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Electrons in Atoms1 Looking back at Electrons in Atoms Chemistry documented materials to restore health (pharmacy). Atoms and elements were recognized during 16th to 18th centuries. The discovery of electrons in 1897 showed that there were more fundamental particles present in the (Dalton) atoms. Fourteen years later, Rutherford discovered that most of the mass of an atom resides in a tiny nucleus whose radius is only 1 / of that of an atom. In the mean time, Max Planck ( ) theorized that light beams were made of photons that are equivalent to particles of wave motion. Explain waves and particles

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Electrons in Atoms2 Announcement Please enroll Chem123 and related physics using Quest by doing the following: First screen - add LAB class No. - DO NOT PRESS CONTINUE! Press INSERT CLASS (Again) - add LECTURE class No. - then press CONTINUE You will see two boxes. Update your attributes (add your tutorials where applicable) and then SUBMIT. This should allow you to enrol in the lecture and lab, which are co-requisites.

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Electrons in Atoms3 Discovery of Electrons J.J. Thomson ( ) determined the charge to mass ratio for electrons in Robert Millikan’s oil- drop experiment determined the charge of electrons. Thus both the charge and mass are known Review Chapter 2

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Electrons in Atoms4 1-, 2-, & 3-dimensional Waves Demonstrate a single wave movement Explain continuous set of 1-dimensional waves Water wave and drum-skin movement as 2-dimensional waves 3-dimensional waves: sound waves seismic waves Explain wave motions

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Electrons in Atoms5 Wavelength, Frequency, and Speed Electromagnetic waves are due to the oscillations of electro- and magnetic-fields. Wavelength ( : the diagram shows one whole wave, and note the wavelength (in m, cm, nm, pm). Frequency ( is the number of waves passing a single point per unit time (s –1 or Hz). Speed of light Be able to apply: = c / ; c c = = 3.0e8 m s –1

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Electrons in Atoms6 Frequency, Wavelength & Wave-numbers A typical red light has a wavelength of 690 nm. What is its frequency? Solution : c 3e8 m s –1 = = = __________s –1 690e–9 m By the way, the wave_number is the number of waves per unit length. 1 wave_number = --- = __________ m –1

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Electrons in Atoms7 Momentum of Photons For a particle with restmass m o, its relative mass m when moving with a velocity of u relative to the speed of light c is m o m = [1 – ( u/c ) 2 ] Light particles, photons, have zero rest mass, travel at the speed c. From this relationship, the momentum of the photon, p, (Text p309) is h p = where h is the Planck’s constant, and is the wavelength. Momentum in any collision is conserved.

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Electrons in Atoms8 Superposition of Waves A sine function y 1 = sin x is a typical wave function. Plot y 1 vs. x Plot y 2 = – sin x. When the two waves combine, what happens? y 1 – y 2 y 1 + y 2

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Electrons in Atoms9 Interference Combination or superposition of waves is called interference.

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Electrons in Atoms10 Radiation spectra There are three types of spectra Continuous spectrum (generated by hot solid) Line spectrum (generated by hot gas, atoms) Absorption spectrum (continuous spectrum with black lines) Give the name of the spectrum from a known source

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Electrons in Atoms11 The Electromagnetic Spectrum

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Electrons in Atoms12 Hydrogen Emission Spectrum The visible spectra of H consists of red (656.3 nm) green, (486.1 nm), blue, (343.0 nm), indigo (410.1 nm), and violet (396.9 nm) lines. Variation of wavelength follow this formula = R H (----–----) ( R H = m n 2 in wave_number ) This is the Balmer series c 1 1 = ---- = R’ H (----–----) ( R’ H = e15 s -1, in frequency ) 2 2 n 2

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Electrons in Atoms13 Atomic Spectroscopy The study of light emitted by or absorbed by atoms is called atomic spectroscopy (AA). It offers qualitative and quantitative analysis of samples, because each element has a unique set of lines. The simplest form is identify elements by flame color. Atomic spectroscopy can be divided into emission and absorption spectroscopy, (AES and AAS).

