals

56. three p orbitals in each p subshell..
Chemistry Lecture 19
p Orbitals
The p orbitals are those for which l = 1. All p orbitals are "dumbbell" or "figure-eight" shaped. There are
three p orbitals in each p subshell..

57. d Orbitals The d orbitals are those for which l = 2.
Chemistry Lecture 19
d Orbitals
The d orbitals are those for which l = 2.
There are five d orbitals in each d subshell.
Four are "four-leaf clovers"; the fifth looks like a p orbital with the addition of a ring around the centre

58. d Orbitals The d orbitals are those for which l = 2.
Chemistry Lecture 19
d Orbitals
The d orbitals are those for which l = 2.
There are five d orbitals in each d subshell.
Four are "four-leaf clovers"; the fifth looks like a p orbital with the addition of a ring around the centre

59. Nodal Planes in Orbitals
Chemistry Lecture 19
Nodal Planes in Orbitals z x y
pz orbital
xz plane
yz plane
xy plane
dxy orbital

60. f Orbitals The f orbitals are those for which l = 3.
Chemistry Lecture 19
f Orbitals
The f orbitals are those for which l = 3.
There are seven f orbitals in each f subshell.
Each has 8 lobes

61. Textbook Questions From Chapter # 7
Chemistry Lecture 19
Textbook Questions From Chapter # 7
EM Radiation: 22, 28, 30
Photoelectric Effect: 32
Bohr Atom: , 38, 40
Matter Waves: 42
Quantum Mechanics: 46, 48, 50, 57, 58, 60
General & Conceptual 68, 72, 77, 79, 91