Atomic Physics

By C K Cheung Atomic PhysicsMatter ----- composed of atoms and molecules. Atom ----- composed of ??????J.J. Thomson Experiment (1897 )Thomson’s ‘ plum – pudding ‘model of atoms: electron 10 – 10 m +ve charged matter spread uniformly over the entire regionRutherford Scattering Experiment (1910)Thin gold foil ~10 –7mPb shield < 100 counter RadiumPrediction ( Base on Thomson's model )1/F+ ~ F- 2/1 By C K CheungF+ ~ F+’ F- ~ F-‘Hence, no matter where the  particles hit, they should be deflected only by a small angle. Experimental results: ..\powerpoint\atomic structure\Rutherford's expt..pptThin gold foil ~10 –7mPb - < 100 counter Radium very few ( 1 in 8000 )1/ many ( > 99.5 % ) of the  particles go straight through the foil and were deflected through very small angles ( < 100 ).2/ very few ( ~ 1 in 8000 ) suffered deflections of more than 900.Note: 1. gold is ductile 2. thin gold foil to avoid multiple collisions inside the foil.2 By C K Cheung Rutherford’s model of atom1/ Due to large-angle scattering of very few  particles  a small massive +ve core or nucleus.2/ The –ve charge at relatively large distances from the nucleus so that their negative charge did not act as a shield to the +ve nucleus charge when the  particles penetrate the atom.3/ The –ve charge in circular motion to balance the electrostatic attraction.Rutherford’s model of atoms (1911 )3 By C K Cheung Rutherford’s estimate of nucleus size r b Z u v = 0 R KEPEZ = atomic no. ..\powerpoint\atomic structure\size of atom.ppt = no. of proton inside nucleus  KE = PE1 2 1 (2e)(Ze)  mu  2 40 rZe2  r = 2 0muFor closest approach: r = bZe2  b = 2 0muTake : Z = 79 e = 1.6x10 – 19 - 12 0= 8.86x10 – 27 m=6.7x 10 u = 2x10 7 ms - 1 b ~ 10 –14 m b R ~ ~ 10 – 15 m 24 By C K CheungLimitations of Rutherford’s model1/By classical theories, all accelerating charged particles will emit radiation ( energy ). If Rutherford’s model is correct, the accelerating electrons will lose energy continuously, then they will be adhered to the nucleus and all atoms will finally be collapsed!2/ Rutherford’s model does not predict the existence of discrete energy levels in atoms.3/ Rutherford’s scattering experiment founded the high energy physics.5 By C K CheungBohr’s Theory of atoms (1913)Consider an H atom: e p r1 (e)(e) mv 2  2 = 4 0 r r e2  v2 = ………………(1) 4 0mrFirst Theory: Only those orbits occur for which the angular momenta of the planetary electron are integral h h multiples of or n( ) 2 2 h  mvr = n( ) 2 h  v = n( )…………………(2) 2mr(2)  (1) n 2 h 2  e2  2 2 2  = 4 m r  4 0mr h2  r = 0 (n2 )  n2 me2 r is in discrete values.6 By C K CheungFor H-atom, r = 0.529( n2) x 10 – 10 mIf n = 1,  r1 = 0.529 AIf n = 2,  r2 = 2.11 A = 4 r1Also, the energy of the system:1 (e)(e) PE = 4 0 r1 1 e2 KE = mv2  m( ) 2 2 40mr e2 Total energy = E = PE + KE = - 80r me4 1  E = - 2 2 2  - ( 2 ) 8 0 h n n e.g.For max. E  n = ______For min. E  n = ______7 By C K CheungConsider: n5 E5 n4 E4 n3 E3 n2 E2 n1 E11 1 3 E12  - (  ) = 22 12 4 5 E23  36 E12 > E23 > …………….As n   E  0  continuousHence:EFormula:1 En = - 13.6 ( ) eV n28 By C K Cheung1 eV = ______J Second TheoryNo electron radiates energy so long as it remains in one of the orbital energy states, and that radiation occurs only when an electron goes from a higher energy state to a lower one, the energy of the quantum of radiation = hf. n2E = En2 – En1 = hf n1 me4 1 1 c (  )  hf  2 2 2 2 = h( ) 8 