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Chapter2: Compton Effect Professor Mohammad Sajjad Alam University at Albany September 28, 2010 Adapted from Web Adapted from the Web.

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Presentation on theme: "Chapter2: Compton Effect Professor Mohammad Sajjad Alam University at Albany September 28, 2010 Adapted from Web Adapted from the Web."— Presentation transcript:

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2 Chapter2: Compton Effect Professor Mohammad Sajjad Alam University at Albany September 28, 2010 Adapted from Web Adapted from the Web

3 2 Compton effect Another experiment revealing the particle nature of X-ray (radiation, with wavelength ~ nm) Compton, Arthur Holly ( ), American physicist and Nobel laureate whose studies of X rays led to his discovery in 1922 of the so-called Compton effect.X rays 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.electromagnetic radiationelectronsquantum theory

4 3 Compton’s experimental setup A beam of x rays of wavelength 71.1 pm is directed onto a carbon target T. The x rays scattered from the target are observed at various angle  to the direction of the incident beam. The detector measures both the intensity of the scattered x rays and their wavelength 

5 4 Experimental data Although initially the incident beam consists of only a single well-defined wavelength ( ) the scattered x-rays at a given angle  have intensity peaks at two wavelength ( ’ in addition), where ‘>      

6 5 Compton shouldn’t shift, according to classical wave theory of light Unexplained by classical wave theory for radiation No shift of wavelength is predicted in wave theory of light

7 6 Modelling Compton shift as “particle- particle” collision Compton (and independently by Debye) explain this in terms of collision between collections of (particle-like) photon, each with energy E = h  pc, with the free electrons in the target graphite (imagine billard balls collision) E 2 =(mc 2 ) 2 +c 2 p 2 E  2 =(m  c 2 ) 2 +c 2 p 2 =c 2 p 2

8 7 Part of a bubble chamber picture (Fermilab'15 foot Bubble Chamber', found at the University of Birmingham). An electron was knocked out of an atom by a high energy photon.

9 8   Scattered photon, E ’ =hc/ ’, p ’ =h/ ’ Scattered electron, E e,p e Initial photon, E=hc/, p=h/ Initial electron, at rest, E ei =m e c 2, p ei =0 2: Conservation of momentum:p = p’ + p e (vector sum) 1: Conservation of E: cp + m e c 2 = cp ’ + E e y x

10 9 p = p ’ + p e (vector sum) actually comprised of two equation for both conservation of momentum in x- and y- directions p’sin  = p e sin  Conservation of l.mom in y- direction Conservation of l.mom in x-direction p = p’cos  + p e co s  Conservation of momentum in 2-D

11 10 Some algebra … Mom conservation in y : p’sin  = p e sin  (PY) Mom conservation in x : p - p’ cos  = p e cos  (PX) Conservation of total relativistic energy: cp + m e c 2 = cp’ + E e (RE) (PY) 2 + (PX) 2, substitute into (RE) 2 to eliminate , p e and E e (and using E e 2 = c 2 p e 2 + m e 2 c 4 ):  ≡ ’- = (h/m e c)(1 – cos  )

12 11 Compton wavelength  e = h/m e c = 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  ≡ ’  = e (1 - cos  ) as a function of the photon scattered angle Be reminded that the relationship is derived by assuming light behave like particle (photon)

13 12 X-ray scattering from an electron (Compton scattering): classical versus quantum picture

14 13  ≡ ’  = (h/m e c)(1 - cos  ) Notice that  depend on  only, not on the incident wavelength,.. For  = 0 0  “grazing” collision =>  = 0  Consider some limiting behaviour of the Compton shift:   ’ ’= nm

15 14  =180 o For   “head-on” collision =>  =  max   photon being reversed in direction  max = max ’  = (h/m e c)(1 – cos 180  ) = 2 e =2( nm) initially  After collision ’ max  +  max


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