Chapter 7: Atomic Structure Chemistry-140 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

Electronic Structure & Electromagnetic Radiation Chemistry-140 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 2.9979 x 108 m/s (speed of light, c). Includes visible light, radio waves, microwaves, infrared (heat),ultraviolet, X-ray, and g-ray radiation….

Electromagnetic Radiation speed of light = (wavelength) x (frequency) Chemistry-140 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

Chemistry-140 Lecture 17

Electromagnetic Radiation Chemistry-140 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?

Answer: Since we know: then n = = c = ln = 5.09 x 1014 s-1 Chemistry-140 Lecture 17 Answer: Since we know: then n = = = c = ln 5.09 x 1014 s-1

Quantum Effects & Photons where h = Planck's constant: Chemistry-140 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 = 6.626 x 10-34 J-s. and E = hn

Quantum Effects & Photons Chemistry-140 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 ?

Step 1: Determine the energy of one quantum (photon). E = hn Chemistry-140 Lecture 17 Answer: Step 1: Determine the energy of one quantum (photon). E = hn = (6.63 x 10-34 J-s) (4.69 x 1014 s-1) = 3.11 x 10-19 J

Step 2: Determine how many quanta are in a laser pulse. Chemistry-140 Lecture 17 Step 2: Determine how many quanta are in a laser pulse. Number = = 4.18 x 1016 quanta

Chemistry-140 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.

Chemistry-140 Lecture 17

Photoelectric Effect EK = EP - EB Chemistry-140 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

Photoelectric Effect Question: Chemistry-140 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?

Step 1: Convert threshold frequency to binding energy. Eb = hn Chemistry-140 Lecture 17 Answer: Step 1: Convert threshold frequency to binding energy. Eb = hn = (6.63 x 10-34 J-s) (5.57 x 1014 s-1) = 3.69 x 10-19 J

Step 2: Determine the photon energy of 510 nm light. Chemistry-140 Lecture 17 Step 2: Determine the photon energy of 510 nm light. EP = = 3.90 x 10-19 J

Step 3: Determine the kinetic energy of the emitted electrons Chemistry-140 Lecture 17 Step 3: Determine the kinetic energy of the emitted electrons EK = EP - EB = (3.90 x 10-19 J) - (3.69 x 10-19 J) = 2.10 x 10-20 J

Step 4: Calculate the velocity of the emitted electrons Chemistry-140 Lecture 17 Step 4: Calculate the velocity of the emitted electrons EK = mv2 = 2.10 x 10-20 J v = = 2.15 x 105 m/s

Bohr’s Model of the Hydrogen Atom Chemistry-140 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,...¥

Chemistry-140 Lecture 17

Chemistry-140 Lecture 17

Bohr’s Model of the Hydrogen Atom Chemistry-140 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

Bohr’s Model of the Hydrogen Atom Chemistry-140 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 = 2.179 x 10-18 J En = -RH

Transitions in the Bohr Hydrogen Atom Chemistry-140 Lecture 17 Transitions in the Bohr Hydrogen Atom

Transitions and the Rydberg Equation Chemistry-140 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

Transitions in the Bohr Hydrogen Atom Chemistry-140 Lecture 17 Transitions in the Bohr Hydrogen Atom The radius of a Bohr orbit is given by: r = n2(5.30 x 10-11 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

Transitions in the Bohr Hydrogen Atom Chemistry-140 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?

Step 1: Use the Rydberg equation with ni = 4 and nf = 2. Chemistry-140 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

Step 2: Convert to wavelength of light Chemistry-140 Lecture 17 Step 2: Convert to wavelength of light l = = = 4.86 x 10-7 m c = ln 486 nm

Chapter 7: Atomic Structure Chemistry-140 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

Bohr’s Model of the Hydrogen Atom Chemistry-140 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 = 2.179 x 10-18 J En = -RH

Transitions and the Rydberg Equation Chemistry-140 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

Transitions in the Bohr Hydrogen Atom Chemistry-140 Lecture 18 Transitions in the Bohr Hydrogen Atom

Transitions in the Bohr Hydrogen Atom Chemistry-140 Lecture 18 Transitions in the Bohr Hydrogen Atom The radius of a Bohr orbit is given by: r = n2(5.30 x 10-11 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

Transitions in the Bohr Hydrogen Atom Chemistry-140 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?

Step 1: Use the Rydberg equation with ni = 4 and nf = 2. Chemistry-140 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

Step 2: Convert to wavelength of light Chemistry-140 Lecture 18 Step 2: Convert to wavelength of light l = = = 4.86 x 10-7 m c = ln 486 nm

Dual Nature of the Electron Chemistry-140 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 =

Dual Nature of the Electron Chemistry-140 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 10-28 g)

Use de Broglie's equation for matter waves. Chemistry-140 Lecture 18 Answer: Use de Broglie's equation for matter waves. = = 1.22 x 10-10 m l = 0.122 nm

Chemistry-140 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

Chemistry-140 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.

Wave Functions & Probability Chemistry-140 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.

The Wave Equation & Orbitals Chemistry-140 Lecture 18 The Wave Equation & Orbitals Solutions to the wave equation are called orbitals…..

Chapter 7: Atomic Structure Chemistry-140 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

The Wave Equation & Orbitals Chemistry-140 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.

y n, l, m (r, q, f) = Rnl(r)Clm(q, f) Chemistry-140 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)

Chemistry-140 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,...¥

Chemistry-140 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

Quantum Numbers ml = -l,... -1, 0, 1, ...l Chemistry-140 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

1s, 2s, 2p, 3s, 3p 3d, 3f etc Electronic Shells & Sub-shells Chemistry-140 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

Chemistry-140 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.

Probability Density in Orbitals Chemistry-140 Lecture 19 Probability Density in Orbitals

Radial Distribution in Orbitals Chemistry-140 Lecture 19 Radial Distribution in Orbitals

three p orbitals in each p subshell.. Chemistry-140 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..

d Orbitals The d orbitals are those for which l = 2. Chemistry-140 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

d Orbitals The d orbitals are those for which l = 2. Chemistry-140 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

Nodal Planes in Orbitals Chemistry-140 Lecture 19 Nodal Planes in Orbitals z x y pz orbital xz plane yz plane xy plane dxy orbital

f Orbitals The f orbitals are those for which l = 3. Chemistry-140 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

Textbook Questions From Chapter # 7 Chemistry-140 Lecture 19 Textbook Questions From Chapter # 7 EM Radiation: 22, 28, 30 Photoelectric Effect: 32 Bohr Atom: 35, 38, 40 Matter Waves: 42 Quantum Mechanics: 46, 48, 50, 57, 58, 60 General & Conceptual 68, 72, 77, 79, 91