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Electrons in Atoms14 Max Planck’s Photon Max Planck (1858 – 1947) proposed that light consists of little quanta of energy, and he called them photons. The energy of the photon E, is proportional to its frequency. E = h The proportional constant h is now known as Planck constant, h = e-34 J s –1, a universal constant. His proposal or assumption was made while studying the radiation from hot (black) body. Know relationships among frequency,, wavelength,, wave number, speed, c, and energy E. Given one, be able to calculate the others.

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Electrons in Atoms15 Wavelength Frequency, Wave-number and Energy of Photon The red line in Balmer series of hydrogen has a wavelength = nm. Calculate the frequency, wave number, and energy of the photon. Solution : frequency = = = 4.57e14 s –1 energy E = h = 6.626e –34 J s –1 * 4.57e14 s –1 = _______ wave number = = _______ E = h c / c 3e8 m s – e-9 m 1 See slide 12 and complete the calculation

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Electrons in Atoms16 The Photoelectric Effect In 1888, Hertz discovered that electrons are emitted when light strikes a metal surface – photoelectric effect. In 1905, Einstein observed and explained * electrons are emitted only when light frequency exceeds a particular value o, he called these values threshold * number of electrons emitted is proportional to the intensity of light * kinetic energies of emitted electrons depend on the frequency,, not the intensity of light See science.uwaterloo.ca/~cchieh/cact/c120/quantum.html for photoelectric effect

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Electrons in Atoms17 Einstein’s Experiment Kinetic energy of electron (½ m u 2 ) is measured by retarding potential V s. ½ m u 2 = e V s V s is proportional to the light frequency, but unrelated to light intensity. The frequency must be greater than certain threshold o, which is metal dependent. V s = k ( – o ) = k – k o = k – V o (substitute V o for k o ) e V s = e k – e V o = ½ m u2 = h – e V o ( h = e k, the Planck constant) h = e ( V s + V o )

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Electrons in Atoms18 Graph Einstein’s Result V s = ½ m u 2 kinetic energy of electron 0 threshold e V s = e k – e V o h = e (V o + V s )

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Electrons in Atoms19 A typical problem Radiation with wavelength of 200 nm causes electron to be ejected from the surface of a metal. If the maximum kinetic energy of electrons is 1.5e-19 J, what is the lowest frequency of radiation that can be used to dislodge electrons from the surface of nickel? Solution : Energy of the photon E = h c / = e-34 J s * 3e8 m / 200e-9 m = _ E 1 _ J Threshold energy E o = E 1 – 1.5e-19 J Threshold photon wavelength, = h c / E o = __please calculate __ m

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Electrons in Atoms20 Significance of Einstein’s Result Max Planck’s assumption is true, a proof. Light indeed consists of photons (quanta of light, not continuous) Quantity of energy in photon E = h (energy of photon) Photochemical reactions O 2 + h O + O O 2 + O O 3 (formation of ozone) Be able to calculate energy of photons, E, threshold, o, and kinetic energy of electrons, in photoelectric experiment.

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Electrons in Atoms21 The Bohr Atom Bohr tried to interpret the hydrogen spectrum by applying Planck’s quantum hypothesis and Rutherford’s atom, and he postulated: The e revolves around the nucleus (Rutherford’s atom) The electron has a set of allowed orbits (angular momentum = n h /2 , where n is an integer) that are stable. Electron changes from one state to another by absorbing or emitting a photon. From Newton’s physics, he showed the energy level of the electron to be E n = – R H n 2

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Electrons in Atoms22 Bohr’s Energy Levels of Electrons in H R H n 2 E n = – – R H – 1 / 4 R H – 1 / 9 R H – 1 / 16 R H – 1 / 25 R H 1 / R H = 0 R H = 2.179e-18 J State transitions – 1 / 36 R H Given R H and state of transition, n f, n i, be able to calculate E,, &, of transition.

