Wavelength, A(nm)

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1 1 Lecture 2 Atomic Theory II Components of a Wave Wavelength (A) Waves and Light Quantum Theory 1 51 Ionization Energy The Shell Model Photoelectron Spectroscopy (PES) The Shielding Effect Components of a Wave The Formula Low Frequency Wave C=AV S d f I " tht./ / Frequency (s or Hz) pee 0 [g)--- 3 x 108mls Wavelength (m) I High Frequency Wave Frequency (v}- the number of times a wave repeats itself per second Ex) Find the frequency of a green light that has a wavelength of 545 nrn. Continuous Spectrum of Light 10-2 Wavelength, A(nm) y-rays x-rays UV;: IR microwaves radio waves 400nm 750nm Every wavelength oflight is represented in the continuous spectrum. 1

2 2 Quantum Theory and Planck Quantum Theory and Planck Max Planck (1900) Hypothesized that the energy radiated from a heated object, such as a stove element or a lightbulb filament, is emitted in discrete units, or quanta. We assume that energy increases in a continuous stream.... Intensity > But it actually increases in discrete units. It.: increases by a full quantum, or not at all. Intl!!lSity ::> Quantum Theory and Planck Planck L-- The height of each step is equal to one quantum of energy for a certain frequency.> electromagnetic radiation. '- Intensity Energy per Quantum (1)<, / E=hv / Planck's Constant (6.63 x Js) Frequency (s') The Photoelectric Effect The science world knew that certain clean metal surfaces would shed electrons when certain frequencies of light were shined on them. :oo : e-, Metal Collector Plate Plate W Lamp Battery The Photoelectric Effect 1st Fact: Highly intense low frequency light does not eject any electrons, even if it shines on the surface fo,,e; Metal Collector Plate Plate OJ Lamp does not tuition Battery 2

3 3 The Photoelectric Effect 2 nd Fact: When the threshold frequency is reached, electrons are ejected immediately. ll 1:PP e-, Metal Collector Plate Plate tumson Battery The Photoelectric Effect 3 nd Fact: Increasing the intensity of the light at a frequency that will cause electrons to eject results in a higher ejection rate. However, all ejected electrons Sh"'. velocily. e:i Metal Collector I Plate Plate Battery ' ' < Lamp son S Current increases The Photoelectric Effect 4th Fact: Increasing the frequency of the light increases the velocity of the ejected electrons. However, all ejected electrons share the same velocity. d e., filc I. Metal Collector Plate Plate Battery LamP...rV'fumson quicker o o o o o Einstein's Theory (1905) A beam of light is a stream of particles called photons. The energy of a photon is related to its frequency according to E = hv. The quantum of Planck is a particle - a photon. If the frequency ofa pboton is below a certain threshold, no electrons are ejected. If the frequency of the photon is at or above a certain threshold, its energy is transferred to the electron. - This causes the electron to overcome the forces of attraction holding it to the metal. - The electron absorbs the photon. Energy per Planck and Einstein Photon(J) / E=hv / Planck's Constant (6.63 X Is) Frequency (S-I) o The Bohr Model of the Atom Forces of attraction between the electrons and the nucleus result from opposite charges. Hydrogen 3

4 4 F= Force of attraction k = Constant Coulomb's Law q = Magnitude of charge associated with a particle - protons or electrons d = Distance between charged particles The Bohr Model of the Atom Electrons move around the nucleus in circular orbits. Only certain orbits are allowed, and each orbit is a fixed distance from the nucleus. An electron within an orbit has a set, quantized, energy The force of attraction decreases as the distance between the outermost electron and the protons increases. Shielding Effect The electrons that are furthest from the nucleus are partly "shielded" by the inner core electrons. This shielding effect (electrostatic repulsion from inner core electrons) reduces electrostatic attractions between the outer electrons and the nucleus. First Ionization Energy The minimum amount of energy that is required to remove the outermost, least tightly held, electron from an atom in the gas phase. A(g) A +(g) + e- Values can be obtained by shining different frequencies of light on a pure sample of neutral atoms in the gas phase, recording the frequency that removes an electron, and using E = hv to calculate the first ionization energy. Q First Ionization Energy 2500 He Ne eo s t.i.l u C "... 0.", es N '2..'3 First Ionization Energy 500 Li 1 r.;: Na K 0 Generally Increases 0 30 Atomic Number I 4

