# Compton Effect and Determination of Compton Wavelength

## Compton Effect

When X-rays fall on a crystal, they are scattered. In 1923, Compton observed that the wavelength of the scattered radiation (位') is always greater than the wavelength of the incident radiation (位)-

位' > 位

The change in wavelength (d位 = 位' − 位) is independent of the wavelength of the incident radiation as well as that of scatterer. This change in wavelength of the scattered radiation is called **compton effect**. Compton explained this effect with the help of Planck's quantum theory. When an incident photon strikes an electron at rest, there will be an elastic collision between the photon and the electron of the scatterer. The photon will transfer K.E. & momentum to the electron and thus the scattered photon will have higher wavelength i.e. lower energy than the incident photon. Thus it provides the most convincing evidence in support of particle or photon or quantum nature of radiation.

## Determination of Compton wavelength

The energy of photon is given by-

E = hc/位

The photon has no rest mass, so its moving mass m_{p} is its total energy, E-

m_{p} = E/c^{2} = hc/位 x 1/c^{2} = h/c位

The velocity of photon is c and hence its momentum i.e. mass times velocity is given by-

M = h/c位 x c = h/位

Let a photon of energy hc/位 and momentum h/位 collide with electron of rest mass, m_{o}. As the electron is taken as at rest before collision, so as per the theory of relativity, its total energy is m_{o}c^{2} and momentum is zero. Following the collision, the photon energy is reduced to hc/位' while its momentum is changed to h/位' at an angle 胃 to the x-axis. The electron gains the energy from the collision and its total energy becomes mc^{2} and momentum hc/位' directed at an angel 蠒 to the x-axis. From the conversion principle, the energy is given by-

hc/位 + m_{o}c^{2} = hc/位' + mc^{2}

or, hc/位 − hc/位' = mc^{2} − m_{o}c^{2} -----Equation-1

Being vector quantity, the momentum has both the magnitude and direction. So in the collision, momentum must be conserved in both x and y-directions. Hence, the x-component of the momentum is given by-

h/位 = h/位' cos胃 + m v cos蠒

or, h/位 − h/位' cos胃 = m v cos蠒

or, h^{2}/位^{2} + h^{2}/位'^{2} cos^{2}胃 − 2h^{2}/位位' cos胃 = m^{2} v^{2} cos^{2}蠒 -----Equation-2

and y-component of the momentum is given by-

0 = h/位' sin胃 − m v sin蠒 -----Equation-3

or, h^{2}/位'^{2} sin^{2}胃 = m^{2} v^{2} sin^{2}蠒 -----Equation-4

On adding equation-1 and equation-2, 蠒 will be removed-

m^{2}v^{2} = h^{2}[1/位^{2} − (2cos胃/位位') + 1/位'^{2}] -----Equation-5

If the velocity of recoiling electron is v, then from the theory of relativity, the mass of electron is given by-

or, m^{2}c^{2} − m^{2}v^{2} = m_{o}^{2}v^{2}

or, m^{2}v^{2} = c^{2}(m^{2} − m_{o}^{2}) -----Equation-6

From equation-5 and equation-6, we have-

c^{2}(m^{2} − m_{o}^{2}) = h^{2}[1/位^{2} − (2cos胃/位位') + 1/位'^{2}] -----Equation-7

Squaring equation-1 and putting in equation-7 we get-

位' − 位 = 螖位 = h/m_{o}c(1 − cos胃) -----Equation-8

where, 螖位 is compton wavelength (i.e. change in wavelength between the original photon and a photon scattered at an angle 胃). The wavelength shift is independent of nature of the substance and the wave length of the incident radiation. It dependents only on the scattering angle 胃. The following three cases may be considered-

**Case-1: 胃 = 0°**

The scattered radiation is parralel to the incident radiation. In this case, cos胃 = 1.

So, 螖位 = h/m_{o}c = 0, that means no wavelength shift.

**Case-2: 胃 = 90°**

The scattered radiation is perpendicular to the incident radiation. In this case, cos胃 = 0.

So, 螖位 = h/m_{o}c = 6.626 x 10^{-34}J/s (9.109 x 10^{-31}Kg)(3 x 10^{8}m/s) = 0.02422 x 10^{-10}m.

In this case, change in wavelength is called compton wavelength.

**Case-3: 胃 = 180°**

The radiation is scattered in a direction opposite to the incident radiation. In this case, cos胃 = -1.

So, 螖位 = 2h/m_{o}c = 0.0484 x 10^{-10}m.

This is twice the value of the compton wavelength.