4.德布罗意波、波粒二象性 De Broglie wave Wave-Particle Dualism The dual nature of light, particle and wave 光的波粒二象性 particle wave Cenergy) E v( frequency) (momentum)P h n (wavelength) These two natures are connected by h Eb=hv…() 2 E E+(P) hy h P=E/C (2)
The dual nature of light, particle and wave 光的波粒二象性 particle wave （energy） （frequency） （momentum） （wavelength） E P h These two natures are connected by h. (1) E = h / (2) h C h P = E C = = 2 2 0 2 E = E + (P c) 4.德布罗意波、波粒二象性 （De Br glie wave .Wave ö -Particle Dualism ）
De broglie assumed that the wavelength of predicted matter waves was given by the same relationship that held for light namely, or 2=h/p 德布罗意,L.V E ttt t wave tlit Y H particl which connects the wavelength of a light wave with the momentum of the associated photons De broglie wave
De Broglie assumed that the wavelength of predicted matter waves was given by the same relationship that held for light namely, or λ= h / P u Z X Y E H particle wave which connects the wavelength of a light wave with the momentum of the De Broglie wave associated photons
De broglie predicted that the wavelength of matter waves would also be given by 2=h/p where p would now be the momentum of the particle of matter. De broglie wave Example l what wavelength is predicted by De broglie wave for a bullet whose mass is mo=0.05kg and velocity is v=300m/s? Solve°V<<C h663×10-34 44×103(m)=44×1024(A) mn0.05×300
De Broglie predicted that the wavelength of matter waves would also be given by λ= h / P where P would now be the momentum of the particle of matter. De Broglie wave Example 1. What wavelength is predicted by De Broglie wave for a bullet whose mass is m0=0.05kg and velocity is v=300m/s? Solve, ∵V << C 4.4 10 ( ) 0.05 300 6.63 10 3 5 3 4 0 m m V h − − = = = = 4.410-24(Å)
Example 1. What wavelength is predicted by De broglie wave for a bullet whose mass is mo=0.05kg and velocity is v=300m/s? olve V<<C h663×10-34 44×10-3(m)即44×1024A 0.05×300 Exp 2. What wavelength is predicted by De broglie wave for a beam of electrons whose kinetic energy is eu and velocity n isv? (V<<C) B From the law of KE mo y conservation of energy 发射电 e -mv=el 子阴级U加速电极
Example 1. What wavelength is predicted by De Broglie wave for a bullet whose mass is m0=0.05kg and velocity is v=300m/s? Solve, ∵V << C 4.4 10 ( ) 0.05 300 6.63 10 3 5 3 4 0 m m V h − − = = = 即4.410-24Å U B K 发射电 子阴级 加速电极 e m0 V Exp 2. What wavelength is predicted by De Broglie wave for a beam of electrons whose kinetic energy is eU and velocity is v ? （VC） m V = eU 2 0 2 1 From the law of conservation of energy
B 加速电极 mo y QeL KE 发射电 e 子阴级U De broglie wavelength follows from h 12.3° P my v2emU VU 12.3 12.3 or|2===×10(m)or A 12.3 When u=100,元 √U 12.3(A)
U B K 发射电 子阴级 加 速 电 极 0 2 m eU V = P h = em U h 2 0 = = A U 12.3 m V h 0 = e m0 V or 10 ( ) 12.3 1 0 m U − = or U 12.3 = Å When U=100(V), 12.3 12.3 = = U (Å) m V = eU 2 0 2 1 De Broglie wavelength follows from
Demonstration of De broglie wave 1)C J. Davisson and L. h Germer Test 戴维逊-革末实验 Figure shows the apparatus of Davisson and germer 实验装置: B 加速电极 G q K 发射电 O不N单晶 电流计 子阴级U
G Ni单晶 电 流 计 Demonstration of De Broglie wave 1) C. J. Davisson and L. H. Germer Test 戴维逊--革末实验 Figure shows the apparatus of Davisson and Germer. 实验装置： U M I B K 发射电 子阴级 加 速 电 极
1)C J Davisson and L H Germer Test 实验装置 B 加速电极 G q K 发射电 MN单晶 电流计 子阴级U N单晶 =50° √U a=0.215nmd=0.0908nm U=54(V)
1）C. J. Davisson and L. H. Germer Test 实验装置： U I a=0.215nm d=0.0908nm U=54(v) U G Ni单晶 电 流 计 M I B K 发射电 子阴级 加 速 电 极 Ni单晶 d a φ=50°
实验装置: B 加速电极 1 suppose electrons K多 电 are waves 发射电 MN单晶 流 子阴级 计 De broglie wavelength of electron waves would be given by h 12.3° Experi 4…( g mental value QemU gIven by From Bragg's law 2d sin =kn ki p=500. h 2dsin =k (10 This is emu excellent agreement 9=51°
suppose De Broglie wavelength of electron waves would be given by (9) 12.3 2 0 = = A em U U h From Bragg’s law 2d sin = k (10) 2 2 sin 0 em U h d = k k =123. φ=51o 实验装置： G Ni单晶 电 流 U 计 I B K 发射电 子阴级 加 速 电 极 electrons are waves. Experimental value is given by φ=50o . This is excellent agreement. M
2)G P. Thomson electron diffraction test 1927年汤姆逊(GP· Thomson)以600伏慢电子 0=0.5A)射向铝箔,也得到了像X射线衍射一样的衍 射,再次发现了电子的波动性。 1937年戴维逊与GP汤姆逊共获当年诺贝尔奖 GP· Thomson为电子发现人 J.J. Thomson的儿子 尔后人们又发现了质子、中子的衍射 (b 图 劳厄相 a)x射线在单晶上形成的衍射图样 b)中子在单晶上形成的衍射图样
2) G.P. Thomson electron diffraction test 1927 年汤姆逊（G·P·Thomson）以600伏慢电子 (=0.5Å)射向铝箔，也得到了像X射线衍射一样的衍 射，再次发现了电子的波动性。 1937年戴维逊与GP汤姆逊共获当年诺贝尔奖. G·P·Thomson为电子发现人J·J·Thmson的儿子. 尔后人们又发现了质子、中子的衍射
(a) (b) 德拜相a波长为071nm的x射线通过铝箔 所得到的衍射环 b波长为005mm的电子束通过铝 箔所得到的衍射环 c中子通过铜箔所得到的衍射环