@"@P(147): 53280,12: 53281,0F@Z 11)"***********************"j@d 11)"* *"@n 11)"* ATOMIC STRUCTURE *"@x 11)"* PART II *"@ 11)"* *"@ 11)"***********************"A::&A 12)" A PHYSICS PROGRAM":?A 16)"DEVELOPED BY"EAbA 14)"PAUL W. MCDANIEL"~A 14)"4295 WARREN WAY"A 14)"RENO, NV 89509":""A14)"COPYRIGHT 1988":""A12)"ALL RIGHTS RESERVED": 5A(147)QB6)"THE PRECEEDING PROGRAMS IN THIS SERIES OF STUDY PROGRAMS IN MODERN PHYSICS ARE":{B10)"(1) INTRODUCTION TO PHYSICS":B10)"(2) MATTER IN BULK":B10)"(3) ATOMIC STRUCTURE PART I":SC"6)"THESE PROGRAMS CAN BE DOWNLOADED ON Q-LINK, OR OBTAINED DIRECTLY FROM THEAUTHOR BY SENDING $10.00 FOR EACH PROGRAM TO":""C,12)"DR. PAUL W. MCDANIEL":12)"4295 WARREN WAY":12)"RENO, NEVADA 89509": 20C6(147):""C@10)" HINT TO THE STUDENT ":gDJ6)"WHEN THE SCREEN ACTION IS FROZEN AND THE QUESTION MARK ? AND FLASHING CURSOR APPEAR PRESS 'RETURN' TO CONTINUE":DT6)"TWENTY SECONDS ARE ALLOWED TO VIEWEACH GRAPHIC PRESENTATION. NO ACTION IS REQUIRED BY THE READER."D^:Dh(147):""Er8)"ALPHA PARTICLE SCATTERING ":5E|(147):""E6)"SIR ERNEST RUTHERFORD DEVELOPED A FORMULA TO DESCRIBE THE SCATTERING OF ALPHA PARTICLES BY THE ATOMS OF THIN ".F"FOILS OF GOLD. THE STUDENT IS ADVISED T0MASTER RUTHERFORD'S DERIVATION BECAUSE IT CONTAINS SEVERAL IDEAS THAT ARE "hF"IMPORTANT TO MANY ASPECTS OF MODERN PHYSICS.":tF(147)F6)"RUTHERFORD ASSUMED THAT THE ALPHA PARTICLE AND THE GOLD NUCLEUS FROM WHICH"BG"IT IS SCATTERED ARE BOTH SMALL ENOUGH TOBE CONSIDERED AS POINT MASSES AND THAT THE ONLY SIGNIFICANT FORCE ACTING"G"BETWEEN THE TWO PARTICLES IS THE ELECTROSTATIC REPULSIVE FORCE BETWEEN THE POSITIVELY CHARGED ALPHA PARTICLE"?H"AND THE POSITIVELY CHARGED GOLD NUCLEUS.(IF THE STUDENT HAS ANY DOUBTS THAT THE ATTRACTIVE GRAVITATIONAL FORCE IS SO "H"SMALL AS TO BE NEGLECTED HE MAY SATISIFYHIMSELF BY MAKING SOME CALCULATIONS OF THE MAGNITUDES OF THE TWO FORCES AT ""PARAMETER B. THE ANGLE MADE BY THE ASYMPTOTIC DIRECTION OF APPROACH OF THE ALPHA PARTICLE AND THE ASYMPTOTIC "QH"DIRECTION IN WHICH IT RECEDES IS CALLED THE SCATTERING ANGLE A. LET US FIRST FIND THE RELATIONSHIP BETWEEN B AND A.":RR(147):1,1R\1,2,2,"THE RESULT OF THE FORCE F ACTING ON THE ALPHA PARTICLE IN THE TIME INTERVAL DT IS AN IMPULSE":Rf1,120,72,10,10,270,10:1,100,72,10,10,90,190Rp1,16,9,"F DT":1,16,8,""2Sz1,2,12,"AS A RESULT OF THIS IMPULSE THE MOMENTUM OF THE ALPHA PARTICLE CHANGES BY"QS1,12,16," "rS1,12,17,"DELTA P = P - P "S1,12,18," 2 1"S 1,200,170,10,10,270,10:1,180,170,10,10,90,190:1,25,20,""S1,25,21,"F DT":1,20,21,"=": 20 T 0:(147)T6)"NOW IF THE GOLD NUCLEUS REMAINS STATIONARY DURING THE TRANSIT OF THE ALPHA PARTICLE, THE MAGNITUDE OF THE "T"MOMENTUM OF THE ALPHA PARTICLE ALSO REMAINS CONSTANT. HENCE":""T12)"P = P = M V" U12)" 1 2 A":""U6)"THE GEOMETRIC RELATIONSHIPS IN RUTHERFORD SCATTERING CAN BETTER BE UNDERSTOOD BY EXAMINING THE FOLLOWING DIAGRAM":U(147):1,1U1,2,0,"RUTHERFORD SCATTERING OF ALPHA PARTICLES BY GOLD NUCLEI"MV1,2:1,20,134 300,134:2:1,230,112,5,4:1,230,112,1: 1,230,160,10,8:1,230,160,1V1,230,160 160,40 166,43 157,45 160,40:1,16,4,"DELTA P":1,16,3," "W2:1,40,124 80,123 120,123 160,122 200,118 230,112 260,100 280,88 300,75W$1,200,155 330,55@W.1,3,20,"- - - - - - - - - - - -"hW81,8,17,"":1,8,19,"":1,8,18,"B"WB1,230,160 230,40 227,43 233,43 230,40:1,30,5,"F":1,30,4,""WL1,230,165 330,85:1,300,80 325,95:1,38,11,"B"dXV1,245,130,20,16,20,100:1,31,15,"A":1,210,135,40,32,270,340:1,15,12,"(PI - A)":1,15,13,"":1,17,14," 2"X`1,230,160,100,80,335,358:1,26,11,"P": 20Xj 0:(147):1,1Xt1,2,2,"MOMENTUM CHANGE IN THE ALPHA PARTICLE"LY~1,60,120 200,120 240,60 240,64 235,60:1,197,123 200,120:1,200,120 196,117 196,125 200,120Y1,200,120 100,60 105,60 100,65 100,60:1,60,120 100,60 240,60Y1,10,13,"A":1,60,120,50,40,30,90:1,200,120,50,40,270,305:1,180,115 190,135GZ1,17,16,"P":1,24,16,"(PI - A)":1,24,17,"":1,24,18," 2":1,17,17," 1"}Z1,17,9,"DELTA P":1,29,11,"P":1,29,12," 2": 20Z 0:(147)Z6)"FROM THE LAW OF SINES WE HAVE":""Z12)"DELTA P MV"Z12)" = "[12)" SIN A ( - A)9[12)" SIN "W[12)" 2"a[""u[2)"NOW SINCE"[ 12)" 1 A"[12)"SIN ( - A) = COS"[12)" 2 2"[(2)"AND" \212)" A A")\<12)"SIN A = 2 SINCOS"K\F12)" 2 2":_\P(147):""q\Z2)"WE HAVE"\d12)" A"\n12)"DELTA P = 2 MV SIN"\x12)" 2":\(147):1,1]1,2,2,"NOW THE IMPULSE"N]1,200,17,10,10,270,10:1,180,17,10,10,90,190:1,26,2,"F DT":1,26,1,""]1,2,6,"IS IN THE SAME DIRECTION AS THE MOMENTUM CHANGE DELTA P, SO"]1,200,80,10,10,270,10:1,180,80,10,10,90,190:1,13,10,"F DT =",^1,100,80,10,10,270,10:1,80,80,10,10,90,190:1,25,10,"F COS P DT":^1,4,12,"WHERE P IS THE INSTANTANEOUS ANGLE BETWEEN F AND DELTA P ALONG THE PATH OF THE ALPHA PARTICLE."