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Chapter 6 — Atomic Structure 1 Electromagnetic Radiation Electromagnetic Radiation or ―Light‖ is composed of two orthogonal vectors: Chapter 6 An electric wave and a magnetic wave. The Structure of Atoms Jeffrey Mack California State University, Sacramento Electromagnetic Radiation • Light is characterized by its wavelength and frequency. Electromagnetic Radiation wavelength The intensity of Amplitude light is a function of the wave’s amplitude. wavelength Node Ultraviolet radiation Electromagnetic Radiation • The frequency of light is represented by the Greek letter ―nu‖, • has units of ―cycles per sec‖ or Hertz (s-1) • All radiation obeys the relationship: = c • Long wavelength, low frequency • Short wavelength, high frequency • Wavelength and frequency are inversely proportional. A point of zero amplitude is called a ―node”. Electromagnetic Radiation Wavelength has units of length: m… m… nm… pm Frequency has units of inverse time: s-1 or Hz (hertz) (m) (s–1) = c (m s–1) c, the speed of electromagnetic radiation (light) moving through a vacuum is: 2.99792458 108 m/s n= c l or l= c n Chapter 6 — Atomic Structure 2 Electromagnetic Radiation Electromagnetic Radiation Long wavelength = low frequency Short wavelength = high frequency increasing frequency The visible region of the electromagnetic spectrum is only a small portion of the entire spectrum. increasing wavelength Electromagnetic Radiation Quantization of Energy Max Planck (1858-1947) proposed that light waves existed as discrete packets of energy, ―quanta‖ in order to account for the ―ultraviolet catastrophe‖ predicted by classical physics. The ―ultraviolet catastrophe‖ arises from the classical theory for the energy emitted by an ideal blackbody governed by the Rayleigh-Jeans law. According to classical physics, the intensity of emitted light approaches infinity as the wavelength of the light approaches zero, hence the term catastrophe. Problem: Visible red light has a wavelength ( calculate the frequency. 685 nm 1 frequency ( ) c 1m 109 nm 3.00 6.85 6.85 m s 10-7 m of 685 nm, 10 7 m 108 4.38 1014 sec 1 Quantization of Energy An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA. A quanta of energy is the smallest unit of energy that may be exchanged between oscillators or emitted as radiation. It is too small to be observed in the classical world in which we live. Energy of radiation is proportional to frequency E = h· h = Planck’s constant = 6.6262 x 10-34 J·s Quantization of Energy E = h· Light with long Light with a short (low ) has a low Energy (high ) has a high Energy Chapter 6 — Atomic Structure 3 Planck’s Law E = h´ n n= and h´c E= l yields: Energy of Radiation c l • As the frequency of light increases, the energy of the photon increases • As the wavelength of light increases, the energy of the photon decreases. Blue Light, (higher frequency) has more energy than Red Light, with a lower frequency. Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. The energy of a photon in units of kJ/mol can be determined by converting units of wavelength to frequency to energy using Planck’s law and Avogadro's number via dimensional analysis. Energy of Radiation Energy of Radiation Problem: Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Ephoton = h ´ n = h´c l Ephoton = h ´ n = Ephoton = Energy of Radiation h´c l hc = l Energy of Radiation Problem: Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Ephoton = h ´ n = Ephoton = hc = l 6.626 10 34 h´c l J s ´3.00 ´ 108 m s photon 685 nm ´ 1m 109 nm Ephoton = h ´ n = Ephoton = hc = l h´c l J s ´3.00 ´ 108 m s 6.022 ´ 1023 photons photon ´ mole 1m 685 nm ´ 9 10 nm 6.626 10 34 = 1.75 ´ 105 J mole of photons Chapter 6 — Atomic Structure 4 Energy of Radiation Photoelectric Effects Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. 1.75 ´ 105 J 1kJ kJ ´ = 175 mole of photons 103 J mole The units cancel leaving J/mol of photon. This is in the range of energies that can break chemical bonds! Photoelectric Effect • Classical theory suggests that energy of an ejected electron should increase with an increase in light intensity. • This however is not experimentally observed! • No ejected electrons were observed until light of a certain minimum energy is applied. • Number of electrons ejected depends on light intensity so long as the light is above a minimum energy. (This ―minimum energy‖ is also the ionization energy of the metal.) • Certain metals will release (eject) electrons when light strikes the metal surface. • The energy of the light must exceed a minimum or ―threshold energy‖ for this to occur. • Any excess energy beyond this minimum goes into the kinetic energy of the ejected electron. (They fly away with greater velocity) A. Einstein (1879-1955) Photoelectric Effect Experiment demonstrates the particle like nature of light. A. Einstein (1879-1955) Photoelectric Effect • Conclusion: There is a one-to-one correspondence between ejected electrons and light waves. • This can only occur if light consists of individual units called “PHOTONS” . • A Photon is a packet of light of discrete energy. Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7 10-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Chapter 6 — Atomic Structure 5 Photoelectric Effect Photoelectric Effect Problem: Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7 10-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7 10-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Solution: Ephoton = h ´ n = Determine if the energy of the light is greater than the minimum threshold energy. If so, then electrons will be ejected, if not, then the switch will not work. Photoelectric Effect h´c l Photoelectric Effect Problem: Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7 10-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7 10-19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Ephoton = h ´ n = h´c l 3.7 10-19J < 6.7 10-19J Conclusion: m 6.626 ´ 10 J × s ´3.00 ´ 10 s m 540 nm ´ 9 10 nm -34 Ephoton = 8 = 3.7 10 19J Atomic Line Emission Spectra and Niels Bohr Niels Bohr (1885-1962) Bohr is credited with the first modern model of the hydrogen atom based on the ―line spectra‖ of atomic emission sources. He proposed a ―planetary‖ structure for the atom where the electrons circled the nucleus in defined orbits. In this model, the attractive electrostatic forces of the electron and nucleus were balanced by the centripetal forces of the orbiting electron. The energy of the light is below the minimum threshold. No ejection of electrons will occur. The incident light must have a 297 nm to eject electrons. (Confirm this on your own.) Spectrum of White Light When white light passes through a prism, all the colors of the rainbow are observed. Chapter 6 — Atomic Structure Spectrum of Excited Hydrogen Gas 6 Line Emission Spectra of Excited Atoms • Excited atoms emit light of only certain wavelengths • The wavelengths of emitted light are unique to each individual element. When the light from a discharge tube containing a pure element (hydrogen in this case) is passed through the same prism, only certain colors (lines) are observed. Recall that color (wavelength) is related to energy via Planck’s law. Atomic Spectra & Bohr Model Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states called orbits. Atomic Spectra & Bohr Model Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states called orbits. The orbits or energy levels are ―quantized‖ such that only certain levels are allowed. n = 1, 2, 3... + + The Bohr Model: rn = n2ao ao = Bohr radius (53 pm) Origin of Line Spectra The lines (colors) corresponded to ―jumps‖ or transitions between the levels. Line Spectra of Other Elements The ―Balmer‖ series for the hydrogen atom is in the visible region of the spectrum. A ―series‖ of transitions end with a common lower level. • Each element has a unique line spectrum. • The lines indicate that the electrons can only make ―jumps‖ between allowed energy levels. • Knowing the color (wavelength) on can determine the magnitude of the energy gaps using Planck's Law. Chapter 6 — Atomic Structure Line Emission Spectra of Excited Atoms 7 The Bohr Model of the Atom The energy of each level is given by: High E Short High Low E Long Low En = R = Rydberg constant (1.097 107 m-1) h = Planck’s constant (6.626 10-34Js) Visible lines in H atom spectrum are called the BALMER series. Each shell can be thought of as a step on a ladder c = speed of light (2.997 108 ms-1) n = the quantum level of the electron (1, 2, 3… ) The sign of En is negative because the potential energy between the electron and the nucleus is attractive. Energy Levels •The location of electrons in an energy level is indicated by assigning a number n. The value of n can be 1, 2, 3, 4, etc. •The higher the n value, the higher is the energy of the “shell” that that the electrons occupy. Rhc n2 Energy Levels The spacing between adjacent levels is given by: n= DE = En+1 - En between n = 1 and 2: DE = n=3 n=2 n=1 3Rhc = 0.75 ´ Rhc 4 n= virtual continuum of levels n=4 n=3 n=2 between n = 2 and 3: DE = 5Rhc = 0.14 ´ Rhc 36 n=1 (as n increases, the levels get closer together) Energy Levels Key terms and Vocabulary: Ground State: The lowest energy level (n = 1) Excited State: A subsequently higher energy level. n = 2 is the ―first excited state‖ and so on. Absorption: An electron moving from a lower energy level to a higher energy level via