m(g) / M = q(c) / N V Q

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Homer Gray
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1 CHAPTER 3 THE PARTICLE NATURE of MATTER ELECTROLYSIS Faraday established the law of electrolysis in After passing a current I=dq/dt through an electrolytic solution ( he used molten NaCl ), a know amount of ion species was deposited or liberated at the electrode. The charge q is related to mass of the species deposited. A charge of Q = Ú Idt = 96500C is needed to deposit 1 mole M of substance for a monovalent electrolyte. Thus one can conclude that the current was not continuous but was somehow associated with transferring a known molecular weight, like a bucket brigade! m(g) / M = q(c) / N V Q A Na Cl J. J. THOMPSON EXPERIMENT on the e/m RATIO J.J.Thompson claims discovery of the electron in He measures the deflection of the electron beam coming from the cathode in a Crooke s tube. From measurements of the charge from electrochemistry earlier in the century he was able to calculate the mass. The cathode ray was deflected by an electric field E showing that it was negatively charged. The deflection could be cancelled by applying a crossed magnetic field B. Thompson used this ExB technique could be used to calculate the cathode ray velocity Vx..

2 L Anode (high) Dy Vx Vy E x x x x x B x x x x x d Cathode (low) x x x x x V Vy = ay t t = L/Vx F E =e E F B =e Vx B ay = F E /m = e E/m E = V/d kinematics equation transit time Force due to electric field Force due to magnetic field vertical acceleration from Newton s Law definition of electric field in terms of voltage gradient V and plate spacing d Vy = (e E/m) (L/Vx) tan q = Vy/Vx = (e/m) (V/d) (L/Vx 2 ) Velocity selector F E = F B = ee = evx B or Vx = E/B tan q = (e/m) (V/d) (L/Vx 2 ) with Vx = E / B e/m = V tan q / B 2 L d Thompson measured e/m = 1.0 x C. The e/m ratio Thompson calculated was much larger than previous electrolysis measurements on Hydrogen. Thompson realized that this implied that the mass of the cathode ray particles were 1/1000 smaller than the atom of Hydrogen. Scientist were then convinced that the cathode rays emitted were a beam of very small negative particles they called electrons.

3 MILLIKAN OIL DROP EXPERIMENT MEASUREMENT OF the ELECTRIC CHARGE Millikan measured the charge on the electron by balancing charged oil droplets. An oil droplet falls under the force of gravity and reaches terminal velocity because of a counteracting drag force. An external electric field E was used to keep the charge stationery. The drag force always opposes the velocity! E q bv mg m dv/dt = mg bv dv/dt = g (b/m)v dt = dv/ [g (b/m)v] t (b/m) Ú 0 dt = Ú 0 dv/ [(m/b)g v] v (b/m) t = dv/ [(m/b)g v] Ú 0 v b = 6 p a h a =radius h = viscosity v(t) = (mg/b) ( 1 e bt/m ) After a long time the falling body will reach a terminal velocity vo. t >>0 v(t>>0) = vo Typical Measurement Measure vo for a falling droplet with the field off. Then turn on the field and allow the droplet to achieve another terminal velocity v1. m g b vo = 0 q1 E mg b v1 = 0 Solving: field off field on q1 = (mg/e) [(vov1)/vo] During the measurement the droplet will gain or loose charge. q2 = (mg/e) [(vov2)/vo] The ratio q1/q2 = (vo v1)/(vov2) will reveal a series of fractions revealing the integral nature of the deposited charges.q = N Dq. To find the absolute value of the charges and unit of charge Dq the droplet mass must be determined from the fieldoff condition.

