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Quantum Physics. Why do the stars shine?why do the elements exhibit the order that’s so apparent in the periodic table? How do transistors and microelectronic.

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Presentation on theme: "Quantum Physics. Why do the stars shine?why do the elements exhibit the order that’s so apparent in the periodic table? How do transistors and microelectronic."— Presentation transcript:

1 Quantum Physics




5 Why do the stars shine?why do the elements exhibit the order that’s so apparent in the periodic table? How do transistors and microelectronic devices work?why does copper conduct electricity but glass doesn’t?

6 Quantum Physics Particle Theory of Light Particle Theory of Light De Broglie Hypothesis Photo- electric effect Photo- electric effect Compoton effect Compoton effect Hydrogen spectrum Hydrogen spectrum Uncertainty Wave Function Wave Function Schrodinger equation and application

7 Quantum Blackbody Radiation Planck’s hypothesis The photoelectric effect The particle theory or light X-rays Diffraction Photons Electromagnetic Key words

8 The wave properties of particles The uncertainty principle The scanning tunneling microscope Atomic Spectra The Bohr Theory of Hydrogen Key words

9 27.1 Blackbody Radiation and Planck’s Hypothesis Classical theory Experimental data Wavelength Intensity Ultraviolet Catastrophe

10 Planck’s Hypothesis In 1900 Planck developed a formula for blackbody radiation that was in complete agreement with experiments at all wavelengths. Planck hypothesized that black body radiation was produced by submicroscopic charged oscillators, which he called resonators. The resonators were allowed to have only certain discrete energies, E n, given by

11 27.2 The Photoelectric Effect and The Particle Theory of Light 1) saturated current is proportion to intensity of light is fixed,I is proportion to voltage,but there’s a saturated current which is proportion to intensity of light A V pow - I K A -V a V I o IsIs I l higher I l loweer 1.The Photoelectric Effect

12 inverse voltage is applied , I c =0. V a is called cutoff voltage or stopping voltage A V pow - I K A -V a V I o IsIs I l higher I l loweer

13 2) initial kinetic energy is proportion to the frequency of incident light,has no relation with the intensity of light 3) there’s a cutoff frequency (threshold frequency) to a metal,only when > o, there’s a current 4) photo current produce immediately,delay time is not more than 10 -9 s 。

14 2.Einstain’s theory 1)the hypothesis of Einstian Light has particle nature Intensity of light 2) Einstian equation A work function

15 While < A/h 时, no photoelectric effect c. instantaneously effect Cutoff frequency b. Notes: a. the initial kinetic energy is linearly proportion to the frequency of incident light

16 3.the wave-particle duality Mass: 2) the energy, momentum and mass 1) Light has dual nature energy momentum M 0 =0

17 1.oq 2.op 3. op/oq 4. qs/os Example :the experiment result is as follows,find h p o q s eU a  solution : 3

18 Example: the cutoff o =6500Å, the light with =4000Å incident on the metal (1)the velocity of photoelectrons? (2)stopping voltage? Solution:  =6.5×10 5 (m/s) (2) c = : V a =1.19 (V) h= 6.63×10 -34

19 Example:incident frequency is fixed ; we’v experimental curve (solid line),and then with fixed intensity of light,increase the incident frequency , the experimental curve is as dot line,find the correct answer in the following graphs V I o (A) V I o (B) V I o (C) V I o (D)

20 27.3 The Compton Effect light go through medium and propagate in different direction 。 from classic view:the scatter wavelength is same with incident wavelength 。 but in graphite experiment,we found a change in wavelength,this is called compoton scatter 。 1.scattering 

21 2.principle suppose: photos collide with electrons without loss of energy   x y mm Energy conservation h o +m o c 2 = h +mc 2 Momentum conservation : x:x: y:y: c=

22  =0 , min = o ;  =180°, max = o +2 c Compton wavelength : Å   x y mm

23 1) photon pass some energy to electrons 2) compton effect is strong when the photon act on the atom with small number atom The explanation to compton effect

24 3) meaning : photon theory is correct show the duality nature of light 4) micro particle obey the conservative law 5) photoelectric effect, compton effect

25 Example: with o =0.014Å X ray in compton experiment, find maxmum kinetic energy in electrons ? solution:fron energy conservation in fact max = o +2 c ,  , E max =1.1×10 -13 J

26 Example: o =0.1Å X ray in experiment 。 In the direction of 90°,find wavelength? Kinetic energy and momentum of electron? solution:  =90° = o +  =0.1+0.024=0.124Å =3.8×10 -15 J

27 x:x: y:y:  =90° =8.5×10 -23 (SI)   x y mm From momentum conservation

28 example: o =0.03Å X ray in experiment, the velocity of recoil electron  =0.6c, find (1)the rate of the scattering energy of electron to its rest energy (2)the scattering =? And scattering  =? solution: (1)the scattering energy =0.25m o c 2 (2)  = 0.0434Å for so  =63.4°

29 1.the atomic hydrogen spectrum Rydberg constant 2.the empirical formula of Balmer 27.4 Line Spectra and Bohr Model

30 Baschen infrared Lyman series ultraviolet Balmer series visible The line spectrum system General form:

31 Physicist: Rutherford


33 3.Bohr model 1) stationary hypothesis: electrons can be in some certain stable orbits 3) quantization of angular momentum 2) quantum transition:  to hydrogen atom  hypothesis

34 4.ionization: Notes: 1.Ground state n=1 2.excited state n>1 3.n=2,the first excited state 5. explanation of hydrogen spectrum