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Electrons in Atoms23 H-spectra & Energy Levels Ionization energy

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Electrons in Atoms24 Excitation and Ionization of H R H n 2 E n = – – R H – 1 / 4 R H – 1 / 9 R H – 1 / 16 R H – 1 / 25 R H 1 / R H = 0 R H = 2.179e-18 J Excitation and ionization of an atom differs from those of a molecule – 1 / 36 R H Absorption of a suitable h excites an atom Absorption of a photon with energy equal or greater than R H results in ionization of H atom

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Electrons in Atoms25 A typical problem The electron in a hydrogen atom undergoes a transition from 4s to one of the 5p orbitals when the atom absorbs a single photon. What is the frequency of the absorbed photon? Solution : E ni – nf = E i – E f = R H (---- – ----) R H = 2.179e-18 J = h c / E ni – nf h = ee-34 J s c = 3e8 m s n i 2 n f 2

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Electrons in Atoms26 Photons & Transition

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Electrons in Atoms27 Wave-Particle Duality When in 1920 I resumed my studies... what attracted me... to theoretical physics was... the mystery in which the structure of matter and of radiation was becoming more and more enveloped as the strange concept of the quantum, introduced by Planck in 1900 in his researches into black-body radiation, daily penetrated further into the whole of physics. E = h = m c 2 = m c = p = h c h E – energy of photon and particle h – Planck constant h/, m v, m u, or p – momentum c, u, v – velocity of photon or particle – wavelength of photon or particle A particle with momentum p = m v is a wave with wave length h = p = m v Louis de Broglie ( ) Nobel laureate 1929

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Electrons in Atoms28 Wavelength of Electrons Estimate the velocity and wavelength of electrons with kinetic energy of 100 eV. Solution : (data look up and background information required) Mass of e – m e = 9.1e-31 kg; h = 6.626e-34 J s E = ½ m v 2 1 J = 1 N m = 1 kg m 2 s –2 1 eV = 1.6e-19 J; 100 eV = 1.6e-17 J v = (2 E / m ) ½ = (2 * 1.6e-17 kg m 2 s –2 / 9.1e-31 kg) ½ = 5.9e6 m s –1 m v = 9.1e-31 kg * 5.9e6 m s –1 = 5.4e-24 kg m s –1 = h / p = 6.626e-34 1 kg m 2 s –1 / 5.4e-24 kg m s –1 = 2.23e-10 m (approximately the diameter of atoms) Be able to calculate momentum and wavelength of particle when its speed is given. Estimate p & when an electron travels at 50% c

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Electrons in Atoms29 Validity of Particle-Wave Duality Example of a LEED pattern from the Si(111)7×7 surface. Electrons are usually considered particles. In 1927, a Davisson and Germer observed electron diffraction by Ni surface. Low energy electron diffraction (LEED) uses a beam of 30-to- 300 eV electrons to bombard a sample; a diffraction pattern is shown.

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Electrons in Atoms30 The Heisenberg Uncertainty Principle When electrons are considered particles, we should be able to measure their positions ( x ) and momenta ( p ) accurately, but Heisenberg showed that is not the case. The arguments seem complex, but the result is simple. The uncertainty of position x and uncertainty in momentum p has this relationship: x p > h4h4 The implications: the position (location) is fussy if we know the energy accurately. We are concerned with the energy more than we are with location probability of finding the electron correlates to orbital (not orbit) Read the arguments for the uncertainty principle.Heisenberg at 22

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Electrons in Atoms31 Quantities and the Uncertainty Principle If the uncertainty of an electron is 1e-10 m (100 pm), what is the uncertainty of the momentum? p = h / (4 * x) = = 5.3e-25 kg m s –1 6.63e–34 kg m 2 s –1 4*3.1416*1e-10 m Rest mass of electron = 9.1e-31 kg Speed of e with p = 5.3 e-25 kg m s –1 v e = 5.3e-25 kg m s –1 / 9.1e-31 kg = 5.8e5 m s –1 Energy = ½ * 9.1e-31 kg * (5.8e5 m s –1 ) 2 = 1.5e-19 J(recall 1 eV = 1.6e-19 J) Ionization energy of H is –13.6 eV; Energy = 0.9 / 13.6 = 7% Discuss physical meaning of results. If ionization energy of H is 7% accurate, the e – is some-where within 100 pm, size of H atom. More accuracy will result in an electron in larger volume.

