Compton Scattering

Arthur H. Compton is an American scientist who conducted experiment to test the prediction, of the wave-model of Maxwell and photon-model of Einstein, about how light behaves when undergoing scattering by a single electron.

In his experiment, he used a beam of light and aimed a certain solid target. After the light approached the target, the wavelength of the radiation scattered from the target is then measured. Compton then discovered that the light scattered from the target has longer wavelength than the incident light. Since the scattered light has longer wavelength, it also follows that is has smaller frequency. Another observation from the experiment is that the change in the wavelength of the light is dependent on the angle through which the light is scattered. This explains the prediction of the photon-model of light which states that compared to the incident light, the scattered light has a longer wavelength and lower frequency. This process is what called as the Compton scattering.

Supposed the light is scattered from a given angle $$\phi$$ with respect to the incident light, Compton scattering can be represented by the equation:

$$\lambda’ - \lambda = {h \over mc} (1 - \cos \phi)$$     (1)

where

$$\lambda’$$ is the wavelength of the scattered light

$$\lambda$$ is the wavelength of the incident light

$$h$$ is the Planck’s constant

$$m$$ is the electron rest mass

$$c$$ is the speed of light in a vacuum

$$\phi$$ is the scattering angle

and $${h \over mc}$$ has a numerical value of $$2.426 \times 10^{-12}\;m$$.

Equation (1) shows that $$\lambda’$$ is greater than $$\lambda$$.

By using the photon-model of Einstein and the principles of the conservation of energy and conservation of momentum, Compton provided a clear explanation of the results of his experiment. To show this explanation, the relativistic energy-momentum relationships is used given the following variables:

For the incident photon:

$$\vec p$$ is the momentum

$$p$$ is the magnitude of the momentum

$$pc$$ is the energy.

For the scattered photon:

$$\vec{p’}$$ is the momentum

$$p’$$ the magnitude of the momentum

$$p’c$$ is the energy.

Since the electron is initially at rest, its initial momentum is zero and its initial energy is the rest energy $$mc^2$$. The final momentum of the electron is $$\vec {P_e}$$ with a magnitude $$P_e$$. The final energy of the electron is given as $$E_e^2 = (mc^2)^2 + (P_ec)^2$$. From the conservation of energy, we obtain the relationship

$$pc + mc^2 = p’c + E_e$$     (2)

Rearranging equation (2), we have

$$(pc – p’c + mc^2)^2 = E_e^2 = (mc^2)^2 + (P_ec)^2$$     (3)

Using the conservation of momentum, the momentum of the electron $$\vec {P_e}$$ can be eliminated. Thus,

$$\vec{P_e} = \vec{p} - \vec {p’}$$      (4)

Now, taking the scalar product of each side of equation (4) by itself, we obtain

$$P_e^2 = p^2 + p^{’2} – 2pp’ \cos \phi$$     (5)

After algebraic processes and substituting equation (5) to equation (3), we get

$${mc \over p’} – {mc \over p} = 1 - \cos \phi$$     (6)

Substituting $$p’ = {h \over \lambda’}$$ and $$p = {h \over \lambda}$$ , and multiplying by $${h \over mc}$$, equation (1) is then obtained.