4 Dq = e = 1.6 x C RUTHERFORD SCATTERING and the DISCOVERY of the ATOMIC NUCLEUS Rutherford set up an experiment in which he scattered alpha particles 4 He 2 from gold foils. He expected the energetic a particles to penetrate the Au foil with ease undergoing a small deflection from the atomic charges spread throughout the atom. To his great surprise he recorded backward scattered a particles. This could only happen if the Au atom had a concentration of positive charges in a dense region at it s center. Thus Rutherford proposed that an atom had a dense nucleus of positive charge with orbiting elecrtrons. This was called the Rutherford Model ofthe atom. A B Forward scatter a Gold atom Electron Cloud Backscatter Nucleus Ze Consider an a particle of kinetic energy KEo. We can determine the distance of closest R approach to the Au nucleus for backscatered events from conservation of energy. KE A PE A = KE B PE B KEo 0 = 0 kqq/r At B the a particle comes to rest with KE B =0 and PE B = kqq/r, q=2e and Q=79e (Since no rest masses change in the problem we can ignore them and use the classical definition of energy conservation. This may not always be true!) R ~ Nuclear Radius = kqq/keo Rutherford discovered that R~ m. The atomic radius was estimated to be about R ATOM ~ m = 1 Angstrom, a large descrepancy. Henceforth the atom could only be understood as a negative electron cloud orbiting the a dense central region of the positive chargecalled the NUCLEUS.

5 SPECTRAL LINES Fraunhoffer in the early 1800 s noted that dark lines exist in the continous spectrum of of the Sun. Bunsen and Kirchoff in the mid 1800 s introduce absorption spectroscopy as an experimental tool. H gas Continuous spectra Absorption Or Line Spectra The line spectra of H gas yielded a simple relationship between the wavelength l of the lines and numerology. Lyman, Balmer, Paschen, Brackett. Pfund measured spectra in different regions of wavelength. All could be described by the relations below. 1/l n m = R ( 1/n 2 1/m 2 ) m>n R = x = Rydberg constant 1/l m = R ( 1/1 2 1/m 2 ) m = 2, 3, 4,5 Lyman Series uv 1/l m = R ( 1/2 2 1/m 2 ) m = 3, 4, 5, 6 Balmer Series visuv 1/l m = R ( 1/3 2 1/m 2 ) m = 4, 5, 6, 7 Paschen Series IR 1/l m = R ( 1/4 2 1/m 2 ) m = 5, 6, 7, 8 Brackett Series IR 1/l m = R ( 1/5 2 1/m 2 ) m = 6, 7, 8, 9 Pfund Series IR

6 BOHR MODEL OF THE ATOM Classically speaking Rutherford s Model of the atom had a serious problem. As the electrons orbited in it s cloud they would radiate light and eventually loose their energy. Bohr introduced the radical idea, that the electrons were not able to radiate their energy continuously, but only in quantum jumps! These jumps corresponded to the spectral lines discussed above. Consider an atom with one electron and nucleus of charge Ze. e Ze r E = 1/2 m v 2 kze 2 / r energy of the electron (1) mv 2 / r = k(e) (Ze)/ r 2 centripetal force balance (2) 1/2 m v 2 = k Ze 2 / 2 r solving (2) E = k Ze 2 / r substitution in to (1) (3) All classical so far and not that negative energy indicates a bound orbital state. Bohr introduced the quantum idea that angular momenum must be quantized in units of Planck s constant. Phase space ( x vs p ) is quantized in cells of area = h! m v r = n h n=1,2,3 (4) h Dx Dp=D(mv) Phase Space x p

7 Quantum Radius By solving (2) and (4) we can define the quantum radius of the electron. r n = n 2 h 2 / k m Ze 2 n=1,2,3 a O = r 1 = h 2 / k m e 2 =.0529 nm is called the Bohr Radius. Z=1 r n = (n 2 /Z) a O (5) Quantum Energy Substituting (5) in to (3) E n = k Z e 2 / 2 n 2 a O E n = 13.6 (Z 2 / n 2 ) ev (6) We can calculate the velocity in each quantum state by using the angular momentum Equation m v n r n = n h v n = =nh / m r n = k Z e 2 / n h = (Z/n) (c/137) The velocity of the electron in the ground state is BUT v0 = Z x 2.2 x 10 6 m/s << c b =.007 and we can use nonrelativistic theory!! But when Z = 80 for example b =0.58 and our nonrelativistic theory will not yield true results.!!!