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Electrons in Atoms32 Announcement International Exchange Information Night Nov 5, 2003, 5:30-6:30pm in DC 1301 (Fishbowl) Pizza and Beverages will be served. This information session is geared mainly toward students in their 1st or 2nd year who are interested in International Academic Exchanges. Criteria to be accepted for an International Exchange Program: Completed two years of University and maintained an overall accumulative average of 70% Proficient in the language of desired country of exchange (ie. must be proficient in French to go to France)

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Electrons in Atoms33 Standing Waves A traveling wave can be of any wavelength. The boundaries restrict a standing wave to some integer times the half-wavelength as illustrated by the 1-dimensional diagram. Note that the points at the boundary are fixed. www2.biglobe.ne.jp/~norimari/science/JavaEd/e-wave4.html

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Electrons in Atoms34 Wavelength in Standing Waves For standing waves with both ends fixed in a length L, the wavelength is limited to (where n is an integer) = ; n = 1, 2, 3, …. In quantum mechanics, electrons in atoms are treated as standing waves confined by the electric field due to the atomic nuclei. Thus, the electrons are represented by wave functions. The wave, energy etc are called state of the electron, and with spin, each state accommodate 2 electrons. A state is called an orbital (not orbit) 2 L n

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Electrons in Atoms35 Particle in a 1- Dimensional Box of Length L The wave representing a particle in a box of length L (variable x ) can be represented by ( x ) = ( ) sin ( ), n = 1, 2, 3, … n x L 2L2L Please work out the following values for n = 1 and n = 2 yourself n = 1 n = 2 (0) = ___0____0_ (boundary) ( L /4) = __________ ( L/2 ) = __________ (3 L /4) = __________ ( L ) = ___0____0_ (boundary)

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Electrons in Atoms36 An electron viewed as a wave in an atom, imagination goes a long way Animation by Naoki Watanabe

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Electrons in Atoms37 Energy of Waves Kinetic energy of a particle with speed u E k = ½ m u 2 = ( m u ) 2 / 2 m = p 2 / 2 m de Broglie’s relationship p = h /, E k = = = = p 2 2 m h 2 2 m 2 L n Recall = h 2 2 m (2 L / n ) 2 n 2 h 2 8 m L 2 The lowest energy of a (wave) particle is when n = 1, the zero point energy

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Electrons in Atoms38 Energy Level of a Particle in a Box E n = n 2 h 2 8 m L 2 n = 1 n = 2 n = 3 In a 3-dimensional space E ( n x, n y, n z ) = ( ) h 2 8 m n x 2 L x n y 2 L y n z 2 L z What is the expression for E ( n x, n y, n z ) in a 3-D cube?

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Electrons in Atoms39 Wavefunction of H atom The wave function of hydrogen atom satisfies the Schrodinger equation: – ( ) – = E Solutions of this equation is beyond the scope of this course, but a few points can be made. This second order DE has many solutions and by implying the boundary conditions in here, these solutions are characteristic of three quantum numbers n, l and m n – the principle q.n. (dominate energy) l – the orbital angular momentum q.n. (orbital momentum) m – the magnetic q.n. h 2 8 m 2 x22 x2 2 y22 y2 2 z22 z2 e2 r2e2 r2

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Electrons in Atoms40 Properties of Quantum Numbers Restrictions of quantum numbers are due to physical and mathematical reasons. Thus, n = 1, 2, 3, 4, … (integer) l = 0, 1, 2, 3, … ( n – 1) m = - l, - ( l –1), - - ( l – 2) … 0 … ( l –2), - ( l –1), l The consequency: Subshells are named according to value of l l = 0 ( s-subshell ) (1 state, m = 0) l = 1 ( p-subshell ) (3 states, m = -1, 0, 1) l = 2 ( d-subshell ) (5 states, m = -2, -1. 0, 1, 2) l = 3 ( f-subshell ) (7 states, m = -3, -2, -1. 0, 1, 2, 3) …

