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Compton effect

Compton effect. Another experiment revealing the particle nature of X-ray (radiation, with wavelength ~ 10 -10 nm). Compton, Arthur Holly (1892-1962), American physicist and Nobel laureate whose studies of X rays led to his discovery in 1922 of the so-called Compton effect.

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Compton effect

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  1. Compton effect • Another experiment revealing the particle nature of X-ray (radiation, with wavelength ~ 10-10 nm) Compton, Arthur Holly (1892-1962), American physicist and Nobel laureate whose studies of X rays led to his discovery in 1922 of the so-called Compton effect. The Compton effect is the change in wavelength of high energy electromagnetic radiation when it scatters off electrons. The discovery of the Compton effect confirmed that electromagnetic radiation has both wave and particle properties, a central principle of quantum theory.

  2. Although initially the incident beam consists of only a single well-defined wavelength (l) the scattered x-rays have intensity peaks at two wavelength (l’ in addition), where l’ > l. Wavelength shift Beam of x-ray with sharp wavelength l falls on graphite target. For various angle q the scattered x-ray is measured as a function of their wavelength Unexplained by classical wave theory for radiation No shift of wavelength is predicted in wave theory of light

  3. Compton (and independently by Debye) explain this in terms of collision between collections of (particle-like) photon, each with energy E = hn, with the free electrons in the target graphite (imagine billard balls collision)

  4. Scattered photon, E’=hc/l’, p’=h/l’ q f 1: Conservation of E: hc/l + mec2= hc/l’ + Ee Initial photon, E=hc/l, p=h/l Initial electron, at rest, Eei=mec2, pei=0 y x Scattered electron, Ee,pe 2: Conservation of momentum:p = p’ + pe (vector sum)

  5. p’sinq = pesinf p = p’cosq + pecosf Conservation of momentum in 2-D p = p’ + pe (vector sum) actually comprised of two equation for both conservation of momentum in x- and y- directions Conservation of l.mom in y-direction Conservation of l.mom in x-direction

  6. Some algebra… Mom conservation in y: p’sinq = pesinf (PY) Mom conservation in y: pecosf= p - p’cosq (PX) Conservation of total relativistic energy: E + mec2 = E’ + Ee (RE) (PY)2+ (PX)2, substitute into (RE)2 to eliminate Ee and pe: (E + mec2 - E’)2 = c2(p2 – 2pp’cosq+p’2)+ mec2, leading to   Dl≡ l’- l = (h/mec)(1 - cosq)

  7. Compton wavelength • le = h/mec = 0.0243 Angstrom, is the Compton wavelength (for electron) • Note that the wavelength of the x-ray used in the scattering is of the similar length scale to the Compton wavelength of electron • The Compton scattering experiment can now be perfectly explained by the Compton shift relationship Dl≡ l’- l= le(1 - cosq) as a function of the photon scattered angle • Be reminded that the relationship is derived by assuming light behave like particle (photon)

  8. For q = 00 “grazing” collision =>Dl = 0 l  l’ l l l’= l+Dlmax Dl≡ l’- l= (h/mec)(1 - cosq) Notice that Dl depend on q only, not on the incident wavelength, l. Consider some limiting behaviour of the Compton shift: q0 • For q = 1800 “head on” collision • Dl= Dlmax = 2 lc =2( 0.00243nm) (photon being reversed in direction), Dl is maximum Dlmax 180o l’=0.1795 nm

  9. Example • X-rays of wavelength 0.2400 nm are Compton scattered and the scattered beam is observed at an angle of 60 degree relative to the incident beam. • Find (a) the wave length of the scattered x-rays, (b) the energy of the scattered x-ray photons, (c) the kinetic energy of the scattered electrons, and (d) the direction of travel of the scattered electrons

  10. solution • l’= l + lc (1 - cosq) = 0.2400 nm + 0.00243 nm (1– cos 60o) = 0.2412 nm • E’ = hc/l’ • = 1240 eV∙nm /0.2412 nm • = 5141 eV

