By C K Cheung Atomic Physics Matter ----- composed of atoms and molecules. Atom ----- composed of ?????? J.J. Thomson Experiment (1897 ) Thomson’s ‘ plum – pudding ‘model of atoms: electron +ve charged matter spread uniformly over the entire region 10 – 10 m Rutherford Scattering Experiment (1910) Thin gold foil ~10 –7m Pb shield < 100 counter Radium Prediction ( Base on Thomson's model ) 1/ F+ ~ F- 1 By C K Cheung 2/ F+ ~ 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..ppt Thin gold foil ~10 –7m Pb Radium - < 100 counter 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 atom 1/ 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 u v=0 Z R KEPE ..\powerpoint\atomic structure\size of atom.ppt Z = atomic no. = no. of proton inside nucleus KE = PE 1 1 (2e)( Ze) mu 2 2 40 r r= Ze 2 0 mu 2 For closest approach: r = b b= Ze 2 0 mu 2 Take : Z = 79 e = 1.6x10 – 19 0= 8.86x10 - 12 m=6.7x 10 – 27 u = 2x10 7 ms - 1 b ~ 10 –14 m R~ b ~ 10 – 15 m 2 4 By C K Cheung Limitations of Rutherford’s model 1/ 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 Cheung Bohr’s Theory of atoms (1913) Consider an H atom: e p r mv 2 (e)( e) = 40 r 2 r 1 v2 = e2 40 mr ………………(1) First 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) n2h2 e2 2 2 2 = 40 mr 4 m r 0h2 2 (n ) r= me2 n2 r is in discrete values. 6 By C K Cheung For H-atom, r = 0.529( n2) x 10 – 10 m If n = 1, r1 = 0.529 A If n = 2, r2 = 2.11 A = 4 r1 Also, the energy of the system: PE = (e)( e) 40 r KE = 1 2 1 e2 mv m( ) 2 2 40 mr 1 Total energy = E = PE + KE = - E= - me4 2 8 0 h 2 n 2 e2 80 r -( 1 n2 ) e.g. For max. E n = _____________ For min. E n = _____________ 7 By C K Cheung Consider: n5 E5 n4 E4 n3 E3 n2 E2 n1 E1 E12 - ( 1 2 1 2 )= 3 4 2 1 5 E23 36 E12 > E23 > ……………. As n E 0 continuous Hence: E Formula: 8 By C K Cheung En = - 13.6 ( 1 ) n2 eV 1 eV = ________________J Second Theory No 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. n2 E = En2 – En1 = hf n1 c me4 1 1 ( 2 2 ) hf = h( ) 2 2 8 0 h n1 n2 me4 1 1 wave number ( 2 3 )( 2 2 ) 8 0 h c n1 n2 1 Rydberg constant R, R=1.0973x107m-1 9 By C K Cheung e.g. The diagram shown gives some levels of an atom. In the unexcited state the levels above eV are unoccupied. ( Given: h = 6.6x10 - 34 Js, C = 3x108 ms-1, e = 1.6x10 -19C) - 10.4 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 eV 10 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. E = 13.6 ( 1 1 )eV 12 4 2 1 1 2 ) x1.6 x10 19 C 2 1 4 -18 = 2.04x10 J = 13.6 ( 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 n=2 possible number of spectral lines = 6 n=1 11 By C K Cheung Evidence of energy levels 1/ 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 levels 2/ X-ray spectra Photoelectric Effect Source of variable & known frequency (f) vacuum A Emitter collector V Results 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 = constant I’/Wm -2 Wave theory: the incident light energy is uniformly distributed amongst the free 12 By C K Cheung electrons in the emitter, we should predict time delays of ~ 103 seconds. 2/ The KE of the emitted electrons varies from 0 to a maximum value. This definite depends only on the frequency of the light, and not on its intensity. ..\powerpoint\atomic structure\photoelectric effect1.ppt maximum I /A f = constant 2I’ I’ V/V Vs 0 Vs: stopping potential 1 2 mvmax eVs 2 Wave 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 The value of fo varies from metal to metal. threshold value fo. I/A I’ = constant A fo B f/Hz fo’ Wave theory: cannot predict the existence of fo! Explanation Planck’s Theory (1900) When radiation was emitted or absorbed, the emitting oscillator always showed discrete sudden change of E, where E = hf. a Einstein’s Theory (1905) Extended Planck’s original idea by suggesting that E.M. wave could exhibit particle ( called photon ) behavior 1/ Electron emission is the result of a direct collision between an electron and a photon, is no time delay before emission starts ..\powerpoint\atomic structure\photoelectric effect2.ppt so there 14 2/ By C K Cheung Light intensity number of photon arriving per m per second but not energy of individual photon . 2 I’ , f 2I’ , f 3/ fo depends on the material because each material requires a certain minimum energy work function , W ) to free an electron. ( called W = hfo 15 By C K Cheung Formula: hf 1 mv 2 2 W 1 mv 2 2 hf = W + hf = hfo + 1 mv 2 2 hf = hfo + eVs This formula was verified by Millikan in 1916 Vs/V slope = fo’ h e f/Hz 16