35 n=4 n=3 n=2 n=1 r =a 1 r =4a 1 r =9a 1 r =16a 1 lymanBalmer Paschen

36 E n =  h c R/n 2  h c R/25  h c R/16  h c R/9=-1.51eV  h c R/4=-3.39eV  h c R=13.6eV 1 2 6  5 3 4 lyman Balmer Paschen Brackett T=R/n 2 109677cm -1 2741cm -1 12186cm -1 6855cm -1 4387cm -1 Energy level diagram

37 Example:find the energy for hydrogen atom giving longest wavelength in lyman series? 1)1.5ev 2)3.4ev 3)10.2ev 4)13.6ev solution:3 n=2-1

38 Example:with 913A violent light,hydrogen atom can be ionized,find the wavelength expression of lyman series solution : 4

39 Example:with visible light,can we ionized the first excited state of hydrogen atom? Solution: Needed enrgy no

40 Example: hydrogen atom in third excited state,find the number of its line after transition,name its series ? solution: -13.6 -3.4 -1.51 -0.85 1 2 3 4 lyman: 3 Balmer: 2 Pachen : 1

41 Physicist: De Broglie

42 Particle nature of light: chapter 42 Quantum Mechanics 27.5 Wave Nature of Particles Notes:1) 1. De Broglie wavelength

43 Notes:1) example : a bullet with m=0.01kg , v=300m/s h is so small,the wavelength is so small It’s difficulty to measure,behave in particle nature On the atomic scale,however,things are quite different M e =9.1*10 -31,v=10 6, =0.7nm This wavelength is of the same magnitude as interatomic spacing in matter,and in diffraction experiment the phenomena is evidence.

44 Standing wave 2) quantization condition of angular momentum of Bohr is the showing of de broglie wave

45 2. experiment diffraction by electrons slit,double slit diffraction ( Thomoson1927 )

46 M  Experiment Davision and Germer experiment G A   d

47 example: (1)the kinetic energy of electron E k =100eV ; (2)momentum of bullet p=6.63×10 6 kg.m.s - 1, find 。 solution:for the small kinetic energy,with classic formula =1.23Å (2)bullet h= 6.63×10 -34 = 1.0×10 -40 m conclusion:the wave nature is only obvious in microparticle,to macroparticle,you can’t detect the effect

48 Example:with 5×10 4 V accelerating voltage,find the solution:with relativity effect  =1.24×10 8 (m/s) =10×10 -31 (kg) =0.0535Å m o =9.11×10 -31 (kg)

49 Example : suppose , kinetic energy equal to its rest energy , Example:the first bohr radius a,electron move along n track , solution : solution : 2

50 27.6 Heisenberg’s Uncertainty Relation 1.uncertainty relation It states that measured values can’t be assigned to the position and momentum of a particle simultaneously with unlimited precision. Notes: 1) represent the intrinsic uncertainties in the measurement of the x components of and even with best measure instruments. 2) small size of planck’s constant guarantees uncertainty relation is important only in atomic scale

51 1)the result of dual nature 2)give the applied extent of classic mechanics , if h can be ignored,the uncertainty relation have no use. 3 ) the intrinsic of uncertainty is that there’s a uncertain action between observer and the object, It can’t be avoided. 2.the meaning of uncertainty


53 example:estimate  v x in hydrogen atom solution:  x=10 -10 m(the size of atom), From  x  p x  h, so big,velocity and coordinate can’t be determined at the same time 。

54 example: a bullet m=0.1kg,  x =10 -6 m/s , find  x 。 solution:  x  p x  h so small,we can consider it with classic view (to macro object)

55 example: =5000Å ,  =10 -3 Å , find  x solution: from  x  p x  h,

56 Physicist: De Broglie

57 1.Schrodinger equation 27.7 Wave Function, Schrodinger Equation 2.the statistic explanation of wave function wave view particle view Bright fringe:  (x,y,z,t)  2 big, possibility big; Dark fringe:  (x,y,z,t)  2 small, possibility small 。  (x,y,z,t)  2 is proportion to possibility density in this point.

58 x x s 2 s 1 p o D 2a r 2 r 1... K=0 K=1 K=2 The meaning of wave function:it’s related to the probability of finding the particle in various regions Character: single value, consecutive limited, be one in whole space

59 E=E k +E p =p 2 /2m+u non relativity form From the Solution of equation, E  can only take special value  2 express possibility density Quantum condition can be got in natural way,this related with atomic state Notes on Schrodinger equation

60 2 、 the application of Schrodinger equation on hydrogen atom 1) quantum energy 2) quantum of angular momentum 3) if in magnetic field Angular momentum quantum number

61 example : l=1, l=2, z L 0 z 0  Example:n=3,the possible value of L. and L z Solution 1) 3 2) 5

62 4) possibility distribution of electron Solution:from schudigder equation: Ψ nl (r, ,  ) =R nl (r)  l (  )Φ (  ) Possibility density: Ψ nl (r, ,  ) 2 for example: to 1S electron , possibility density

63 aoao r p 1s 图 20-6

64 1921 , (O.Stern) and (W.Gerlach) prove:except the orbital motion,there’s a spin z component of s Spin magnetic number Spin angular momentum 27.8 Electron Spin Four Quantum Number

65 z 0 

66 (1)principal quantum number : n=1,2,3,… 。 determine the energy of atom 。 (2)angular quantum number l=0,1,2,…,(n-1) 。 determine the angular momentum 。 (3)magnetic quantum number m l =0,±1,±2,…,±l 。 determine L z , that’s space quantum property 。 determine z component of spin angular momentum (4)spin magnetic quantum number in summary:the status of atom is determined by four quantum number 。

67 Physicist: Plank

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