^1,4,16,"COMBINING THESE EQUATIONS WE HAVE"^1,6,19," A"^1,6,20,"2 MV SIN ="_1,6,21," 2"g_1,200,160,10,10,270,10:1,180,160,10,10,90,190:1,23,22,"O":1,25,18,"INF"_1,27,20,"F COS P DT": 20_ 0:(147):""`6)"LET US CHANGE THE VARIABLE FROM T TO P. THIS CHANGES THE INTEGRATION LIMITS TO -1/2 ( - A) AND +1/2 ( - A)"y`"CORRESPONDING TO THE VALUE OF P AT T = 0AND T = INF RESPECTIVELY, AND SO WE HAVE":`"(147):1,1`,1,6,9," A"`61,6,10,"2 MV SIN ="`@1,6,11," 2"3aJ1,180,80,10,10,270,10:1,22,12,"-1/2(PI-A)":1,160,80,10,10,90,190:1,22,8,"+1/2(PI-A)"maT1,26,10,"F COS P DP":1,31,9," DT":1,33,11,"DP"a^1,6,14,"THE QUANTITY DP/DT IS THE ANGULAR VELOCITY W OF THE ALPHA PARTICLE ABOUT THE GOLD NUCLEUS."Xbh1,6,17,"SINCE THE FORCE ACTING BETWEEN THEALPHA PARTICLE AND THE NUCLEUS IS DIRECTED ALONG THE RADIUS VECTOR JOINING"br1,0,20,"THEM THE ALPHA PARTICLE HAS CONSTANT ANGULAR MOMENTUM. HENCE": 20b| 0:(147)b12)" 2"b12)"MWR = CONSTANT":c12)" 2 DP"c12)" = MR "5c12)" DT":Mc12)" = MVB":pc6)"FROM WHICH WE SEE THAT":c14)" 2"c14)"DT R"c14)" = "c14)"DP VB":d6)"SUBSTITUTING THIS EXPRESSION FOR DT/DP WE HAVE":d1,1:(147)*d1,6,5," 2 A"Dd1,6,6,"2MV B SIN ="\d1,6,7," 2"d&1,170,52,10,10,270,10:1,150,52,10,10,90,190d01,20,8,"-1/2(PI-A)":1,22,4,"+1/2(PI-A)"d:1,26,5," 2"dD1,26,6,"F R COS P DP"keN1,6,10,"NOW F IS THE ELECTROSTATIC FORCE EXERTED BY THE NUCLEUS OM THE ALPHA PARTICLE. THE CHARGE ON THE ALPHA "eX1,0,13,"PARTICLE IS 2E AND THE CHARGE ON THE GOLD NUCLEUS IS ZE. THEREFORE THE FORCE F IS"Ufb1,12,18,"F = ":1,13,17," 1 2ZE":1,13,16," 2":1,13,20," 4(PI)E R":1,13,19," 2"tfl1,13,21," 0": 20fv(147):1,1f1,2,1,"AND"f1,4,3," 2"f1,4,4,"4(PI)E MV B"f1,4,5," 0 A"g1,4,6," SIN = COS P DP"6g1,4,7," ZE 2"hg1,215,52,10,10,270,10:1,195,52,10,10,90,190g1,27,4,"+1/2(PI-A)":1,24,8,"-1/2(PI-A)"g1,20,12,"= 2 COS ":1,20,11," A":1,20,13," 2"3h1,4,15,"THE SCATTERING ANGLE A IS RELATED TO THE IMPACT PARAMETER B BY THE EQUATION"h1,10,20,"COT = B":1,13,21," 2":1,14,19,"A 0":1,18,18,"2(PI)E MV"h1,22,22,"ZE ":1,14,21,"2 2" :1,10,17," 2"h 20: 0h(147)ji 6)"FROM AN EXAMINATION OF THE PRECEEDING EQUATION IT IS CLEAR THAT A SMALL VALUE FOR B (A NEAR MISS) IS "i"REQUIRED FOR A SUBSTANTIAL DEFLECTION. THERE IS, OF COURSE, NO WAY THAT THE IMPACT PARAMETER B CAN BE DIRECTLY"bj "MEASURED. HOWEVER ALL ALPHA PARTICLES APPROACHING A GOLD TARGET NUCLEUS WILLBE SCATTERED THROUGH AN ANGLE OF A OR "j*"MORE. THIS MEANS THAT AN ALPHA PARTICLE THAT IS INITIALLY DIRECTED ANYWHERE"j426)"2":"WITHIN THE AREA ()B":mk>"AROUND A NUCLEUS WILL BE SCATTERED THROUGH AN ANGLE OF A OR MORE. THIS AREAIS CALLED THE CROSS SECTION FOR THE INTERACTION."skHkR6)"THE GENERAL SYMBOL FOR CROSS SECTION IS THE GREEK LETTER SIGMA.":k\(147):""Qlf6)"IN A THIN GOLD FOIL OF THICKNESS TWHICH CONTAINS N GOLD ATOMS PER UNIT VOLUME THE NUMBER OF GOLD NUCLEI PER"lp"UNIT AREA IS NT , AND AN ALPHA PARTICLEBEAM INCIDENT UPON AN AREA A WILL ENCOUNTER NTA NUCLEI.";mz6)"THE AGGREGATE CROSS SECTION FOR SCATTERINGS OF ALPHA PARTICLES THROUGH AN ANGLE OF A OR MORE IS THE NUMBER OF"m"GOLD NUCLEI (NTA) MULTIPLIED BY THE CROSS SECTION FOR SUCH SCATTERING PER NUCLEUS (SIGMA) OR,":m14)"NTA(SIGMA)":m(147)=n6)"NOW THE FRACTION OF THE ALPHA PARTICLES THAT ARE SCATTERED BY AN ANGLEOF A OR MORE IS GIVEN BY":en8)"NUMBER SCATTERED BY A OR MORE"n4)"F = "n" NUMBER OF INCIDENT ALPHAS":n4)" AGGREGATE CROSS SECTION" o4)" = "*o4)" TARGET AREA":Co4)" NTA(SIGMA)"\o4)" = "ro4)" A":o 4)" 2"o 4)" = NTB "o o$ (147)o. 6)"SUBSTITUTING THE VALUE FOR B FROM (EQ A) WE HAVE":p8 8)" 2"!pB 8)" 2 "BpL 8)" ZE 2 A"ipV 8)"F = NT COT (A)p` 8)" 4E T' 2"pj 8)" 0 "pt p~ (147)p ""p 14)" PROBLEM ":""_q 6)"WHAT FRACTION OF A BEAM OF 7.7 MEVALPHA PARTICLES IS SCATTERED THROUGH ANGLES OF MORE THAN 45 DEGREES WHEN "q " -7"q "INCIDENT UPON A GOLD FOIL 3 X 10 METERSTHICK?":q (147):q 14)" SOLUTION "q :Tr 6)"THE NUMBER OF GOLD ATOMS PER UNIT VOLUME IN THE FOIL CAN BE FOUND FROM THERELATIONSHIP":r " ATOMS (ATOMS/KMOL) * (MASS/VOLUME)"r 2)" = "r 2)"VOLUME MASS/KMOL"r r 10)" N D" s 10)" 0"&s 10)" N = "@s( 10)" W":fs2 "WHERE N IS AVAGADRO'S NUMBER "vs< " 0"sF " D IS THE DENSITY OF GOLD, AND"sP " W IS THE ATOMIC WEIGHT OF GOLD":sZ (147) td 2)" 26 4 3";tn 2)"6.02 X 10 ATOMS/MOL X 1.93 X 10 KG/M"jtx "N="t 2)" 197 KG/KMOL":t 3)"OR 3"t 10)"N = 5.91 ATOMS OF GOLD/METER"t :gu 6)"NOW THE ATOMIC NUMBER Z OF GOLD IS 79, THE ANGLE A IS 45 DEGREES AND THE KINETIC ENERGY T OF A 7.7 MEV ALPHA "u " -12"u "PARTICLE IS 1.23 X 10 JOULE, SO WE HAVEFROM EQUATION (A) ABOVE"u 