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Electrons in Atoms41 Wave (electron density) of Some Orbitals Know the shape of 1s, 2s, 2p, 3s, 3p, & 3d orbitals – for your bonding lessons in the future. Know the sign of the orbitals in various regions. Explain the significance of vs. r plots, 2 vs. r plots, r2 2 vs. r plots etc. Explain nodes, know how numbers of nodes related to n and l. Animations are used during the lecture to illustrate all the above points, and these are subjects of tests and final exams.

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Electrons in Atoms42 Energy Levels of H-atoms Solutions to the Schrodinger equation results in expressions for the energy, which is essentially the same as the one derived by Bohr, E n = – = – 13.6 eV The Z eff is the effective atomic number (modified atomic number). Z eff 2 m e e 4 8 0 h 2 n 2 Z eff 2 n 2 1s –– 2s –– 2p –– –– –– 3s –– 3p –– –– –– 3d –– –– –– –– –– Energy levels of H orbitals 4s –– 4p –– –– –– 4d – – – – – 4f - -

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Electrons in Atoms43 Energy Levels of Many-electron atoms For many electron atoms, the energy levels of subshells change slightly due to Z eff E n = – = – 13.6 eV Z eff 2 m e e 4 8 0 h 2 n 2 Z eff 2 n 2 1s –– 2s –– 3s –– Energy levels of many electron atoms 4s –– 2p –– –– –– 4p –– –– –– 3d –– –– –– –– –– 4d –– –– –– –– –– 4f ––

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Electrons in Atoms44 Electron Spins Stern-gerlach experiment In a magnetic field, a beam of electrons splits into two beams, and a beam of atoms are also splits into two beams. The interpretation of this observation came after some years is due to the spin of electrons, thus a fourth quantum number s. For an applet animation of this experiment see

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Electrons in Atoms45 Electronic Configurations of Atoms Electrons go to the lowest possible energy levels (minimize the energy of the atom). Pauli’s exclusion principle: No two electrons in an atom may have all four quantum numbers alike Hund’s rule: electrons occupy singly in orbitals of identical energy (degenerate orbitals: p – 3, d – 5, f – 7 etc.) The aufbau (build-up) process: illustrate this process during lecture and urge students to do it. Understand the energy level is the key.

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Electrons in Atoms46 The Modern Periodic Table 1s 1 2s 1-2 3s 1-2 4s 1-2 5s 1-2 6s 1-2 7s 1-2 1s 2 2p 1 – 2p 6 3p 1 – 3p 6 4p 1 – 4p 6 5p 1 – 5p 6 6p 1 – 6p 6 7p 1 – 7p 6 3d 1 – – – 3d 10 4d 1 – – – 4d 10 5d 1 – – – 5d 10 6d 1 – – – 6d 10 4f 1 – – – – – 4f 14 Th Pa U – – – 5f 14 Quantum mechanical theory led to the modern periodic table, which correlates chemical properties of elements nicely! Work out the filling order of orbitals and the electronic configurations from the periodic table

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Electrons in Atoms47 Writing Electronic Configuration Z= s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 He Ne Ar Kr Xe Rn

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Electrons in Atoms48 The Wavefunctions 1s & 2s 1s1s 2s2s URL: optoele.ele.tottori-u.ac.jp/~abe/hyd/

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Electrons in Atoms49 The Wavefunctions, 2ps 2pz2pz 2px2px 2py2py

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Electrons in Atoms50 Atomic orbitals See Table 9.1 of your Text for A table of wavefunctions of atomic orbitals of the hydrogen atom. Please identify the wave function for the following orbitals: 1 s 2 s, 2 p x, 2 p y, 2 p z 3 s, 3 p x, 3 p y, 3 p z 3 d xy, 3 d yz, 3 d x 2 – y 2, 3 d z 2 The Orbitron is a British website that gives wonderful views of the atomic orbital, and its URL is www. shef.ac.uk/chemistry/orbitron/