0 h n1 n2 1 me4 1 1  wave number  ( )(  )  2 3 2 2  8 0 h c n1 n2Rydberg constant R, R=1.0973x107m-19 By C K Cheung e.g. The diagram shown gives some levels of an atom. In the unexcited state the levels above - 10.4 eV are unoccupied. ( Given: h = 6.6x10 - 34 Js, C = 3x108 ms-1, e = 1.6x10 -19C)0 eV- 1.6eV- 3.7eV energy in eV- 5.5eV- 10.4eV a/ What is the ionization energy of the atom?Ionization energy = 0 – (-10.4) = 10.4 eV b/ What change is taking place if radiation of wavelength 141 nm is emitted?Ans.: - 1.6eV to - 10.4 eV10 By C K Cheung e.g. a/ Find the energy required to excite a hydrogen atom from the ground state to the n = 4 state. 1 1 E = 13.6 (  )eV 12 4 2 1 1 = 13.6 (  )x1.6x10 19 C 12 4 2 = 2.04x10-18 J b/ How many different possibilities of spectral line emission are there for the atom when the electron goes from n = 4 to the ground state ?Note: n=4 n=3  possible number of spectral n=2 lines = 6 n=111 By C K Cheung Evidence of energy levels1/ Optical line spectra (optical ~ visible region) A particular line ( or a set of lines ) in an optical emission spectrum indicates the presence of a particular frequency emitted in a transition of e ( directly or in stages ) between energy levels2/ X-ray spectraPhotoelectric Effect Source of variable & known frequency ( f ) vacuum A Emitter collectorVResults of Experiment ( Laws of Photoelectric Emission )1/ When the incident light is monochromatic, the number of photoelectrons emitted per second ( current I ) is proportional to the light intensity ( I’). Such an emission occurs effectively instantaneously. I/ A V, f = constantI’/Wm -2Wave theory: the incident light energy is uniformly distributed amongst the free electrons in the emitter, we should predict time delays of ~ 103 seconds.12 By C K Cheung2/ The KE of the emitted electrons varies from 0 to a maximum value. This definite maximum depends only on the frequency of the light, and not on its intensity. ..\powerpoint\atomic structure\photoelectric effect1.pptI /A f = constant2I’I’V/V Vs 0Vs: stopping potential 1 mv 2  eV 2 max sWave theory: we feel hotter in sunlight than that in moonlight,  K.E. of the emitted e should depend on light intensity.13 By C K Cheung 3/ Electrons are not emitted when the light has a frequency lower than a certain threshold value fo. The value of fo varies from metal to metal.I/A I’ = constant A B f/Hz fo fo’Wave theory: cannot predict the existence of fo!ExplanationPlanck’s Theory (1900) When radiation was emitted or absorbed, the emitting oscillator always showed a discrete sudden change of E, where E = hf.Einstein’s Theory (1905) Extended Planck’s original idea by suggesting that E.M. wave could exhibit particle ( called photon ) behavior1/ Electron emission is the result of a direct collision between an electron and a photon, so there is no time delay before emission starts ..\powerpoint\atomic structure\photoelectric effect2.ppt14 By C K Cheung 2/ Light intensity   number of photon arriving per m2 per second  but not energy of individual photon .I’ , f2I’ , f3/ fo depends on the material because each material requires a certain minimum energy ( called work function , W ) to free an electron.W = hfo15 By C K Cheung Formula: hf1 mv 2 2W1  hf = W + mv 2 2 1 hf = hfo + mv 2  2  hf = hfo + eVsThis formula was verified by Millikan in 1916Vs/V h slope = e f/Hz fo’16

Atomic Physics

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