12)" -8"v 12)"F = 1.2 X 10 ":bv 6)"THUS A FOIL OF GOLD OF THIS THICKNESS IS QUITE TRANSPARENT TO ALPHA PARTICLES!":zv (147):""v 12)"ATOMIC SPECTRA ":v (147):""+w 6)"WHEN AN ATOMIC GAS OR VAPOR IS SUITABLY EXCITED BY THE PASSAGE OF AN ELECTRIC CURRENT THROUGH IT, IT EMITS A"w "CHARACTERISTIC RADIATION SPECTRUM WHICH CONTAINS CERTAIN DISCRETE WAVELENGTHS. THE NEXT FIGURE ILLUSTRATES A TYPICAL"w "LABORATORY ARRANGEMENT FOR OBSERVING SUCH ATOMIC SPECTRA.":w" (147):1,1[x, 1,1,1,"IONIZED GAS EXCITED BY AN ELECTRICAL DISCHARGE EMITS CHARACTERISTIC SPECTRA"x6 1,8,3,"WHICH CAN BE SEPARATED BY A":1,12,4,"PRISM AND PROJECTED ON A":1,16,5,"SCREEN AS IMAGES OF THE":1,18,6,"SLIT, CALLED 'LINES'"y@ 1,40,40,20,16,270,90:1,40,120,20,16,90,270cyJ 1,20,40 20,120:1,60,40 60,120:1,25,50 55,50:1,40,50 40,20yT 1,25,110 55,110:1,40,110 40,150:1,80,60 80,140:1,80,60 120,40 120,120 80,140z^ 1,140,110 160,110 150,85 140,110:1,150,85 170,75zh 1,93,98 108,92Lzr 1,170,75 180,95 160,110:1,17,15,"PRISM"z| 2:1,108,92 175,92 280,130 285,125 275,135 270,140z 1,100,95 168,95 265,135 275,125 260,140z 1,93,98 160,98 250,140 265,1258{ 1,200,120 320,120 300,145 180,145 200,120:1,30,19,"SCREEN"c{ 1,25,12,"R":1,25,13,"Y":1,24,14,"B"s{ 1,90,90,1:{ 1,11,13,"SLIT"{ J 1 50: I 1 20:1,30J,80I 30J,80I: I: J{ 1,5,11,"GLOW"| 1,4,21,"AN IDEALIZED EMISSION SPECTROMETER": 20+| 0:(147):""| 6)"WHEN AN ELECTRICAL DISCHARGE IS PASSED THROUGH AN ATOMIC GAS OR VAPOR THE CURRENT EXCITES THE ATOMS IN SUCH A"%} "WAY THAT RESULTS IN THE EMISSION OF RADIATION WHICH CONTAINS ONLY CERTAIN DISCRETE WAVELENGTHS CHARACTERISTIC OF"} "THE ATOMIC SPECIE CONTAINED IN THE GAS. THE NEXT FIGURES SHOW PORTIONS OF THE EMISSION SPECTRA OF SOME ELEMENTS.":} (147): 1,1:1,2} 1,0,4,"EMISSION SPECTRA OF HYDROGEN AND HELIUM"~ 1,50,170 290,170(~& I 6 36 5:1,I,20,"": I~0 1,6,22,"7" :1,10,22,"6.5":1,16,22,"6":1,20,22,"5.5":1,26,22,"5":1,30,22,"4.5":1,36,22,"4"~: 1,5,19,"RED":1,17,19,"YELLOW":1,32,19,"VIOLET"~D 1,40,140 290,140:1,40,120 290,120WN 1,80,120 80,140 82,140 82,120:1,220,120 220,140:1,280,120 280,140:1,260,120 260,140gX 1,1,16,"H"b 1,40,100 290,100:1,40,80 290,80:1,70,80 70,100:1,140,80 140,100 142,100 142,80l 1,216,80 216,100:1,214,80 214,100:1,220,80 220,100:1,250,80 250,100Bv 1,240,80 240,100:1,1,11,"HE" 1,12,24,"ANGSTROMS X 10":1,12,23," 3": 20 0:(147):"" 6)"EVERY ELEMENT HAS A UNIQUE LINE SPECTRUM WHEN A SAMPLE OF IT IS EXCITED IN A SPECTROMETER. THUS SPECTROSCOPY CAN"_ "BE USED TO DETERMINE THE COMPOSITION OF AN UNKNOWN SAMPLE.":w (147):"" 6)"WHEN WHITE LIGHT IS PASSED THROUGHA GAS, THE GAS ABSORBS LIGHT OF CERTAIN WAVELENGTHS PRESENT IN ITS EMISSION "h "SPECTRUM. THE RESULTING 'ABSORPTION LINESPECTRUM' CONSISTS OF A BRIGHT BACKGROUND CROSSED BY DARK LINES" "CORRESPONDING TO THE MISSING WAVELENGTHS": (147):""/ 6)"AN EXAMPLE OF ABSORPTION LINE SPECTRA CAN BE FOUND IN THE SPECTROGRAM PRODUCED BY OBSERVING THE LIGHT FROM THE" "SUN. THE SURFACE OF THE SUN RADIATES WHITE LIGHT BECAUSE IT IS A HOT OBJECT (ABOUT 5,800 DEGREES FAHRENHEIT). THE"+ "SURFACE IS SURROUNDED BY AN ENVELOPE OF COOLER GASES WHICH ABSORB LIGHT OF CERTAIN WAVELENGTHS ONLY. ACCORDINGLY " "IF WE LOOK AT A SPECTROGRAM PRODUCED BY SUNLIGHT WE CAN OBSERVE THE DARK LINES OF THE ELEMENT SODIUM OCCUPYING THE SAME" "POSITIONS ON THE PHOTOGRAPHIC PLATE THATWOULD BE OCCUPIED BY THE BRIGHT LINES OFTHE SODIUM EMISSION SPECTRUM"U 6)"THIS IS ILLUSTRATED IN THE NEXT FIGURE.":k (147):1,1:1,1 1,50,60 250,60 250,80 50,80 50,60…* 1,50,60 50,80:1,250,60 250,804 1,2:1,50,100 250,100 250,120 50,120 50,100d> 1,60,100 60,120:1,64,100 64,120:1,165,100 165,120:1,178,100 178,120:1,190,100 190,120ȆH 1,1:1,60,60 60,80:1,64,60 64,80:1,165,60 165,80:1,178,60 178,80:1,190,60 190,80R 1,2: I 6 30 5:1,I,20,"": I\\ 1,6,22,"6" :1,11,22,"5":1,16,22,"4":1,21,22,"3":1,26,22,"2":1,15,24,"ANGSTROMS X 10":1,30,23,"3"f 1,3,19,"YELLOW":1,13,19,"VIOLET":1,20,19,"ULTRAVIOLET"p 1,40,170 290,170z 1,3,2,"A = ABSORPTION SPECTRUM OF NA VAPOR":1,3,8,"A"3 1,2:1,3,5,"B = EMISSION SPECTRUM OF NA VAPOR":1,3,13,"B": 20O 0:(147):""q 12)"SPECTRAL SERIES ":} (147) 6)"DURING HIS STUDY OF THE VISIBLE PORTION OF THE SPECTRUM OF HYDROGEN J.J.BALMER IN 1885 DISCOVERED A RELATIONSHIP"u "BETWEEN THE SPECTRAL LINES OF HYDROGEN BEGINNING AT THE LINE WITH THE WAVELENGTH, 6,565 ANGSTROM, TO WHAT" "SEEMED TO BE A LIMIT AT 3,646 ANGSTROMS.BALMER FOUND THAT THE SPECTRAL LINES WERE FOUND CLOSER TOGETHER AND WEAKER IN"u "INTENSITY UNTIL THE LIMIT WAS REACHED, BEYOND WHICH THERE WERE NO FURTHER DISCRETE LINES BUT RATHER