# Modern Physics: Photoelecric effect and Compton scattering

1) plancks constant and work function

2) photon energy, electron kinetic, direction

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Answer:

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

or

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Answer:

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

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

or

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

(b)

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.

(C)

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.

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Please check the numeric calculations. Thanks.

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