5 5 First Ionization Energy H and He First Ionization Energy Hand Li JE 2500 IE,., = 1312 kj/mol IEII< = 2372 kj/mol 2500 H = 1312 kj/mol JE Li = 520 kj/mol He He "0 Ne "0 IEHc is almost twice JE H Ne leu is less than half as large as JE H IIf the most loosely held e eetro inli ThO increase can be ''''';;-1:t'''' was about the same distance fro the OJ Coulomb's Law, as He has nric as OJ e c: Ul many protons as H. tn c: 0 0. " nucleus as those in H and He, w would expect JE Li to be abou ee Zn I : times greater than JE H. ;:.s F = k q]q2.s I " I The most loosely held electron in Li '6 500 Li d Li must be much further away from the Na K I Na K nucleus than those in H and He 0 0 F =k Q,Q Atomic Number Atomic Number d 2 The Shell Model of the Atom Ionization data suggests that electrons are arranged in shells. The Shell Model of the Atom Ionization data suggests that electrons are arranged in Hydrogen Helium Lithium Boron Carbon Neon The Shell Model of the Atom Shell Model and the Periodic Table 2500 Ionization data suggests that electrons are He arranged in shells. The I" row of the periodic table has 2 elements and n = I can hold 2 electrons. The 2 nd row of the periodic table has 8 elements and n = 2 can hold 8 el ctrons. The 3,d row of the periodic table has 8 elements and n = 3 can hold 8 electrons. Sodium Argon 500 Li Na o L...:::::::::=-- o 18 Atomic Number 5

6 6 Shell Model and the Periodic Table The I'" row of the periodic table has 2 elemenis and n I can hold 2 electrons. Inner Core and Valence Electrons Tnner core electrons are contained in the inner shells. The 2 nd row of the periodic table has 8 elemenis and n 2 can hold 8 electrons. The 3'J row of the periodic table has 8 elements and n 3 can hold 8 electrons- Sodium Valence electrons are contained in the outer shell. Argon Photoelectron Spectroscopy (PES) In these experiments, high energy photons remove electrons from any shell, not just the outer shell. KE of ejected electrons is determined Frequency of photons is recorded Electrons Ionization energy for any electron is calculated using: Magnetic Field IE=hv-KE X-R'l) 0' UVlight H '"g He 0 "0; '- 0 o.0 E;::l s:: Li ":> Be. 0; B Photoelectron Spectra 11.5 A 6.26 A237 Ionization Energy (MJ/mol).A 090 A1.36 A. o : 0 o c A n = 1 contains one subshell - 1s - L72- (Is can hold a maximum of2 electrons)..a Photoelectron Spectra Subshells t(i '" A A '"2 N u 39.6 "<l ' " 52.6 s:: 3 s:: F. Ne A 67.2 A A" // Iomzati«Energy (Ml/rnol) n = 2 contains two subshells - 2$ and 2p (2s can hold a maximum of2 electrons) (2p can hold a maximum of 6 electrons) n = 3 contains three subshells - 3s, 3p, and 3d (3s can hold a maximum of2 electrons) (3p can.hold a maximum of 6 electrons) (3d can hold a maximum of 10 electrons) 6