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Electrons in Atoms51 Excitation and de-excitation of electrons of the hydrogen atom. Animation by Naoki Watanabe

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Electrons in Atoms52 Fun with waves An electric current viewed as waves Just for fun Animation by Naoki Watanabe cms.phys.s.u- tokyo.ac.jp/~naoki/ research/review/indexe.html Animations are done by computational method for large-scale and long-term fisrt- principles numerical solutions of time-evolving quantum electronic states. The base equation for this study is the time-dependent Schroedinger equation. In principle, by solving this partial differential equation numerically, it would be possible to analyze many kinds of quantum dynamic phenomena.

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Electrons in Atoms53 Quantum review 0 The longest wavelength of radiation that will cause the emission of electrons from gold surface is 257 nm. What is the enrgy per photon of this radiation? (photoelectric effect) The threshold photons for gold have a wavelength of 257 nm, what is the threshold energy? E = h c / Which of the following transitions for the H atom produces radiation of the shortest wavelength? - n from 2 to 3; 3 to 2; 5 to 6; 6 to 5; 2 to 1(energy level diagram)

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Electrons in Atoms54 Quantum Review 1 Draw an energy level diagram according to Bohr’s atom. Identify the transitions for red (656.3 nm) green, (486.1 nm), blue, (343.0 nm), indigo (410.1 nm), and violet (396.9 nm) lines of hydrogen. ( = c / ; E = h ) Evaluate, wave_number, and energy of the red and violet lines. Evaluate the wavelength of the transition from n = 2 to n = 1 and compare with that from n = 10 to n = 1 in the Lyman Series. (transition and energy level diagram)

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Electrons in Atoms55 Quantum Review 2 What is the wavelength, frequency and energy of the photon in the Balmer series for n = 6? Calculate these for the Lyman series. If traveling at equal speed, which of the following particles has the longest wavelength, and why electron, proton, neutron, alpha particle (He 2+ )? (de Broglie’s theory) What are the quantum numbers n, l, & m for the states 2s, 3p, 4d, 5f? Write the electronic configuration for uranium U. Review questions No. 24 of Chapter 9 in General Chemistry, 8 th ed. How many photons are emitted per second by an IR lamp consuming 95 W if 14% of the power is converted to photons of wavelength of 1525 nm?

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Electrons in Atoms56 Quantum Review 3 The work function of mercury is 435 kJ / mole (energy required to remove a mole of electrons from Hg surface). (Photoelectric effect) What is the threshold energy in eV per electron? What is the wavelength, frequency and wave number for the threshold photon? What is the kinetic energy of electron if the light used has a wavelength of 215 nm? Energy per electron = 4.35e5 J / 6.023e23 = 7.22e-19 J / e – * 1 eV / 1.6e-19 J = 4.51 eV Frequency = 7.22e-19 J / 6.63e-34 J s = 1.09e21 s –1 = h c / E; E = h = h c / = c / Kinetic energy of electron = h ( – o ) = h c (1/ – 1/ o ) = extra energy after threshold.

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Electrons in Atoms57 Quantum review 4 Sketch the shape of these atomic orbitals according to their electron density: 1 s, 2 s, 3 s, 2 p, 3 p, 3 d, and what are the signs of the wave functions in various lobs of the atomic orbitals? What is the meaning of and 2 ? What do the plots of 2 in a three-dimensional space represent? What do the plots of 2 and 4 r2 2 against r represent? How many nodal shells are there in these atomic orbitals, 1 s, 2 s, 3 s, 2 p ? How many nodal planes are there in these atomic orbitals: 2 p, 3 p, & 3 d ? How many atomic orbitals are there for n = 5 and l = 3? What are the possible values of m l for these orbitals? How many electrons do these orbitals accommodate?

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