A FAINT " "CONTINOUS SPECTRUM.": (147):""% 6)"SOMEWHAT LATER FOUR OTHER SERIES OF HYDROGEN SPECTRA WERE OBSERVED AND MEASURED. THESE FIVE SERIES ARE CALLED" "LYMAN, BALMER, PASCHEN, BRACKETT AND PFUND SERIES, NAMED FOR THE SCIENTISTS WHO DISCOVERED THEM." 6)"IT WAS FOUND THAT THE WAVELENGTHS OF THE EMITTED SPECTRAL LINE IN EACH SERIES COULD BE SPECIFIED BY A SIMPLE"+"EMPIRICAL FORMULA."9:(147)|$4)"BALMER'S FORMULA FOR THE WAVELENGTHS OF THIS SERIES IS":.10)" "810)"1 1 1 "B10)" = R - N=2,3,4,.."L10)"W 2 2"V10)" 2 N "9`10)" ":hj"WHERE W IS THE WAVELENGTH OF THE LINE":t10)" AND":ύ~" R, THE RYDBERG CONSTANT HAS THE VALUE OF":10)" 7 -1"10)"R = 1.097 X 10 M ":-10)" -3 -1"L10)" = 1.097 X 10 A":d(147):""ݎ6)"USING THIS FORMULA ONE CAN DETERMINE THAT THE VARIOUS LINES OF THE BALMER SERIES SHOULD BE AS FOLLOWS": I 3 150:W 1(1.097E3 (14 1I2))- (W),"ANGSTROMS": I:E(147):""6)"THE FIRST NINE LINES BY THIS FORMULA ARE AT 6563, 4861, 4340, 4102, 3970, 3889, 3835, 3797, AND 3770 "?"ANGSTROMS, CORRESPONDING ALMOST EXACTLY TO THE POSITIONS OF THESE LINES AS OBSERVED IN A SPECTROMETER. THE LINES""RAPIDLY DRAW NEARER TO EACH OTHER AS N INCEASES UNTIL THE LIMIT AT 3646 ANGSTROMS IS REACHED, BEYOND WHICH ONLY""A FAINT CONTINUOUS SPECTRUM IS OBSERVED.": (147)v6)"THE BALMER SERIES CONTAINS ONLY THOSE WAVELENGHTS IN THE VISIBLE PORTIONOF THE EMISSION SPECTRUM OF HYDROGEN.""USING A SPECTROMETER CAPABLE OF DETECTING ULTRAVIOLET RADIATION ONE CAN OBSERVE THE LYMAN SERIES.":+(6)"THE FORMULA FOR THE WAVELENGTHS OFTHE LYMAN SERIES IS":H210)" "e<10)"1 1 1 "F10)" = R - N=2,3,4,.."P10)"W 2 2"ɒZ10)" 1 N "d10)" ":n"WHERE W IS THE WAVELENGTH OF THE LINE":3x10)" AND":"R, THE RYDBERG CONSTANT, HAS THE SAME VALUE HERE AS NOTED IN THE FORMULA FOR THE BALMER SERIES":(147)26)"A MORE GENERAL FORMULA WHICH DESCRIBES THE FIVE SERIES OF THE HYDROGEN EMISSION SPECTRUM IS THE FOLLOWING":O10)" "l10)"1 1 1 "10)" = R - "10)"W 2 2"Ô10)" N N "10)" 1 2"10)" ":,6)"LYMAN SERIES: N =1, N = 2,3,4,.."O6)" 1 2"{6)"BALMER SERIES N =2, N = 3,4,..."6)" 1 2"ʕ6)"PASCHEN SERIES: N =3, N = 4,5,..."6)" 1 2""6)"BRACKETT SERIES:N =4, N = 5,6,..."<,6)" 1 2"h66)"PFUND SERIES: N =5, N = 6,7,..."@6)" 1 2":J(147):"" T6)"THE STUDENT SHOULD RECALL THAT THESE FORMULAS ARE BASED ON EMPIRICAL RELATIONS BETWEEN THE POSITIONS OF THE"^"SPECTRAL LINES AND ARBITRARY VALUES OF RYDBERG'S CONSTANT AND THE NUMERICAL INDICES N AND N "h" 1 2":r(147):""5|6)"THIS IS A REMARKABLE COINCIDNCE BETWEEN ARBITRARY FORMULA AND THE OBSERVED ATOMIC SPECTRA OF HYDROGEN."6)"THE EXISTENCE OF SUCH REMARKABLE REGULARITIES IN THE HYDROGEN SPECTRUM, TOGETHER WITH SIMILAR REGULARITIES IN"."THE SPECTRA OF MORE COMPLEX ELEMENTS LEDTO AN EXAMINATION OF THE POSSIBLE EXPLANATION OF THIS PHENOMENUM IN TERMS"q"OF THE STRUCTURE OF THE ATOMS OF THE VARIOUS ELEMENTS.":(147):""6)"THE BEST INFORMATION WE HAVE IS THAT THE HYDROGEN ATOM CONSISTS OF A NUCLEUS WITH A SINGLE PROTON SURROUNDED ""BY A SINGLE ELECTRON. IT IS CLEAR THAT IF THE ELECTRON IS OBLIGED TO WHIRL AROUND THE NUCLEUS LIKE A PLANET AROUND""THE SUN, THAT THE PRINCIPLES OF CLASSICAL PHYSICS WOULD REQUIRE THAT THEELECTRON RADIATE ELECTROMAGNETIC ENERGY ""CONTINUOUSLY. IT IS NECESSARY TO LOOK INTO QUANTUM CONCEPTS TO FIND AN EXPLANATION FOR THE EMISSION OF LIGHT ":(147):""12)" DE BROGLIE WAVES ":ٛ(147):""V6)"LOUIS DE BROGLIE SUGGESTED IN 1924THAT WAVES MIGHT BE ASSOCIATED WITH PARTICLES, THE CONVERSE OF THE DISCOVERY"Ҝ"IN 1905 THAT PARTICLES CAN BE ASSOCIATEDWITH WAVES. THE ACTUAL EXISTENCE OF THE DE BROGLIE WAVES WAS DEMONSTRATED IN"Q"1927. THE EXISTENCE OF THESE WAVES WAS THE STARTING POINT FOR SCHROEDINGER'S DEVELOPMENT OF HIS QUANTUM MECHANICS.":](147)՝6)"AS WE HAVE SEEN IN OTHER PARTS OF THIS SERIES ON MODERN PHYSICS A PHOTON OF LIGHT HAS NO REST MASS AND HAS A"&"MOMENTUM OF":012)" H(NU)":12)"P = "+D12)" C" :RN6)"WHERE H IS PLANCK'S CONSTANT"X6)"(NU) IS THE FREQUENCY OF THE LIGHT PHOTON, AND"b6)"C IS THE VELOCITY OF LIGHT":l6)"SINCE C/(NU) IS THE WAVELENGTH (LAMBDA) OFTHE LIGHT WE HAVE":v12)" H(NU)"112)"P = "J12)" LAMBDA" :6)"SINCE THE MOMENTUM OF A PARTICLE OF MASS M AMD VELOCITY C IS "12)"P = MV , WE HAVE":Ɵ(147):""6)"THE DE BROGLIE WAVELENGTH OF A PHOTON OF FREQUENCY (NU) IS":)12)" H"D12)"LAMBDA = "^12)" MV"ʠ"":6)"THE STUDENT SHOULD NOTE HERE THAT THE MASS M OF THE PHOTON IS THE RELATIVISTIC MASS"12)" M"12)" O"12)"M = "312)" 2 2 1/2"P 12)" (1- V /C )":h(147):"" 12)" WAVE FUNCTION ":*(147):""46)"BEFORE WE CAN FULLY UNDERSTAND DE BROGLIE WAVES WE MUCH DEFINE THE TERM 'WAVE FUNCTION', DENOTED BY