7 7 Ionization Energies in MJ/mol obtained from PES Ffruu-n t 1, Na Mg "!p I- 3p "1., AI 15t Si P CI A< I.SZ K OA2 Ca Sc 43;3 485 ; Shielding Effect Within a Shell An electron in 2s spends more of its time further from the nucleus than one in 2p, but the electron density near the nucleus is greater for 2s electrons due to the first maximum. Distance from nucleus 2p electrons experience a greater shielding effect from the Is electrons than do the 2s electrons. c 6Shielding Effect Within a Shell -;;;:: s In all shells, electf.on density -;;; l. Et near the nucleus Distance from nucleus. decreases with..eachsubsequnt :0.g 3 subshell that 15 :.e P added. 0. I Distance from nucleus Thus, the.g. 3d increases for Sh.ieldin g efti.ect a. each subshell Distance from nucleus that is added. I Shielding Effect Within a Shell d JS > d Jp > d Jd 1E3s > IE3p > 1E3d Although the average distance between the I nucleus and electrons decreases as subsequent I subshells are added within a shell, the ionization energy decreases due to an increase in the shielding effect. Copyright 2013,2011,2009,2008 AP Chem Solutions. All rights reserved. 7

8 1 Lecture 2 Atomic Theory II Worksheet 1) Which colour of light has the highest frequency: red or green? 2) Which colour of light has the longest wavelength: green or violet? 3) Hydrogen emits light with a wavelength of 410 run. Find the frequency of this light 4) A radio station broadcasts at a frequency of 1310 khz. What is the wavelength of this radio signal in run? 5) Use two analogies to describe quantized change. 6) Use two analogies to describe continuous change. 7) Can prolonged exposure to highly intense infrared light cause electrons to be ejected from a clean metal surface? Explain. 8) Explain how Einstein was able to use experimental evidence related to the photoelectric effect to conclude that quantized energy must be contained by individual particles. 9) The following questions pertain to the element potassium. a. Write the equation for the first ionization of potassium. b. Draw the shell model that represents the potassium atom. c. Identify the electron that has the lowest ionization energy in the shell model that you drew. d. Use Coulomb's Law to explain why this electron has the lowest ionization energy. e. Use the 'shielding effect' to explain why this electron has the lowest ionization energy. 10) Why is the first ionization energy for lithium less than that of neon? (Discuss both atoms in your response.) IHJ,, _ I; :;u /rl'ifn'''' ;<.lumr Copyright e 2019, 2011, 2009, 2008 AP Chem Solutions. All rights reserved.

9 2 11) The following questions pertain to the element aluminum. a. Draw a photoelectron spectrum for aluminum, which includes all of the peaks but does not include exact ionization energies. b. Label each peak: with numbers and letters that indicate its associated shell and subshell. c. Indicate the number of electrons that are contained within each sublevel on your photoelectron spectrum for aluminum. d. Which subshell contains the electron with the lowest ionization energy? Justify your answer. 12) The following questions pertain to the photo electronic spectrum below. Ionization ncrgy (MJ mol) a. Identify the element that would produce the above spectrum. b. Label each peak with numbers and letters that indicate its associated shell and subshell. c. Indicate the number of electrons that are contained within each sublevel on the photoelectron spectrum. d. Which subshell contains the highest energy electron(s)? Justify your answer. e. Which subshell contains the lowest energy electron(s)? Justify your answer. 13) Photoelectron spectrometry (PES) data indicates that the ionization energy for an electron in the 2s orbital of calcium is 42.7 MJ/mol and the ionization energy for an electron in the 3s orbital of calcium is only 4.65 MJ/mol. Provide an explanation that accounts for this data. 14) Describe the modifications to the Bohr model that are required due to the experimental PES data that was provided in the lecture.

10 Ap Chemistry: PES Sample Items (available for classroom use, formative, or summative assessments): 1. Which element could be represented by the complete PES spectrum below? 10 Binding Energy (MJlmol) a. Li b. B c. N d. Ne 2. Which of the following best explains the relative positioning and intensity of the 2s peaks in the following spectra?... 'Vi e>1 OJ 1: A u Binding Energy (MJ/mol) 2 o ClJ of.' >1 of.' ';:1) C s:: Be Binding Energy (MJ/mol) 2 o a. b. c. d. Be has a greater nuclear charge than Li and more electrons in the 2s orbital Be electrons experience greater electron-electron repulsions than Li electrons Li has a greater pull from the nucleus on the 2s electrons, so they are harder to remove Li has greater electron shielding by the 1s orbital, so the 2s electrons are easier to remove Ii:>2013 Tlie COllege BOard.

