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Λc Is the Compton Wavelength, = 2.4263102389 X 10 M Where H Is

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Λc Is the Compton Wavelength, = 2.4263102389 X 10 M Where H Is Compton effect Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: January 13, 2012) 1. Overview By the early 20th century, research into the interaction of X-rays with matter was well underway. It was known that when a beam of X-rays is directed at an atom, an electron is ejected and is scattered through an angle θ. Classical electromagnetism predicts that the wavelength of scattered rays should be equal to the initial wavelength;[3] however, multiple experiments found that the wavelength of the scattered rays was greater than the initial wavelength. In 1923, Compton published a paper in the Physical Review explaining the phenomenon. Using the notion of quantized radiation and the dynamics of special relativity, Compton derived the relationship between the shift in wavelength and the scattering angle: h ' (1 cos ) mc c is the Compton wavelength, h = 2.4263102389 x 10-12 m c mc where h is the Planck's constant, m is the mass of electron, and c is the velocity of light. 1 g' l' q g l e f pe' e' 2. Compton scattering A photon γ with wavelength λ is directed at an electron e in an atom, which is at rest. The collision causes the electron to recoil, and a new photon γ′ with wavelength λ′ emerges at angle θ. Let e′ denote the electron after the collision. From the energy conservation, we have E E e E ' E e' Compton postulated that photons carry momentum; thus from the conservation of momentum, the momenta of the particles should be related by p p ' pe' assuming the initial momentum of the electron is zero. The photon energies are related to the frequencies by hc hc E h , E h ' where h is the Planck constant, is the frequency of the incident photon and is the wavelength From the relativistic energy-momentum relation, the electron energies are 2 2 4 2 2 Ee mc , Ee' m c c pe' 2 Along with the conservation of energy, these relations imply that 2 2 4 2 2 h mc h ' m c c pe' or 2 2 2 4 2 2 (h mc h ') m c c pe' or 2 2 2 2 2 4 c pe' (h mc h ') m c . (1) _________________________________________________________________________ From the conservation of momentum, pe' p p ' The x component: h h p cos cos e' ' The y component: h p sin sin 0 e' ' 3. Derivation of formula Then by making use of the scalar product, we have 2 2 2 pe' (p p ' ) (p p ' ) p p ' 2 p p ' cos or p 2c2 p 2c2 p 2c2 2c2 p p cos e' ' ' (2) (h )2 (h ')2 2(h )(h ')cos From Eq. (1) and Eq. (2), 3 (h mc2 h ')2 m2c4 (h )2 (h ')2 2(h )(h ')cos or mc2h mc2h ' (h )(h ')(1 cos ) or ( ') 1 1 h (1 cos ) ' ' mc2 Then we get ' c (1 cos ) h where is the Compton wavelength. The CODATA 2006 value for the Compton c mc wavelength of the electron is h = 2.4263102175±33×10-12 m = 2.4263102175±33×10-2 A c mc Note that the scattered photon always has a long wavelength than the incident photon. Dl nm 0.004 l0 =0.00247968 nm 0.003 0.002 0.001 q p p 3 p 0 p 4 2 4 4 ((Note)) The momentum conservation along the y direction h sin p sin ' e' DEe keV 300 250 l0 =0.00247968 nm 200 150 100 50 q p p 3 p 0 p 4 4 2 DEp keV 500 400 l0 =0.00247968 nm 300 200 100 q p p 3 p 0 p 4 2 4 ___________________________________________________________________ 5 DEe keV 1.2 l0 =0.071073 nm 1.0 0.8 0.6 0.4 0.2 q p p 3 p 0 p 4 2 4 DEp keV 17.4 17.2 17.0 l0 =0.071073 nm 16.8 16.6 16.4 16.2 q p p 3 p 0 p 4 2 4 4. Determination of angles and We start with the momentum conservation law, h h p cos cos , (3) e' ' h p sin sin , (4) e' ' Dividing Eq.(4) by Eq.(3); 6 h sin tan ' h h cos ' Using the relation ' c (1 cos ) we get h sin sin tan ' h h cos ' cos ' sin (1 cos ) c (1 cos ) sin ( c )(1 cos ) 2 sin cos 2 2 2( )sin 2 c 2 cos 2 ( )sin ( )tan c 2 c 2 The recoil angle of electron is cot (1 c )tan 2 5. Energy of scattered electron h mc2 h ' m2c4 c2 p 2 e' 7 2 4 2 2 E' m c c pe' h( ') mc2 ' hc mc2 ' hc mc2 ' Then the kinetic energy of the scattered electron is K' E'mc2 ' hc ' hc ' 8 .
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