excitation. Emission: An electron moving from a higher to a lower energy level accompanied by the release of a photon. Energy Absorption/Emission Since the gaps between states get closer and closer together with increasing n, the frequency of the light emitted changes. Chapter 6 — Atomic Structure Problem: 8 Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: En = - Rhc n2 Problem: Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: En = - Rhc n2 En = - DE = Efinal - Einitial Rhc n2 DE = Efinal - Einitial DE = - Rhc æ Rhc ö - - 2 ÷ n2final çè ninitial ø Problem: Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: En = - Rhc n2 DE = Efinal - Einitial DE = - æ 1 Rhc æ Rhc ö 1 ö - - 2 ÷ = -Rhc ç 2 - 2 ÷ n2final çè ninitial ø è nfinal ninitial ø æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ è nfinal ninitial ø Chapter 6 — Atomic Structure 9 Problem: Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ è nfinal ninitial ø æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ è nfinal ninitial ø E= h´c l E= h´c l DE = h´c lphoton Problem: Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ è nfinal ninitial ø æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ è nfinal ninitial ø E= h´c l DE = lphoton (meters) = h´c lphoton E= h´c DE h´c l DE = lphoton (meters) = h´c DE h´c lphoton take the absolute value to assure a positive wavelength. The sign of E tells the direction: (+) indicated absorption (-) indicates emission Problem: Problem: Determination the photon wavelength of a transition between two energy levels: Determination the photon wavelength of a transition between two energy levels: R ´ h ´ c = 2.179 ´ 10-18 J æ 1 1 ö DE = -Rhc ç 2 - 2 ÷ nfinal = 5 è nfinal ninitial ø Watch your math… æ 1 1ö DE = -2.179 ´ 10-18 J ´ ç 2 - 2 ÷ è 5 2 ø energy ninitial = 2 n =5 in DE = 4.576 ´ 10 -19 J n =2 The value of E is positive because this is an absorption. æ 1 1 ö ç n2 - n 2 ÷ è final initial ø = 1 1 - = 0.04 - 0.25 = -0.21 25 4 æ ö 1 ç n 2 - n2 ÷ è final initial ø 1 = +0.048 25 - 4 Chapter 6 — Atomic Structure 10 Particle-Wave Duality: A Prelude to Quantum Mechanics Problem: Determination the photon wavelength of a transition between two energy levels: DE = λphoton (m) h´c DE h´c lphoton = Louis de Broglie in response to Planck & Einstein’s assertion that light was ―particle-like‖ (photon) stated that small particles should exhibit a characteristic wavelength. = 4.576 ´ 10-19 J 6.626 10 34 J s ´ 2.997 ´ 108 m s 4.576 10 19 J L. de Broglie (1892-1987) photon 4.340 10 7 m ´ 109 nm = 1m 434.0 nm Particle-Wave Duality: A Prelude to Quantum Mechanics E = mc 2 L. de Broglie (1892-1987) h h = p or l = l p Uncertainty Principle W. Heisenberg 1901-1976 What is the wavelength associated with an 80g tennis ball (d = 8 cm) moving at 115 miles per hour? hn = mc 2 hn = mc = p (momentum) c n 1 since = c l Conclusion: Light waves have mass, particles have a wavelength. Problem: • According to Heisenberg the limits of quantum mechanics state that it is impossible to determine simultaneously both the position and velocity of an electron or any other particle with any great degree of accuracy or certainty. • Therefore an electron is both a particle and a wave simultaneously. Dx × mDv ³ h l = (tennis ball) = 2 x 10 –34 m l(ball) = 2.5 ´ 10-33 diameter(ball) h p The ball is a bajillion times the size of the wavelength !!! No wonder we can’t see it. Compare that to an electron moving at the same velocity: (electron) = 1.4 x 10 –5 m l(electron) = 3.2 ´ 109 diameter(electron) By golly the electron looks a little like a wave !!! Wave or Quantum Mechanics E. Schrodinger 1887-1961 • Taking on the ideas of Bohr, de Broglie and Heisenberg, Irwin Schrödinger proposed that matter can be described as a wave. • In this theory, the electron is treated as both a wave and a particle. • An electron is described by a Wave Function “ ” that completely defines a system of matter. Chapter 6 — Atomic Structure 11 Wave Functions, Ψ • The mental picture of an electron corresponds to a wave superimposed upon the radial trajectory of a particle orbiting the nucleus. • The position of an electron is found from the solutions to the Schrödinger Wave Equation which predict the ―probability‖ in a region space where the electron is likely to be found. • The orbitals are really Probability Distributions Wave motion: wave length and nodes “Quantization” in a standing wave Types of Orbitals • The solutions to the Schrödinger equation yields the probability in 3-dimensons for the likelihood of finding and electron about the nucleus. • It is these probability functions that give rise to the familiar hydrogenlike orbitals that electrons occupy. Quantum Numbers & Electron Orbitals Quantum Numbers are