THE GREEK"2>"LETTER (PSI).":JH(147):""kR14)"DEFINITION ":""\6)"THE VALUE OF PSI THE WAVE FUNCTIONASSOCIATED WITH A MOVING BODY AT A PARTICULAR POINT (X,Y,Z) IN SPACE AT THE"jf"TIME T IS RELATED TO THE LIKELIHOOD OF FINDING THE BODY THERE AT THAT TIME. PSI HAS NO DIRECT PHYSICAL SIGNIFICANCE.":p(147):""z6)"WHILE (PSI) ITSELF IS THE PROBABILITY OF FINDING THE PHOTON AT A PARTICULAR PLACE, THE AMPLITUDE OF THE"|"PHOTON WAVE CAN BE EITHER POSITIVE OR NEGATIVE, AND OF COURSE, A NEGATIVE PROBABILITY IS MEANINGLESS. HOWEVER THE"ऎ"SQUARE OF (PSI), IS ALWAYS POSITIVE. WE THEREFORE DEFINE THE PROBABILITY DENSITYAS FOLLOWS":(147):""12)"PROBABILITY DENSITY ":6(147):""6)"THE PROBABILITY OF EXPERIMENTALLY FINDING THE PHOTON DESCRIBED BY THE WAVEFUNCTION (PSI) AT THE POINT (X,Y,Z) AT""THE TIME T IS PROPORTIONAL TO THE VALUE OF"" 2"*"(PSI) AT THE POINT (X,Y,Z) AND AT THE TIME T.":A(147):""j8)" BOHR'S FIRST POSTULATE ":(147):""6)"BOHR'S FIRST POSTULATE ON THE STRUCTURE OF THE ATOM WAS THAT AN ELECTRON MOVES IN A STATIONARY ORBIT"9"ABOUT THE NUCLEUS WITH AN ANGULAR MOMENTUM OF":P12)" H"h12)"LAMBDA = "$12)" MV":.(147):""86)"DE BROGLIE HAD ASSUMED THAT PARTICLES,SUCH AS ELECTRONS AND PROTONS,HAD WAVE PROPERTIES AND THAT THE"fB"WAVELENGTHS OF THESE WAVES ARE RELATED TO THE MOMENTUM BY THE SAME EQUATION.":rL(147)V6)"THE WAVES OF AN ELECTRON MOVING INA CIRCULAR ORBIT OF RADIUS R MOVE WITH THE ELECTRON ALONG THIS PATH. IN THE"k`"COURSE OF A FEW REVOLUTIONS OF THE ELECTRON, AND ITS ASSOCIATED WAVES, DIFFERENT PARTS OF THE WAVE WILL OVERLAP"j"AND IF THEY ARE NOT TO INTERFERE DESTRUCTIVELY, THEY MUST PRODUCE A STANDING WAVE. IF THE LENGTH OF THE "Et"ELECTRON PATH, 2R SHOULD BE A WHOLE MULTIPLE OF THE WAVELENGTH LAMBDA, OR IF":c~12)"2R = N (LAMBDA)":6)"THEN WE SHOULD HAVE A STANDING WAVE ALONG THE PATH,":12)" NH"ڪ12)"2R = OR,"12)" MV": (147):""%5)" H"A5)" MVR = N "_5)" 2":"WHICH IS THE BOHR QUANTIZATION CONDITION": 6)"HERE N IS AN INTEGER OF ANY VALUE 1,2,3,... AND H IS PLANCK'S CONSTANT. THUS ACCORDING TO THIS HYPOTHESIS ONLY ""THOSE ELECTRON ORBITS ARE PERMISSIBLE FOR WHICH THE ANGULAR MOMENTUM OF THE ELECTRON IS A WHOLE MULTIPLE OF"14)" H"14)" " 14)" 2"::(147)16)"WE SAW EARLIER THAT THE DEBROGLIE WAVELENGTH OF THE ELECTRON IN ORBIT IN THE HYDROGEN ATOM IS GIVEN BY":L12)"LAMBDA = H/MV":q("WHERE THE ELECTRON SPEED IS":212)" E"<12)"V = "F12)" "˭P12)" 4E MR"Z12)" 0":d"HENCE"n12)" 1x12)" 4E R"R12)" H 0 "t12)"LAMBDA = "12)" E M ":(147)ɮ6)" -11"A6)"BY SUBSTITUTING 5.3 X 10 FOR R IN THIS EQUATION, WE FIND THAT THE ELECTRON DE BROGLIE WAVELENGTH IS":`12)" -11"12)"LAMBDA = 33 X 10 METERS":6)"THIS WAVELENGTH IS EXACTLY EQUAL TO THE CIRCUMFERENCE OF THE ELECTRON ORBIT "12)" -11"!12)"2R = 33 X 10 METERS":6)"THUS THE ORBIT OF THE ELECTRON IN A HYDROGEN ATOM CORRESPONDS TO ONE ELECTRON WAVE JOINED ON ITSELF.":(147):1,11,15,1,"FOR N = 1"ְ1,20,50 300,50 1,100,100,90,72,315,45:1,228,0,90,72,135,225["1,35,60 35,90:1,292,60 292,90:1,5,10,"<----CIRCUMFERENCE OF ORBIT--->"±,1,3,14,"THE ORBIT OF THE ELECTRON CORRESPONDS TO A COMPLETE DEBROGLIE WAVE JOINED ON ITSELF": 20ұ6 0:(147)۱@1,1J1,15,1,"FOR N = 2"T1,20,50 300,50i^1,63,78,50,40,315,45:1,200,78,50,40,315,45:1,130,20,50,40,135,225:1,268,20,50,40,135,225h1,30,60 30,90:1,300,60 300,90:1,5,10,"<----CIRCUMFERENCE OF ORBIT--->")r1,3,14,"THE ORBIT OF THE ELECTRON CORRESPONDS TO TWO COMPLETE DEBROGLIE WAVES JOINED ON THEMSELVES": 209| 0:(147)[12)" NUCLEAR MOTION ":g(147)޳6)"LET US NOW EXAMINE THE EFFECT THATNUCLEAR MOTION HAS ON THE BEHAVOIR OF THE HYDROGEN ATOM. EARLIER WE HAD "\"ASSUMED THAT THE PROTON OF THE HYDROGEN ATOM REMAINED STATIONARY WHILE THE ORBITAL ELECTRON REVOLVED AROUND IT.":ٴ6)"WE HAVE SEEN THAT A STABLE ORBIT IN AN ATOM MUST CONSIST OF AN INTEGRAL NUMBER OF DE BROGLIE WAVELENGTHS, OR ""STATED MATHEMATICALLY":10)"N(LAMBDA) = 2R":M6)"THE DE BROGLIE WAVELENGTH (LAMBDA)IS GIVEN BY"d12)" H"|12)"LAMBDA = "12)" MV""WHERE H IS PLANCK' CONSTANT"::(147)ݵ6)"SO WE MAY WRITE":12)" NH"12)"MVR = "12)" 2":&6)"THE QUANTITY MVR IS CALLED THE ANGULAR MOMENTUM OF THE ELECTRON IN ITS ORBIT. THUS ANOTHER WAY TO EXPRESS"0"BOHR'S FIRST POSTULATE IS THAT THE ANGULAR MOMENTUM OF A HYDROGEN ATOM MUSTBE AN INTEGRAL MULTIPLE OF"::12)"H/2":.D(147):""uN6)"NOW THE ANGULAR VELOCITY W OF THEELECTRON IS EQUAL TO V/R":X6)"SO THE QUANTIZATION RULE CAN BE STATED AS"b12)" 2 NH"l12)"MWV = (2)"v12)" 2": (147)6)"NOW A NUCLEUS OF MASS M AND ITS ORBITAL ELECTRON MUST REVOLVE AROUND A COMMON CENTER OF MASS.THE CENTER OF MASS""OF A BODY IS THE BALANCE POINT. WE