The particle nature of earlier established light as wave only..

Compton experiment

INTRODUCTION 

Arthur Holly Compton (1892– 1962) was a professor at Washington University. Compton developed a theory of the intensity of X-ray reflection from crystals as a means of studying the arrangement of electrons and atoms. In 1918 he started a study of X-ray scattering. Working with X-rays, he perfected his apparatus to measure the shift of wavelength with scattering angle that is now known as the Compton effect.

    Compton experiment

    THE PHOTON MODEL’s CONFIRMATION

    Compton Effect or Compton Scattering is one of the forms of photon interaction ( x-ray and Gamma rays) with free electrons( not bounded or loosely bounded with atoms).

    Compton effect setup

    In this effect , the resultant incident photon is scattered and imparts energy to the recoiled electron. The scattered photon will have a different wavelength ( this is observed ) and energy .

    The Compton effect is a partial absorption process as the original photon has lost energy , known as Compton shift .

    Mathematical visualization of Compton effect

    Assumptions:

    1. This effect is the result of an interaction of an individual particle and free electron of target.
    2. Collison is relativistic and elastic.

    According to the conservation of energy :
    Total energy before collision  = Total energy after collision 
     $h\nu +m_{0}c^{2}$  =  $h\nu' +mc^{2}$

    According to the conservation of momentum :

    In horizontal direction :$\dfrac{h\nu }{c}=\dfrac{h\nu '}{c}\times \cos \phi +p\cos \theta$------(2)
    In vertical direction : $0=\dfrac{h\nu }{c}\times \sin \phi -p\sin \phi$ ------(3)
    Where p=mv
    By eq(2) & eq(3) :$m^{2}v^{2}c^{2}=h^{2}\left[ \nu +\nu' -2\nu \nu' \cos \phi \right]$----(4)
    By eq(1) :$m^{2}c^{4}=\left[ h\left( \nu -\nu \right) +m_{0}c^{2}\right] ^{2}$----(5)
    Now eq(4) - eq(5) gives
    $\dfrac{1}{\nu '}-\dfrac{1}{\nu }=\dfrac{h}{m_{0}c^{2}}\left( 1-\cos \phi \right)$
    Or 
    $\lambda' -\lambda =\dfrac{h}{m_{0}c}\times \left( 1-\cos \phi \right)$

    ⇒$\lambda '-\lambda =\lambda _{c}\left( 1-\cos \phi \right)$

    Where $\lambda _{c}=\cdot \dfrac{h}{m_{0}c}=0.0242A^{0}$is called Compton wavelength

    DETECTION OF RECOILED ELECTRON :

    When the photon interacts with a free electron , photon may give a portion of energy to it. Now, the electron can be ejected form atom , this is called recoiled electron.

    By equation (2) and (3) :

    tan Φ=$\left( \dfrac{\nu '\cdot \sin \phi }{\nu -\nu '\cdot \cos \phi }\right)$----(1)

    But $\nu'$ = $\left( \dfrac{v}{1+2\alpha \sin ^{2}\dfrac{\phi }{2}}\right)$ --- (by equation 6)

    Where $\alpha =\dfrac{h\nu }{m_{0}c^{2}}$

    Then tan $\phi$=$\left( \dfrac{\cot \dfrac{\phi }{2}}{1+\left( \dfrac{h\nu }{m_{0}c^{2}}\right) }\right)$

    This is measured with respect to incident line of photon.

    Energy of recoiled electron 

    KE=$h\nu -h\nu '$=$h\nu \left( \dfrac{2\alpha \sin ^{2}\dfrac{\phi }{2}}{1+2\alpha \sin ^{2}\dfrac{\phi }{2}}\right)$

    So, (KE)max =($\dfrac{2h^{2}\nu ^{2}}{m_{0}c^{2}\left( 1+\dfrac{2n\nu }{m_{0}c^{2}}\right) }$)

    IMPORTANCE OF COMPTON EFFECT

    The collesion of photon and free electron demonstrates that light is not purely wave. There is a presence of particle nature also in light. This effect leads us to think light as a particle also(dual nature).

    WHY WE CAN’T USE VISIBLE LIGHT IN THIS EXPERIMENT ?

    The Compton effect is the chief means by which x-rays lose energy when they pass through matter. The energy difference depends upon change in wavelength. I get the maximum difference between the wavelength of scattered photon and incident photon is two times of Compton wavelength (that is =0.0484 A) corresponding to 180* . This value is too small when I consider the wavelength of visible light . So, if visible light is used in place of X-rays , this is very difficult to observe this effect. Changes of this magnitude or less are readily observable only in x-rays.

    So, X-ray is mainly use to perform this practical.

    WHAT HAPPENS IF THERE IS NO FREE ELECTRON ?

    Now , again I consider the difference in wavelength . If there is no free electron in atom , then the the mass of electron is replaced with mass of atom ( that is so huge then mass of electron). It means the difference of wavelength in scattered photon and incident photon becomes very small or I can say negligible. Means almost no change in wavelength.

    So ,this is not useful for our aim to show scattering.

    OUTCOME

    Compton effect proves that light is not only a wave . It consists particle nature also.