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    Modern Physics: Photoelecric effect and Compton scattering

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    1) plancks constant and work function

    2) photon energy, electron kinetic, direction

    © BrainMass Inc. brainmass.com October 5, 2022, 1:43 pm ad1c9bdddf


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    Phototubes operate on the principle of photoelectric effect. This effect is showing the particle nature of the light. When light of sufficiently small wavelength incident on the metal plate, the emission of electrons from the surface of metal takes place. For every metal there is a cut off wavelength called threshold wavelength above that no emission of electrons takes place and the tube is cut-off.

    The energy of photon of light is inversely proportional to the wavelength of light hence energy of photons of the light decreases with the wavelength. If the wavelength of light is larger then threshold wavelength, the photons are not having sufficient energy to extract electrons from the surface of that metal. This minimum energy required to extract electron from the surface of metal is called work function.

    The Einstein's equation of photoelectric equation is relating the energy of emitted electron the work function of the surface and the energy of incident photon as

    where h is the plank's constant,  is the wavelength of light incident, and (1/2)nv2 is the maximum energy of emitted electrons.

    Substituting the vales given for the two cases in above equation we get
    ------- (1) (1 eV = 1.6*10-19 J)
    and ----- (2)

    Subtracting equation 2 from equation 1 we get

    or Js.

    And substituting this in equation we get




    This is the case of Compton scattering.

    The energy of the incident photon = hn (n is the frequency of incident photon)
    The energy of scattered photon = hn'
    The initial energy of electron = m0c2
    Where m0 is the rest mass of electron and c is speed of light in vacuum.
    The energy of electron after collision = mc2

    Let the initial direction of motion of photon is + x direction and the photon scattered along y direction. Let the electron recoils with speed v at and angle  with x-axis.

    (a) According to the principle of conservation of energy
    hn + m0c2 = hn' + mc2
    Gives mc2 = h (n - n') + m0c2
    Or m2c4 = h2(n2 - 2nn' +n'2) + 2h(n - n') m0c2 + m02c4 ---- (1)

    The momentum of photon is given by hn/c hence applying the law of conservation of momentum along x and y direction respectively we get
    hn/c = mv*cos  ---- (2)
    and hn'/c = mv*sin  ------ (3)

    squaring and adding equation 2 and 3 we get

    or ---- (4)

    Subtracting equation 4 from equation 1 we get

    m2 c2(c2 - v2) = - 2h2nn' + 2h(n - n')m0c2 + m02c4 ------- (5)

    Now the theory of relativity gives

    or ---- (6)
    Hence from equation (5) and (6) we get

    m02 c4 = - 2h2nn' + 2h(n - n')m0c2 + m02c4

    or h2nn' = h(n - n')m0c2

    So the energy of the scattered photon hn' = 1.339*10-14 J = 83.671 KeV.


    The differencee of the energy of the incident and the scattred electron gives the kinetic energy of the recoiling electron and hence it is equal to

    K.E. = 100 KeV - 83.671 KeV = 16.329 KeV = 16329 eV.

    Dividing equation 3 by equation 2 we get the direction of recoiling electron as

    tan  = hn'/hn = 83.671/100 = 0.8367
    or  = 39.90.

    Please check the numeric calculations. Thanks.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 5, 2022, 1:43 pm ad1c9bdddf>