terms that arise from the mathematics of the Schrödinger equation. They describe location of an electron in a particular orbital much like an address. Each electron in an orbital has its own set of three quantum numbers. ―n‖ = 1, 2, 3, 4…up to infinity Principal Quantum Number shell Azimuthal or Angular Quantum Number ―l‖ = 0, 1, 2, 3…up to a maximum of ―n – 1‖ sub-shell Magnetic Quantum Number ―ml‖ ml may take on the value an integer from – l to + l individual orbitals Quantum Numbers & Electron Orbitals Quantum Numbers & Electron Orbitals n defines the Principal energy level ―shell‖ Each “l” within an ―n-level‖ represents a sub-shell. There are n ―sub–shells‖ for each n – level corresponding to l Each “l” sub-shell is divided into ml degenerate orbitals, where ml designates the spatial orientation of each orbital. if “n” equals: n=1 n=2 n=3 “l” can have values of: l=0 l=0&1 l = 0, 1 & 2 Each l is divided into (2l + 1) ml ―orbitals‖ separated by orientation. “ml” can have values of: if “l” equals: l=0 ml = 0 l=1 ml = 0, ±1 l=2 ml = 0, ±1, ±2 l=0 l=1 l=2 l=3 Type of orbital “s” sub-shell (sharp) “p” sub-shell (Principal) “d” sub-shell (diffuse) “f” sub-shell (fine) # of orbitals 1 3 5 7 each subshell contains 2l+1 orbitals Chapter 6 — Atomic Structure Types of Orbitals s orbital p orbital 12 s-Orbitals d orbital Spherical Nodes • l = 0, ml = 0 • 2l+1 = 1 • one s-orbital that extends in a radial manner from the nucleus forming a spherical shape. 1s, 2s, 3s-Orbitals 2 s orbital • All s-orbitals have ―n-1‖ spherical nodes. • A 1s-orbital has none, a 2s has 1 and so on. • These nodes represent a sphere of zero probability for finding an electron p-Orbitals When n = 2, then l = 0 and 1 Therefore, in 2nd shell there are 2 types of orbitals, i.e. 2 subshells For l = 0 ml = 0 (s-orbital) For l = 1 ml = -1, 0, +1 2(1) + 1 or 3 p-orbitals p-Orbitals Each p-orbital is characterized by a “nodal plan” The three degenerate p-orbitals spread out on the x, y & z axis, 90° apart in space. Chapter 6 — Atomic Structure 2px- & 3px-Orbital 13 d-Orbitals When n = 3, l = 0, 1, 2 resulting in 3 sub-shells within the shell. For l = 0, ml = 0 s-sub-shell with a single s-orbital For l = 1, ml = -1, 0, +1 p-sub-shell with 3 p-orbitals For l = 2, ml = -2, -1, 0, +1, +2 d-sub-shell with 5 d-orbitals 2(2)+1 = 5 d-Orbitals 3dxy, 3dxz, 3dyz-Orbital s-orbitals have no nodal planes (l = 0) p-orbitals have one nodal plane (l = 1) d-orbitals therefore have two nodal planes (l = 2) 3dx2- y2 & 3dz2-Orbital f-Orbitals When n = 4, l = 0, 1, 2 & 3 resulting in 4 sub-shells within the shell. For l = 0, ml = 0 s-sub-shell with a single s-orbital For l = 1, ms = -1, 0, +1 p-sub-shell with 3 p-orbitals For l = 2, ms = -2, -1, 0, +1, +2 d-sub-shell with 5 d-orbitals For l = 3, ms = -3, -2, -1, 0, +1, +2, +3 f-sub-shell with 7 f-orbitals 2(3)+1 = 7 Chapter 6 — Atomic Structure f-Orbitals 14 Summary of Quantum Numbers One of the 7 possible f-orbitals. All have 3 nodal planes. Arrangement of Electrons in Atoms • Each orbital can accommodate no more than 2 electrons • Since each electron is unique, we need a way to distinguish the individual electrons in an orbital from one another. • This is done via the 4th quantum number, “ms”. Electron Spin When an atom with one unpaired electron is passed through a magnetic field, the path is altered into two directions. Since there were 2 pathways in the experiment, there must be 2 spins affected by the magnetic field. One spinning to the right, one spinning to the left. Each ―spin state‖ is assigned a quantum number ms = ± ½ + ½ for ―spin up‖ ½ for ―spin down‖ N Source of electrons This means that the electrons have a magnetic moment! Electron Spin Detector S Electron Spin Electron Spin Quantum Number, ms The experiment results indicate that electron has an intrinsic property referred to as “spin.” Two spin directions are given by ms where ms = +1/2 and -1/2. Chapter 6 — Atomic Structure Electron Spin • It is not that the electrons are actually spinning on axis... • Rather it is that the mathematics that describe the electrons ―looks‖ like they are spinning on axis. Measuring Paramagnetism 15 Electron Spin & Magnetism • Diamagnetic Substances: Are NOT attracted to a magnetic field • Paramagnetic Substances: ARE attracted to a magnetic field. • Substances with unpaired electrons are paramagnetic. Electron Quantum Numbers Principal (n) shells Angular (l) sub-shells Magnetic (ml) orbitals Spin (ms) individual electrons When a substance containing unpaired electrons is place into a magnetic field, it is attracted to that magnetic field. The effect is proportional to the number of unpaired electrons. n = 1, 2, 3, 4, ... l = 0, 1, 2, ... n - 1 ml = - l ... 0 ... + l ms = ½