CAN THINK OF THE NUCLEUS AND THE ELECTRON INA HYDROGEN ATOM AS BEING AT THE OPPOSITE"H"ENDS OF A MASSLESS ROD OF LENGTH R. SEE THE NEXT FIGURE":x(147):1,1:53280,0:53281,0: 241,1:1,2ڹ1,1,1,"THE ELECTRON AND PROTON OF THE HYDROGEN ATOM REVOLVE AROUND A COMMON CENTER OF MASS",1,160,95,140,50:1,205,82,8,6:1,205,82,1:1,28,10,"NUCLEUS":1,160,95,60,20m1,200,85 160,95 50,125:1,50,125,4,3:1,4,17,"ELECTRON"1,160,95,3,2:1,22,12,"CM"1,6,19,"R = RADIUS OF ELECTRON"ú1,6,20," E"1,6,21,"R = RADIUS OF NUCLEUS"1,6,22," N"$1,6,23,"R = DISTANCE BETWEEN ELECTRON"G 1,6,24," AND NUCLEUS": 20W 0:(147) 6)"FROM THE DIAGRAM ONE CAN SEE THAT":*12)"R = R + R (3)"412)" E N":>6)"FROM OUR KNOWLEDGE OF MECHANICS WEKNOW THAT":H12)"M R = M R (4)"5R12)" E E N E":\6)"NOW THE TOTAL ANGULAR MOMENTUM OF THE HYDROGEN ATOM IS THE SUM OF THE ANGULAR MOMENTUM OF THE ELECTRON AND THE"ܼf"ANGULAR MOMENTUM OF THE NUCLEUS,":p"OR" *z" 2 2"X"TOT ANG MOMENTUM = M WR + M WR (5)" E E N N":(147) 6)"BOHR'S FIRST POSTULATE REQUIRES THAT THE TOTAL ANGULAR MOMENTUM OF THE ATOM MUST BE AN INTEGRAL MULITPLE OF H/2 , SO WE HAVE":@10)" 2 2 NH"g10)"M WR + M WR = (6)"10)" E E N N 2":6)"FROM EQUATIONS (3) AND (4) WE FINDTHAT":2)" "2)" M M "@2)" N E "q2)"R = R AND R = R (7)"2)" E M + M N M + M "ɿ2)" E N E N"2)" ": (147):""7$6)"SO EQUATION (6) BECOMES":""K.8)" "_88)" M M "|B8)" E N 2 NH "L8)"WR = (8)"V8)"M + M 2 "`8)" N E"j8)"":t(147):""V~6)"LET US NOW LOOK AT THE CLASSICAL DYNAMICS OF THE HYDROGEN ATOM":n(147):""6)"THE NEXT FIGURE ILLUSTRATES THE FORCE BALANCE IN THE HYDROGEN ATOM.":(147):1,1¦1,165,100,10,8:1,7:1,165,100,1:1,2:1,12,12,"PROTON":1,20,12,"+E"^°1,260,120,5,4:1,260,120,1:1,23,16,"ELECTRON":1,34,15,"-E"º1,165,98,100,80:1,1,2,"ELECTRON ORBIT":1,70,30 90,40:1,155,90 105,40:1,18,8,"R=RADIUS"+1,165,100 205,108 203,104 200,110 205,108:1,260,120 220,112 224,116 227,110 220,112c1,24,11,"F":1,25,12,"E":1,30,12,"F":1,31,13,"C"1,4,23,"FORCE BALANCE IN THE HYDROGEN ATOM"1,33,11,"^ V": 20 0:(147)(6)"THE ELECTROSTATIC FORCE OF ATTRACTION BETWEEN THE PROTON AND THE ELECTRON IS GIVEN BY":A12)" 2"Z 12)" 1 E "s12)"F = "12)" E 4E 2"(12)" 0 R":2:6)"AND THE CENTRIPETAL FORCE HOLDING THE ELECTRON IN ITS ORBIT IS":: <12)" 2"F12)" MV"0P12)"F = "HZ12)" C R "::d6)"NOW FOR THE ELECTRON TO BE IN A STABLE ORBIT THESE FORCES MUST BE EQUAL":n(147)x6)"SO WE HAVE":ł12)" 2 2"Ō12)" MV 1 E "Ɩ12)" = "-Ơ12)" R 4E 2"Lƪ12)" 0 R": :ƴ6)"SOLVING THE ABOVE FOR THE ELECTRONVELOCITY V IN TERMS OF ITS ORBITAL RADIUS R WE FIND":ƾ12)" E"12)"V = "12)" Σ"12)" ʹ4E MR")12)" 0"::6)"THE TOTAL ENERGY T OF THE ELECTRONIN ITS ORBIT IN A HYDROGEN ATOM IS THE SUM OF ITS KINETIC ENERGY K AND ITS ""POTENTIAL ENERGY P.":(147):"WHERE"12)" 2" 12)"K = 1/2 MV ":"AND""12)" 2"5,12)" E "L612)"P = - "c@12)" 4E R"}J12)" 0"::T"NOW T = K + P":^12)" 2 2"h12)" MV E "r12)"T = - " |12)" 2 4E R"&Ɇ12)" 0":<ɐ"RECALLING THAT"Wɚ12)" 2 2"rɤ12)" MV 1 E "ɮ12)" = "ɸ12)" R 4E 2"12)" 0 R":(147)"SUBSTITUTING AND SOLVING FOR THE TOTAL ENERGY WE FIND THAT")12)" 2"?12)" E "a12)"T = - (A)x12)" 8E R"12)" 0":6)"WE THUS SEE THAT THE TOTAL ENERGY OF THE ELECTRON IS NEGATIVE JUST AS IT SHOULD BE IN ORDER TO BE BOUND TO THE PROTON"::6)"EXPERIMENTALLY 13.6 ELECTRON VOLTSARE REQUIRED TO SEPARATE A HYDROGEN ATOMINTO A PROTON AND AN ELECTRON. WE THUS "&"SAY THAT THE BINDING ENERGY T OF THE ATOM IS -13.6 ELECTRON VOLTS."10"SUBSTITUTING THIS IN EQUATION (A) ABOVE AND SOLVING FOR R WE FIND THAT":L:12)" -11"oD12)"R = 5.3 X 10 METERS":N"A VALUE WHICH AGREES GENERALLY WITH OTHER ESTIMATES OF THE RADIUS OF THE HYDROGEN ATOM":X(147):""!b4)" QUANTTZATION OF ANGULAR MOMENTUM ":-l(147)v6)"WE HAVE NOW SEEN THAT THE OBSERVEDALLOWED ENERGY LEVELS OF ONE-ELECTRON ATOMS CAN BE JUSTIFIED BY ASSUMING THAT" ΀"THE ANGULAR MOMENTUM OF THE ELECTRON IS QUANTIZED AND CAN HAVE ONLY THE VALUES GIVEN BY"1Ί8)"K = NH/2 N=1,2,3... (1)":|Δ"AS WE HAVE SAID EARLIER N IS CALLED THE PRINCIPAL QUANTUM NUMBER.":Ξ6)"BUT THIS RELATION IS NOT QUITE ACCURATE. IT WAS DERIVED BY ASSUMING CIRCULAR ORBITS FOR THE ELECTRON AND"uϨ"THE CLASSICAL DESCRIPTION OF MOTION. WHEN THE CORRECT THEORY OF QUANTUM MECHANICS IS USED THE ALLOWED VALUES OF "ϲ"THE ORBITAL ANGULAR MOMENTUM ARE GIVEN BY"ϼ12)" "12)"K = L(L+1)H/2 (2)"W"WHERE L = 0,1,2,3,...IS A POSITIVE INTEGER AND IS CALLED THE ANGULAR MOMENTUM QUANTUM NUMBER.":c(147)6)"IN CLASSICAL MECHANICS THE SHAPE OF THE PLANETARY ORBIT VARIES ACCORDING TO THE ANGULAR MOMENTUM OF THE PARTICLE."