Final Exam: Thursday 05/02 7:00 9:00 pm in STEW 183

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1 Final Exam: Thursday 05/02 7:00 9:00 pm in STEW 183 Covers all readings, lectures, homework from Chapters 17 through 30 Be sure to bring your student ID card, calculator, pencil, and up to three onepage (two-side) crib sheet. NOTE THAT FEW EQUATIONS WILL BE GIVEN YOU ARE REMINDED THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED CRIB SHEET YOU WANT TO BRING TO THIS EXAM. 4/14/13 D. Bortoletto PHYS221 1

2 Structure of the Atom: Early Models Plum-Pudding Model 4/14/13 D. Bortoletto PHYS221 2

3 Structure of the Atom: What s inside? Plum-Pudding Model 4/14/13 D. Bortoletto PHYS221 3

4 Structure of the Atom: Early Models 1910 Rutherford, Geiger & Marsden Experiment 4/14/13 D. Bortoletto PHYS221 4

5 The Rutherford experiment Observation: Most of the alpha particles did travel straight through the foil with little or no deviation. A small fraction (about one in ten thousand) rebounded, ending up on the same side of the foil as the incoming beam. Rutherford said that the result of the experiment were "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." Such huge deflections could mean only one thing: some of the alpha particles had run into massive concentrations of positive charge. 4/14/13 D. Bortoletto PHYS221 5

6 Structure of the Atom: Early Models 1910 Rutherford, Geiger & Marsden Experiment Alpha particles are too heavy to be scattered by electrons The mass must be concentrated in smaller areas nucleus Plum model is incorrect Nucleus is m in diameter Electron roam empty space is m across (Å) 6

7 Scale of particle physics λ h/p 4/14/13 D. Bortoletto PHYS221 7

8 Structure of the Atom: Planetary Model Negative electron is attracted to positive nucleus and spins around it as a planet around a star The centripetal force is provided by the electrostatic force of attraction mv ke ke e = v = 2 r r mer 4/14/13 D. Bortoletto PHYS221 8

9 Structure of the Atom: Planetary Model Problem 1: Electron is constantly accelerated According to classical EM theory it should emit EM radiation EM radiation carry energy Electron should lose energy and fall on the nucleus Problem 2: Experiments show that an electron in an atom can accept or release only quantized amount of energy 4/14/13 D. Bortoletto PHYS221 9

10 Structure of the Atom: Bohr Model Electron is allowed to be in only one of a discreet sets of orbits called stationary states Each stationary state has a definite energy- energy level The laws of Newtonian mechanics apply to the motion of the electron The electron can make transition between stationary states through emission or absorption of a single photon ΔE=hf The stationary states are circular orbits in which angular momentum is quantized h L= n = nh, n= 1,2,3... 2π 4/14/13 D. Bortoletto PHYS221 10

11 Radii of Bohr Orbits Bohr radius, minimum radius Radius increases as n 2 L = n h = n 2π n =1,2,3... L = m e vr m e vr n = n ke 2 Using v = canfind r m e r n Radii of orbits in H atom: 10 r = a = = m Angstrom r = n r = n a n r n = n2 2 m e ke 2 n =1,2,

12 Energy Levels of Bohr Orbits The energy on an orbit is the sum of the electron s kinetic and potential energies e 2 e E = K + U = mv k r Substitute v and r and get: v = ke2 m e r r n = n2 2 m e ke 2 n =1,2,3... Energy levels of H atom:: E n = m e k 2 e 4 The energy of the ground state: E1 = 13.6eV We can write 2n 2 2, n =1,2,3... E n = E 1 n 2 n =1,2,

13 Energy Level Diagram of Hydrogen Atom E n E 1, 2, n 1 = n= Energy is zero when the electron is removed from hydrogen Energy is lower when electron is closer to nucleus Hydrogen atom can only absorb or emit light when the energy of the photon is equal to the difference between to levels hc E = = Ei E λ E i =Energy of the electron in the initial state n i E f =Energy of the electron in the final state n f Emission and absorption occurs at the same wavelength f 4/14/13 13

14 Example Consider the transition between the n=2 and n=1 state of the Hydrogen atom. What is the wavelength of the emitted photon when the electron makes this transition? E n = E 1 n 2 n =1,2,3... E 1 = 13.6 ev E 2 = ev = 3.4 ev hf = E 2 E 1 = ΔE f = ΔE λ = hc h ΔE ΔE = E 2 E 1 = ( 3.4eV ) ( 13.6eV ) = E 2 E 1 λ = hc ( ΔE = ev s)( m / s) 10.2eV 122nm ULTRAVIOLET = m = 4/14/13 D. Bortoletto PHYS221 14

15 Hydrogen Energy Levels Infrared Visible Ultraviolet 4/14/13 15

16 Example One wavelength in the infrared part of the hydrogen emission spectrum has wavelength 1.28 µm. What are the initial and final states of the transition that result in this wavelength being emitted. Strategy: The energy of the photon must be the difference between two energy levels. Do we have to solve an equation with two unknowns? We can use the energy level diagram to narrow the choices Solution: The energy of the photon emitted is hc 1240eV nm E = hf = = = 0.969eV λ 1280nm Looking at the energy level we note that the photon must be in the Paschen series. The smallest photon energy in the Balmer series is ( 1.51 ev ) ( 3.40 ev ) = 1.89eV The photon in the Lyman series have even larger energies The largest energy in the Brackett series has energy of 0.85 ev

17 Example Since the photon is in the Paschen series the final state is n=3. Now we can solve for the initial state E = E E photon i f 13.6 ev ev = ( 1.51 ev ) 2 n 13.6eV n = = 5 151eV 0.969eV Note that for an hydrogen spectrum identifying the lower of two energy levels is simplified since the various series do not overlap. All photons in the Lymann series (lower energy level n=1) have larger energies than any of the photons in the Balmers series (lower energy level n=2) All photons in the Balmer series have larger energies than any of the Paschen series (lower energy Level n=3) etc. 17

18 Spectral Analysis We can use gas discharge tube to excite atoms. Electrons collide with atoms and promote their electrons to higher orbitals When electrons in atom return to lower states they emit light Wavelength of this light is defined by the energy difference between the energy levels that are different for different kind of atoms Spectrum of this emission can be analyzed using a spectrographs 4/14/13 D. Bortoletto PHYS221 18

19 Spectral Analysis 4/14/13 D. Bortoletto PHYS221 19

20 ILQ1 What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series? Balmer (A) 0.25 (B) 0.5 (C) 2 (D) 4 Lyman 4/14/13 20

21 Electromagnetic Spectrum 4/14/13 21

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