11 r 13.5 Phosphorus p fh ".'.- e i MJ/mol ill ft \0- Ii> IU.t:I :._ _._..... _...._._. 239 :g,----_ _._-----_.--,----_ _._ ft Sulfur S q) 0:: Binding Energy 0.1 MJ/mol 3. Given the photoelectron spectra above for phosphorus, P, and sulfur, S, which of the following best explains why the 2p peak for S is further to the left than the 2p peak for P, but the 3p peak for S is further to the right than the 3p peak for P? a. S has a greater effective nuclear charge than P, and the 3p sublevel in S has greater electron repulsions than in P. b. S has a greater effective nuclear charge than P, and the 3p sublevel is more heavily shielded in S than in P. c. S has a greater number of electrons than P, so the third energy level is further from the nucleus in S than in P. d. S has a greater number of electrons than P, so the Coulombic attraction between the electron cloud and the nucleus is greater in S than in P. Ir M\!l---.--: ;,-;::;,. <I) I..., t /I o Binding Energy (MJ/mol) 4. Looking at the spectra for Na and K above, which of the following would best explain the difference in binding energy for the 3s electrons? a. K has a greater nuclear charge than Na b. K has more electron-electron repulsions than Na c. Na has one valence electron d. Na has less electron shielding than K Boai<!.

12 lt=================='=======================================-,' I , o Binding Energy (MJ/mol) 5. Looking at the spectra for Na and K above, which of the following would best explain the difference in signal intensity for the 3s electrons? a. K has a greater nuclear charge than Na b. K has more electron-electron repulsions than Na c. Na has one valence electron d. Na has less electron shielding than K - vi..., C :;, eu->-..., 'in c(ij..., C i2p -"\, '1 \. l.... C 15 - A GOO W. Binding energy (ev) Given the photoelectron spectrum above, which of the following best explains the relative positioning of the peaks on the horizontal axis? a. 0 has more valence electrons than Ti or C, so more energy is required to remove them b. 0 has more electron-electron repulsions in the 2p sublevel than Ti and C c. Ti atoms are present in a greater quantity than 0 than C in the mixture. d. Ti has a greater nuclear charge, but the 2p sublevel experiences greater shielding than the 1s sublevel. 3 Q 2013ihe Conege Board

13 l f: O.77-\; \-t1.=====h_a==================a=============================.========\r, 1\ n " II , : Binding Energy (MJjmol) 7. Given the photoelectron spectrum of Scandium above, which of the following best explains why Scandium commonly makes a 3+ ion as opposed to a 2+ ion? a. Removing 3 electrons releases more energy than removing 2 electrons. b. Scandium is in Group 3, and atoms only lose the number of electrons that will result in a noble gas electron configuration c. The amount of energy required to remove an electron from the 3d sublevel is close to that for the 4s sublevel, but significantly more energy is needed to remove electrons from the 3p sublevel. d. Removing 2 electrons alleviates the spin-pairing repulsions in the 4s sublevel, so it is not as energetically favorable as emptying the 4s sublevel completely c' r_ ; QJ r i Binding Energy (Ml/mol) 1 8. On the photoelectron spectrum for magnesium given above, draw the spectrum for aluminum. Board.

Photoelectron Spectroscopy Evidence for Electronic Structure Guided-Inquiry Learning Activity for AP* Chemistry

Introduction Photoelectron Spectroscopy Evidence for Electronic Structure Guided-Inquiry Learning Activity for AP* Chemistry Catalog No. AP7710 Publication No. 7710AS The chemical properties of elements