\"THE SAME RESULT APPLIES IN QUANTUM MECHANICS. THUS EACH ENERGY LEVEL, DETERMINED BY THE PRINCIPAL QUANTUM ""NUMBER N, IS ASSOCIATED WITH SEVERAL ELECTRON ANGULAR-MOMENTUM STATES DETERMINED BY THE VALUES OF L, THE ""ANGULAR MOMENTUM QUANTUM NUMBER.":> 6)"THE VALUES OF L RELATED WITH EACH VALUE OF N ARE":l6)"L = 0,1,2,3,...(N-1) (N VALUES)": 6)"ANGULAR MOMENTUM STATES ARE DESIGNATED BY LETTERS ACCORDING TO THE SCHEME SHOWN NEXT.":*(147):"">410)"ANGULAR MOMEMTUM STATES":"":" "m>6)"L";14)" 0 1 2 3 4":14)""H6)"SYMBOL";14)" S P D F G":14)""R6)"G = 2L+1"14)" 1 3 5 7 9"\" ":yf6)"THIS CODE MAY BE BAFFLING TO THE STUDENT. IT ORIGINATED IN THE EMPIRICAL CLASSIFICATION OF SPECTRA INTO SERIES "p"CALLED SHARP, PRINCIPAL, DIFFUSE, AND FUNDAMENTAL WHICH OCCURRED LONG BEFORE THE THEORY OF THE ATOM WAS DEVELOPED.":z(147):""Մ6)"THUS THE S STATE OF THE ATOM IS ONE WITH ZERO ANGULAR MOMENTUM, A P STATE HAS THE ANGULAR MOMENTUM OF"Վ10)" "՘10)"2 H/2 , ETC":բ(147):""լ8)" ATOMIC STATES DESIGNATION ":ֶ(147)""6)"THE COMBINATION OF THE TOTAL QUANTUM NUMBER N WITH THE LETTER THAT REPRESENTS ORBITAL ANGULAR MOMENTUM IS" "WIDELY USED AS A CONVENIENT NOTATION FORATOMIC STATES. IN THIS NOTATION A STATE IN WHICH N=2,L=0 IS A 2S STATE, AND ONE""IN WHICH N=4,L=2 IS A 4D STATE. THE FOLLOWING TABLE LISTS THE DESIGNATIONS OF ATOMIC STATES IN HYDROGEN THROUGH N=6,L=5.":(147)4)" THE ATOMIC STATES OF HYDROGEN ":8)"S";13)"P";18)"D";23)"F";28)"G";33)"H":A7)"L=0";12)"L=1";17)"L=2";22)"L=3";27)"L=4";32)"L=5":s"":0)"N=1";8)"1S":$0)"N=2";8)"2S";13)"2P":.0)"N=3";8)"3S";13)"3P";18)"3D": 80)"N=4";8)"4S";13)"4P";18)"4D";23)"4F":EB0)"N=5";8)"5S";13)"5P";18)"5D";23)"5F";28)"5G":L0)"N=6";8)"6S";13)"6P";18)"6D";23)"6F";28)"6G";33)"6H":V"":`(147):""j12)" SPACE QUANTIZATION ":t(147)~6)"IN ADDITION TO THE QUANTIZATION OFTHE MAGNITUDE OF THE ORBITAL ANGULAR MOMENTUM L ITS ORIENTATION RELATIVE TO "ڈ"A GIVEN AXIS IS LIMITED TO CERTAIN DIRECTIONS. THIS RESULT IS CALLED SPACE QUANTIZATION.":ڒ(147)aۜ6)"DESIGNATING THE GIVEN AXIS AS Z, WE OBTAIN THE ALLOWED VALUES OF THE Z-COMPONENT OF L AS ":xۦ12)"L = M H/2"۰12)" Z L"ۺ6)"WHERE M IS A POSITIVE OR NEGATIVE "6)" L":"INTEGER, HAVING THE VALUES"(12)"M = 0,+1,-1,+2,-2,....+L(L-1),-L(L-1),+L,-L."L12)"THE UPPER VALUE OF M IS"m12)" L""+L,-L BECAUSE, L CANNOT BE LARGER THAN"" Z" " L."6)"FOR L = 0 (S-STATES) ONLY M =0 IS"6)" L"W("POSSIBLE. FOR L = 1 (P-STATES)WE MAY HAVE M = 0,-1,-1."h2" L":t<(147)F6)"IN GENERAL FOR ANY L VALUE THERE ARE G = 2L+1 POSSIBLE VALUES FOR M ."P6)" L"Z"EACH VALUE OF M GIVES A POSSIBLE "4d" L"Mn"ORIENTATION OF L."x6)"THE NEXT CHARTS WILL ILLUSTRATE THESE ORIENTATIONS.":ނ(147): 1,1ތ1,160,40 160,100 240,110:1,160,100 100,140 ߖ1,160,100 200,60 190,63 202,68 200,60"ߠ1,160,50 195,58Kߪ1,18,8,"L":1,19,9,"Z":1,23,10,"L"ߴ1,20,4,"Z":1,20,13,"0":1,31,14,"Y":1,10,18,"X"߾1,4,20,"ORBITAL ANGULAR MOMENTUM L AND ITS COMPONENT L ALONG THE Z-AXIS":1,15,22,"Z": 20 0:(147):1,1+1,120,20 120,170:1,120,100,80,64,0,180H1,14,1,"Z":1,24,12,">"p1,120,100 200,100:1,120,100,4,31,7,11,"L = 0":1,8,12,"Z":1,7,13,"M = 0":1,8,14,"L"1,120,65,4,3:1,120,135,4,31,4,6,"L = H/2(PI)":1,5,7,"Z":1,3,8," M = +1":1,4,9," L"T1,3,16,"L = -H/2(PI)":1,4,17,"Z":1,3,18,"M = -1":1,4,19,"L"i1,27,11," ""1,27,12,"L=2 H/2(PI)",1,120,100 184,65 120,65:1,120,100 184,135 120,135!61,1,21,"THREE POSSIBLE ORIENTATIONS OF ANGULAR MOMENTUM CORRESPONDING TO L=1 AND TO"5@1,1,23," "WJ1,1,24,"L=2 H/2(PI).": 20qT(147): 0:""^6)"THE STUDENT MAY BE CONFUSED AT THIS STAGE BY THE CONCEPT OF ANGULAR MOMENTUM QUANTIZATION AND MAY THINK THAT"mh"THE IDEA IS UNREAL. ONE MIGHT ASK WHAT POSSIBLE SIGNIFICANCE CAN A PARTICULAR DIRECTION IN SPACE HAVE FOR A HYDROGEN "r"ATOM? HOWEVER THERE IS MUCH EVIDENCE OF THE EXISTENCE OF SUCH QUANTIZATION. THE MOST DIRECT EXPERIMENTS ARE RELATED "C|"TO THE ACTION OF A MAGNETIC FIELD ON THEATOMIC ENERGY LEVELS, THE ZEEMAN EFFECT.":(147):"":10)" THE NORMAL ZEEMAN EFFECT ":(147):""6)"A PHENOMENUM WAS OBSERVED BY THE DUTCH PHYSICIST ZEEMAN IN 1896 WHICH INVOLVED THE 'SPLITTING' OF A SINGLE ""SPECTRAL LINE INTO SEVERAL INDIVIDUAL LINES WHEN THE RADIATING ATOMS WERE PLACED IN A MAGNETIC FIELD, THE SPACING""OF THE LINES WAS FOUND TO BE DEPENDENT UPON THE MAGNITUDE OF THE FIELD.":(147)o6)"THE NORMAL ZEEMAN EFFECT CONSISTS OF THE SPLITTING OF A SPECTRAL LINE OF FREQUENCY F INTO THREE COMPONENTS WITH""FREQUENCIES AS FOLLOWS":2)" EH/2 B E"2)"F = F - = F - B"2)" 1 2M H 4M":""%2)"F = F"42)" 2":] 2)" EH/2 B E" 2)"F = F + = F + B" 2)" 3 2M H 4M":: (147):""D 6)"LET US TRY TO SEE WHY THIS MOST INTERESTING PHENOMENUM CAN OCCUR.":)N (147)AX (147):""b 6)"AN ORBITING ELECTRON IS EQUIVALENTTO AN ELECTRICAL