8 QUANTUM TRANSITIONS E = 0.0 ev E4 = 0.85 ev E3 = 1.51 ev hf = E3E2= D E 32 E2 = 3.40 ev hf = E2E1 D E 21 Conservation of Energy before & after Ei = Ef E1 = 13.6 ev ground state Ei = Ef hf hf = Ei Ef hf Ei = Ef hf =Ef Ei emission absorption Electrons remain in stable orbits briefly (10 8 s) and then decay. A photon of energy hf = Ef Ei is emitted in the decay from state f > I.. The ground state has quantum energy E1 = 13.6eV for hydrogen. A state of zero motion n=0 is not allowed! The Binding Energy of an atom is the energy necessary to just raise the least bound electron to E = 0 The electron remains stable the ground state n=1 until excited by a (1) Photon process, (2) electron process, (3) or collisions. The probablity of an atom being in a quantum state n at temperatur T is Pn = Cn e En/kT see example 3.9

9 EXCITATION by PHOTONS Photnn exitation Ionization by photon hf hf Ef Ef Ei DEfi Ei DEfi hf Ei= Ef hf = EfEi must be satisfied! Binding Energy = energy necessary to eject the least bound electron. hf Ei = Ef hf BE = Ef Ef = hf BE Photoelectric Effect! Here Ef =KEf (PEf=0) EXCITATION BY ELECTRONS Electron exitation Ionization by electron eo ef = eo DE eo ef = eo DE Ef Ef Ei DEfi Ei DEfi eo Ei= ef Ef ef = eo Ei Ef ef = eo DEfi Binding Energy = energy necessary to eject the least bound electron. eo Ei = ef Ef Ef = e0ef Ei Ef = e0ef Ei

10 COLLISIONAL EXCITATION Energy is exchanged in atomic collisions. We can view this energy exchange as occuring through exchange of momentum between atoms. The Coulomb force is responsible for these impulses DP = FDt. atom ao atom ao Atomic View Ef hf =efeo Ei DEfi eo ao Ei= af Ef ef = eo Ei Ef ef = eo DEfi ef AUGER ELECTRON TRNASITIONS Auger realized that the energy released in the DE21 could be reabsorbed by a valence electron KE Auger transition viewed as 2 processes Valence electrons E3 = BE Initial Final state state Ground state hf E2 E1 E2 = E1 hf (1) hf E3 = KE (2) KE =E3 (E2E1) = DE21 BE Internal Photoelectric Effect!!

11 FRANCKHERTZ EXPERIMENT Franck and Hertz perform the first experiment which directly revealed the quantum energy levels of the atom. A beam of electrons are accelerated through a potential V. KE = qedx = e (V/d) dx Near the collection grid electron energy is absorbed by collisions when KE = 4.9 ev, the first energy jump 1>2 in the Hg atom (below). KEo > 4.9 ev At this point the accelerated electrons looses 4.9 ev of energy.and the anode current drops. If the electrons are accelerated to higher potentials a series of dips can be observed each signaling atomic absorption by the 1>2 transition. They were also able to a 4.9 ev pr ~250nm very blue photon emitted in the subsequent 2>1 atomic deexcitation. Cathode KEe ~0 KEe = ev Anode Heater coil Hg Gas grid 1.5V retarding A V I Hg 80 5d 2 6s 2 n=3 n=2 n=1 DE21 =4.9 ev 4.9 V spacing V 78 core electrons Z=80

12 ATOMIC RECOIL Consider an atom of mass M emitting a photon of energy hfo. As the photon is emitted the atom will recoil conserving momentum. P M hf hfo = hf P 2 /2M P = hf/c ignoring the Mc 2 in the energy balance hfo = hf (hf) 2 / 2M hfohf =( hf) 2 / 2M Letting hfo ~ hf on the RHS to 2 nd order!! hf 2 = (hfo d) 2 = hfo 2 2d hfo d 2 Dhf / hfo = hfo / 2M

General Physics (PHY 2140)

General Physics (PHY 2140) General Physics (PHY 140) Lecture 33 Modern Physics Atomic Physics Atomic spectra Bohr s theory of hydrogen http://www.physics.wayne.edu/~apetrov/phy140/ Chapter 8 1 Lightning Review Last lecture: 1. Atomic