  11. p’g pg E’g< Eg q me Initial photon f Eg K pe kinetic energy gained by the scattered electron = energy transferred by the incident photon during the scattering: K = hc/l - hc/l’ = (5167 – 5141)eV = 26 eV Note that we ignore SR effect here because K << rest mass of electron, me = 0.5 MeV

  12. p’g pg E’g< Eg q me Initial photon f Eg K pe By conservation of momentum in the x- and y- direction: pg= p’g cosq + pecosf; p’g sinq = pesin f; tan f = pesin f / pecosf = (p’g sinq)/ (pg - p’g cosq) = (E’g sinq)/ (Eg - E’g cosq) = (5141 sin 600 / [5167-5141 (cos 600] = 0.43 = 1.71 Hence, f = 59.7 degree

  13. Rayleigh scattering and Compton scattering • There are two kinds of electrons in the graphite target: free electron and electron that is bounded to individual atom • The sifted peak corresponds to scattering of stationary free electron with incident photon (for Ke >> biding energy). This is just the Compton effect The unshifted peak corresponds to collision of photon with strongly bounded electron which is not free (incident photon energy << biding energy)

  14. Photon scattering off a bounded electron sees the mass of the whole atom instead of only that of a free electron. Hence the Compton shift, instead being given by Dle= (h/mec)(1 - cosq) the mass term is now give by me Matom (Matom >> me) instead, such that Dl (due to Matom) <<Dl(due to me). Hence, the scattered l’appears unmodified This is called Rayleigh scattering - explainable in classical theory

  15. mass seen by photon = me Compton scattering: le l Dl1= (h/mec)(1 - cosq) le mass seen by photon = Matom Rayleigh scattering: l Dl2= (h/Matomc)(1 - cosq) Dl1<< Dl2because me << Matom (or equivalently, l ~ le >> latom

  16. In the limit l infinity (low incident photon energy), l >> lc, Rayleigh scattering dominates (The effective mass is large as seen by the incident photons) • As llc Compton scattering dominates • Note that Compton scattering occurs at energy scale which is much higher, ~ 0.1 MeV (typical energy of the x-ray photon, with the wavelength ~ Compton wavelength of the electron • In comparison, photoelectricity occurs at a much lower energy scale, ~ eV. • Different mode of interaction happen at different energy scale (or length scale)

  17. Ref: A general reminder 1) The next tutorial will happen in 26 Dec 2003, friday, 5.00 pm. Please make sure to pass up the solution to your tutors latest by 12.00 pm on that day (Friday). You can check out the course schedule for the rest of the semester in http://www.fizik.usm.my/tlyoon/teaching/calander.htm

  18. 2) Students in the tutorial group by the tutor Mr. Sia Chen How (Kumpulan 12), please collect back your tutorial solutions from the box outside his room (room no. 256).3) I have instructed all tutors to provide the password to the tutorial 1 full solution, which u can download from our zct 104 course web page. If you dunt have the password, pls obtain it from your frens or contact me via e-mail.

  19. 4) Please check that your SR lecture notes (in pdf. format) is the latest version. I discover that the version of notes some students has is an old one. It is advised that you should download the latest version (updated on  2 Dec 2003).5) I strongly encourage students to discuss with me any problem they face with the course (e.g. poor english, problem to obtain lecture notes, difficulties in learning or comprehending the course material etc.). This, of course, does not include your love affair problem  lah.=b You may discuss with me in person after lectures or in my room (which i prefer); call me up or write to me.  Please step forward and dont remain silent. I am always willing to listen and to help.

  20. 4) Please take some time to read the extra material I have put up in the course web page, e.g. • Extra reading material • Michelson-Morley experiment: http://www.phys.virginia.edu/classes/109N/lectures/michelson.html • Ether • Relativity • Time • photon • photoelectric effect

  21. Even God could not help those who don’t help themselves

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