CURRENT AND ITS ASSOCIATED MAGNETIC DIPOLE MOMENT. IN"6l "THE ABSENCE OF ANY EXTERNAL MAGNETIC FIELD ATOMS OF HYDROGEN WILL BE ORIENTEDAT RANDOM. WOWEVER, IF AN EXTERNAL "v "MAGNETIC FIELD IS PRESENT THERE WILL BE A TORQUE ON THE MAGNETIC DIPOLES OF THE ATOMS AND THEY WILL THEREFORE ALIGN" "THEMSELVES WITH THE MAGNETIC FIELD DIRECTION.": (147):1,1[ 1,2,1,"A MAGNETIC DIPOLE OF MOMENT K AT AN ANGLE A RELATIVE TO A MAGNETIC FIELD B " 1,100,120 220,60 224,64 104,124 100,120:1,102,120,1:1,210,60 224,60 224,70 210,60 I1 96 8:1,60,(42I) 280,(42I): I J 1 12:1,35,4J,">": J% 1,36,11,"B"C 1,19,10,"K":1,17,14,"A"^ 1,104,124,50,40,50,90 1,4,18,"THE TORQUE T ON A MAGNETIC DIPOLE IN A MAGNETIC FIELD OF FLUX DENSITY B IS" 1,14,21,"T = KB SIN A": 20 0:(147):""q 6)"THE MAGNETIC MOMENT OF THE ORBITALELECTRON IN A HYDROGEN ATOM DEPENDS ON ITS ANGULAR MOMENTUM L. THE MAGNETIC "!"FIELD MAKES A CONTRIBUTION TO THE TOTAL ENERGY OF THE ATOM. THIS CONTRIBUTION TOTHE TOTAL ENERGY DEPENDS ON THE "0 !"MAGNITUDE OF L AND ITS ORIENTATION WITH RESPECT TO THE FIELD.":F!(147):"" !6)"FROM CLASSICAL THEORY AN ELECTRON WHICH MAKES D REVOLUTIONS PER SECOND IN A CIRCULAR ORBIT OF RADIUS R IS "&*!"EQUIVALENT TO A CURRENT -ED (SINCE THE ELECTRONIC CHARGEIS -E),AND ITS MAGNETICMOMENT IS THEREFORE"<4!12)" 2"_>!12)"K = -EDR (A)":H!6)"THE LINEAR SPEED OF THE ELECTRON IS 2DR, SO ITS ANGULAR MOMENTUM IS":R!12)"L = MVR"\!12)" 2"f!12)"L = 2MDR (B)": p!(147):z!6)"COMBINING EQUATIONS (A) AND (B) WEHAVE"P!12)" "f!12)" E "!12)"K = - L (A)"!12)" 2M"!12)" ":!(147):""K!6)"THE QUANTITY (-E/2M), INVOLVING ONLY THE MASS OF THE ELECTRON AND ITS CHARGE, IS CALLED THE GYROMAGNETIC RATIO"!6)"THE MINUS SIGN MEANS THAT THE ELECTRON MAGNETIC MOMENT K IS IN THE OPPOSITE DIRECTION TO ITS ANGULAR MOMENTUL L.":]!6)"A MORE DETAILED CALCULATION BASED ON QUANTUM MECHANICS YIELDS THE SAME RESULTS FOR THE MAGNETIC MOMENT OF THE ELECTRON.":s!(147):""!6)"THE TORQUE ON THE DIPOLE IS A MAXIMUM WHEN THE DIPOLE IS PERPENDICULARTO THE MAGNETIC FIELD AND ZERO WHEN IT "j!"IS PARALLEL TO IT. TO CALCULATE THE POTENTIAL ENERGY P,ESTABLISH A REFERENCECONFIGURATION AT WHICH P IS ZERO BY"!"DEFINITION. A CONVENIENT POSITION IS WHEN K IS PERPENDICULAR TO B, THAT IS P = 0 AT A = 90 DEGREES. THE POTENTIAL "i""ENERGY AT ANY OTHER ORIENTATION IS EQUALTO THE WORK THAT THE FIELD MUSE DO TO ROTATE THE DIPOLE FROM 90 DEGREES TO THE"""ANGLE A. HENCE":"(147):1,1$"1,170,40,12,12,270,15:1,146,40,12,12,90,180."1,21,3,"A":1,18,7,"90"8"1,14,5,"P =":1,22,5,"T DA"4B"1,180,90,12,12,270,15:1,156,90,12,12,90,180uL"1,14,11," =":1,24,11,"SIN A DA":1,22,9,"A":1,20,13,"90"V"1,14,15," = - KB COS A"`"1,6,18,"SUBSTITUTING THE VALUE FOR K FROM EQUATION (A) WE HAVE": 20j" 0:(147)t"12)" "~"12)" E ";"12)"P = LB COS A"P"12)" 2M"e"12)" ""6)"NOW THE MAGNITUDE OF THE ORBITAL ANGULAR MOMENTUM L' OF THE ELECTRON IS QUANTIZED,AND ISGIVEN AS A FUNCTION OF"""ORBITAL QUANTUM NUMBER L BY THE FORMULA")"12)" "I"12)"L' = L(L+1) H/2":"6)"IT IS NOT SURPRISING THEREFORE TO LEARN THAT THE DIRECTION OF L' IS ALSO QUANTIZED WITH RESPEST TO A MAGNETIC "<""FIELD. THIS IS OFTEN REFERRED TO AS 'SPACE QUANTIZATION'. THE MAGNETIC QUANTUM NUMBER M SPECIFIES THE "V"" L"""DIRECTION OF L' BY DETERMINING THE COMPONENT OF L' IN THE FIELD DIRECTION.IF WE LET THE MAGNETIC FIELD DIRECTION"""BE PARALLEL TO THE Z AXIS.": #(147):""P #6)"THE COMPONENT OF L' IN THIS DIRECTION IS GIVEN BY":h#12)"L = M H/2"#12)" Z L":""(#6)"THE POSSIBLE VALUES OF M FOR A GIVEN VALUE OF L RANGE FROM +L THROUGH 0TO -L, SO THAT THE NUMBER OF POSSIBLE"{2#"ORIENTATIONS OF THE ANGULAR MOMENTUM VECTOR L' IN A MAGNETIC FIELD IS 2L+1. THE SPACE QUANTIZATION OF THE HYDROGEN"<#"ATOM IS SHOWN IN THE NEXT FIGURE.":F#(147):1,1P#1,22,2,"SPACE QUANTIZATION"Z#1,22,3,"OF ORBITAL ANGULAR"d#1,22,4," MOMENTUM IN THE"8n#1,22,5," HYDROGEN ATOM"ax#1,40,100,120,96,0,180:1,40,100,3,2#1,40,100 80,10 40,10:1,40,100 132,40 40,40:1,40,100 155,70 40,70:1,40,100 155,130 40,130#1,40,100 132,160 40,160:1,40,100 80,190 40,190Y#1,80,190 84,186 74,187 80,190:1,40,100 158,100:1,19,12,">"#1,2:1,3,12,"0":1,3,8,"1":1,3,4,"2":1,3,1,"3":1,2,16,"-1":1,2,20,"-2":1,2,23,"-3"#1,0,11," H":1,0,12,"":1,0,13,"2PI"c#1,21,12,"0":1,20,16,"-1":1,20,20,"-2":1,20,23,"-3":1,20,8,"+1":1,20,4,"+2":1,20,1,"+3":1,12,1,"M =":1,15,2," L"l# 20#(147): 0:""#6)"THIS PROGRAM IN THE STUDY COURSE IN MODERN PHYSICS, IS CONTINUED ON 'ATOMIC STRUCTURE III' WHICH CAN BE "<#"OBTAINED FROM THE AUTHOR BY SENDING $10.00 TO":""#10)"DR. PAUL W. MCDANIEL":10)"4295 WARREN WAY":10)"RENO, NEVADA 89509"